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Trigonometry

Board Coverage​

BoardPaperNotes
AQAPaper 1, 2Basic trig in P1; compound/double angle, trig equations in P2
EdexcelP1, P2Similar split
OCR (A)Paper 1, 2Includes small angle approximations
CIE (9709)P1, P2, P3Trig functions in P1; identities and equations in P2; further trig in P3
info

All boards provide trigonometric identities in the formula booklet, but not their proofs. You need to know which identities exist and how to apply them.

1. The Unit Circle Definitions​

Definition. Consider the unit circle (radius 1, centre at the origin). For an angle θ\theta measured anticlockwise from the positive xx-axis, the point on the circle is (cos⁡θ,sin⁡θ)(\cos\theta, \sin\theta).

Definition.

cos⁡θ=thex−coordinateofthepointontheunitcircleatangleθ\cos\theta = \mathrm{the } x\mathrm{-coordinate of the point on the unit circle at angle } \theta

sin⁡θ=they−coordinateofthepointontheunitcircleatangleθ\sin\theta = \mathrm{the } y\mathrm{-coordinate of the point on the unit circle at angle } \theta

tan⁡θ=◆LB◆sin⁡θ◆RB◆◆LB◆cos⁡θ◆RB◆(cos⁡θ≠0)\tan\theta = \frac◆LB◆\sin\theta◆RB◆◆LB◆\cos\theta◆RB◆ \quad (\cos\theta \neq 0)

Intuition. These definitions extend the right-triangle definitions (SOH CAH TOA) to all angles, not just those between 0° and 90°. The unit circle makes clear why sin⁡\sin and cos⁡\cos are periodic with period 2π2\pi.

1.1 Radian Measure​

Definition. One radian is the angle subtended at the centre of a circle by an arc equal in length to the radius.

θ(radians)=◆LB◆arclength◆RB◆◆LB◆r◆RB◆\theta\mathrm{ (radians)} = \frac◆LB◆\mathrm{arc length}◆RB◆◆LB◆r◆RB◆

The full circle: 2π2\pi radians =360∘= 360^\circ, so π\pi radians =180∘= 180^\circ.

Theorem (Arc length and sector area). For a sector of radius rr and angle θ\theta (in radians):

Arclengths=rθ\mathrm{Arc length } s = r\theta

SectorareaA=12r2θ\mathrm{Sector area } A = \frac{1}{2}r^2\theta

Proof. By definition, θ=s/r\theta = s/r, so s=rθs = r\theta. The sector is a fraction ◆LB◆θ◆RB◆◆LB◆2π◆RB◆\frac◆LB◆\theta◆RB◆◆LB◆2\pi◆RB◆ of the full circle (area πr2\pi r^2), so A=◆LB◆θ◆RB◆◆LB◆2π◆RB◆⋅πr2=12r2θA = \frac◆LB◆\theta◆RB◆◆LB◆2\pi◆RB◆ \cdot \pi r^2 = \frac{1}{2}r^2\theta. ■\blacksquare

2. Fundamental Identities​

2.1 Pythagorean Identity​

Theorem. For all θ∈R\theta \in \mathbb{R}:

sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1

Proof. The point (cos⁡θ,sin⁡θ)(\cos\theta, \sin\theta) lies on the unit circle x2+y2=1x^2 + y^2 = 1.

cos⁡2θ+sin⁡2θ=1■\cos^2\theta + \sin^2\theta = 1 \quad \blacksquare

Corollary. Dividing by cos⁡2θ\cos^2\theta (where cos⁡θ≠0\cos\theta \neq 0):

1+tan⁡2θ=sec⁡2θ1 + \tan^2\theta = \sec^2\theta

Dividing by sin⁡2θ\sin^2\theta (where sin⁡θ≠0\sin\theta \neq 0):

cot⁡2θ+1=cosec⁡2θ\cot^2\theta + 1 = \cosec^2\theta

3. Compound Angle Formulas​

3.1 Sine of a Sum​

Theorem. For all A,B∈RA, B \in \mathbb{R}:

sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin(A + B) = \sin A \cos B + \cos A \sin B

Proof (using rotation matrices). Consider the rotation of the plane by angle AA followed by rotation by angle BB. The combined rotation is by angle A+BA + B.

The rotation matrix by angle θ\theta is:

R(θ)=(cos⁡θ−sin⁡θsin⁡θcos⁡θ)R(\theta) = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}

Matrix multiplication gives: R(A)R(B)=R(A+B)R(A)R(B) = R(A + B).

(cos⁡A−sin⁡Asin⁡Acos⁡A)(cos⁡B−sin⁡Bsin⁡Bcos⁡B)=(cos⁡(A+B)−sin⁡(A+B)sin⁡(A+B)cos⁡(A+B))\begin{pmatrix} \cos A & -\sin A \\ \sin A & \cos A \end{pmatrix}\begin{pmatrix} \cos B & -\sin B \\ \sin B & \cos B \end{pmatrix} = \begin{pmatrix} \cos(A+B) & -\sin(A+B) \\ \sin(A+B) & \cos(A+B) \end{pmatrix}

The (2,1)(2, 1) entry of the product is:

sin⁡Acos⁡B+cos⁡Asin⁡B=sin⁡(A+B)■\sin A \cos B + \cos A \sin B = \sin(A + B) \quad \blacksquare

3.2 Cosine of a Sum​

Theorem.

cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos(A + B) = \cos A \cos B - \sin A \sin B

Proof. From the (1,1)(1, 1) entry of the matrix product above:

cos⁡Acos⁡B−sin⁡Asin⁡B=cos⁡(A+B)■\cos A \cos B - \sin A \sin B = \cos(A + B) \quad \blacksquare

3.3 Tangent of a Sum​

Theorem.

tan⁡(A+B)=◆LB◆tan⁡A+tan⁡B◆RB◆◆LB◆1−tan⁡Atan⁡B◆RB◆\tan(A + B) = \frac◆LB◆\tan A + \tan B◆RB◆◆LB◆1 - \tan A \tan B◆RB◆

Proof.

tan⁡(A+B)=◆LB◆sin⁡(A+B)◆RB◆◆LB◆cos⁡(A+B)◆RB◆=◆LB◆sin⁡Acos⁡B+cos⁡Asin⁡B◆RB◆◆LB◆cos⁡Acos⁡B−sin⁡Asin⁡B◆RB◆\begin{aligned} \tan(A + B) &= \frac◆LB◆\sin(A + B)◆RB◆◆LB◆\cos(A + B)◆RB◆ \\ &= \frac◆LB◆\sin A \cos B + \cos A \sin B◆RB◆◆LB◆\cos A \cos B - \sin A \sin B◆RB◆ \end{aligned}

Divide numerator and denominator by cos⁡Acos⁡B\cos A \cos B:

=◆LB◆tan⁡A+tan⁡B◆RB◆◆LB◆1−tan⁡Atan⁡B◆RB◆■= \frac◆LB◆\tan A + \tan B◆RB◆◆LB◆1 - \tan A \tan B◆RB◆ \quad \blacksquare

3.4 Difference Formulas​

Theorem.

sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡Bcos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡Btan⁡(A−B)=◆LB◆tan⁡A−tan⁡B◆RB◆◆LB◆1+tan⁡Atan⁡B◆RB◆\begin{aligned} \sin(A - B) &= \sin A \cos B - \cos A \sin B \\ \cos(A - B) &= \cos A \cos B + \sin A \sin B \\ \tan(A - B) &= \frac◆LB◆\tan A - \tan B◆RB◆◆LB◆1 + \tan A \tan B◆RB◆ \end{aligned}

Proof. Replace BB with −B-B in the sum formulas, using sin⁡(−B)=−sin⁡B\sin(-B) = -\sin B and cos⁡(−B)=cos⁡B\cos(-B) = \cos B. ■\blacksquare

4. Double Angle Formulas​

Setting A=BA = B in the compound angle formulas:

sin⁡2A=2sin⁡Acos⁡Acos⁡2A=cos⁡2A−sin⁡2Atan⁡2A=◆LB◆2tan⁡A◆RB◆◆LB◆1−tan⁡2A◆RB◆\begin{aligned} \sin 2A &= 2\sin A \cos A \\ \cos 2A &= \cos^2 A - \sin^2 A \\ \tan 2A &= \frac◆LB◆2\tan A◆RB◆◆LB◆1 - \tan^2 A◆RB◆ \end{aligned}

Theorem. Three equivalent forms of cos⁡2A\cos 2A:

cos⁡2A=cos⁡2A−sin⁡2A=2cos⁡2A−1=1−2sin⁡2A\cos 2A = \cos^2 A - \sin^2 A = 2\cos^2 A - 1 = 1 - 2\sin^2 A

Proof. Using sin⁡2A=1−cos⁡2A\sin^2 A = 1 - \cos^2 A:

cos⁡2A−sin⁡2A=cos⁡2A−(1−cos⁡2A)=2cos⁡2A−1\cos^2 A - \sin^2 A = \cos^2 A - (1 - \cos^2 A) = 2\cos^2 A - 1

Using cos⁡2A=1−sin⁡2A\cos^2 A = 1 - \sin^2 A:

cos⁡2A−sin⁡2A=(1−sin⁡2A)−sin⁡2A=1−2sin⁡2A\cos^2 A - \sin^2 A = (1 - \sin^2 A) - \sin^2 A = 1 - 2\sin^2 A ■\blacksquare

Intuition. The double angle formulas express functions of 2A2A purely in terms of functions of AA. They are the algebraic backbone of many trigonometric manipulations and are essential for integration.

5. Solving Trigonometric Equations​

5.1 Basic Strategy​

  1. Use identities to reduce to a single trigonometric function.
  2. Solve the resulting equation.
  3. Find all solutions in the required interval.

5.2 Key Solutions​

sin⁡θ=a  ⟹  θ=arcsin⁡(a)+2nπorπ−arcsin⁡(a)+2nπcos⁡θ=a  ⟹  θ=±arccos⁡(a)+2nπtan⁡θ=a  ⟹  θ=arctan⁡(a)+nπ\begin{aligned} \sin\theta = a &\implies \theta = \arcsin(a) + 2n\pi \mathrm{ or } \pi - \arcsin(a) + 2n\pi \\ \cos\theta = a &\implies \theta = \pm\arccos(a) + 2n\pi \\ \tan\theta = a &\implies \theta = \arctan(a) + n\pi \end{aligned}
Details

Example Solve 2sin⁡2θ+3cos⁡θ−3=02\sin^2\theta + 3\cos\theta - 3 = 0 for 0≤θ<2π0 \leq \theta < 2\pi.

Using sin⁡2θ=1−cos⁡2θ\sin^2\theta = 1 - \cos^2\theta:

2(1−cos⁡2θ)+3cos⁡θ−3=02(1 - \cos^2\theta) + 3\cos\theta - 3 = 0

−2cos⁡2θ+3cos⁡θ−1=0-2\cos^2\theta + 3\cos\theta - 1 = 0

2cos⁡2θ−3cos⁡θ+1=02\cos^2\theta - 3\cos\theta + 1 = 0

(2cos⁡θ−1)(cos⁡θ−1)=0(2\cos\theta - 1)(\cos\theta - 1) = 0

cos⁡θ=12  ⟹  θ=◆LB◆π◆RB◆◆LB◆3◆RB◆\cos\theta = \frac{1}{2} \implies \theta = \frac◆LB◆\pi◆RB◆◆LB◆3◆RB◆ or ◆LB◆5π◆RB◆◆LB◆3◆RB◆\frac◆LB◆5\pi◆RB◆◆LB◆3◆RB◆.

cos⁡θ=1  ⟹  θ=0\cos\theta = 1 \implies \theta = 0.

Solutions: θ=0,◆LB◆π◆RB◆◆LB◆3◆RB◆,◆LB◆5π◆RB◆◆LB◆3◆RB◆\theta = 0, \frac◆LB◆\pi◆RB◆◆LB◆3◆RB◆, \frac◆LB◆5\pi◆RB◆◆LB◆3◆RB◆.

6. Small Angle Approximations​

Theorem. For small θ\theta (in radians):

sin⁡θ≈θ−◆LB◆θ3◆RB◆◆LB◆6◆RB◆cos⁡θ≈1−◆LB◆θ2◆RB◆◆LB◆2◆RB◆tan⁡θ≈θ+◆LB◆θ3◆RB◆◆LB◆3◆RB◆\begin{aligned} \sin\theta &\approx \theta - \frac◆LB◆\theta^3◆RB◆◆LB◆6◆RB◆ \\ \cos\theta &\approx 1 - \frac◆LB◆\theta^2◆RB◆◆LB◆2◆RB◆ \\ \tan\theta &\approx \theta + \frac◆LB◆\theta^3◆RB◆◆LB◆3◆RB◆ \end{aligned}

These follow from the Maclaurin series expansions (see Differentiation).

For the linear approximations (when θ\theta is very small):

sin⁡θ≈θ,cos⁡θ≈1,tan⁡θ≈θ\sin\theta \approx \theta, \quad \cos\theta \approx 1, \quad \tan\theta \approx \theta

Intuition. Near the origin, the curves y=sin⁡θy = \sin\theta and y=tan⁡θy = \tan\theta are almost indistinguishable from the line y=θy = \theta. The curve y=cos⁡θy = \cos\theta is nearly flat at y=1y = 1.

warning

The small angle approximations are only valid when θ\theta is in radians, not degrees. This is a very common exam error.

7. Half-Angle Formulas​

Theorem. For all θ\theta:

sin⁡◆LB◆θ◆RB◆◆LB◆2◆RB◆=±◆LB◆◆LB◆1−cos⁡θ◆RB◆◆LB◆2◆RB◆◆RB◆\sin\frac◆LB◆\theta◆RB◆◆LB◆2◆RB◆ = \pm\sqrt◆LB◆\frac◆LB◆1 - \cos\theta◆RB◆◆LB◆2◆RB◆◆RB◆

cos⁡◆LB◆θ◆RB◆◆LB◆2◆RB◆=±◆LB◆◆LB◆1+cos⁡θ◆RB◆◆LB◆2◆RB◆◆RB◆\cos\frac◆LB◆\theta◆RB◆◆LB◆2◆RB◆ = \pm\sqrt◆LB◆\frac◆LB◆1 + \cos\theta◆RB◆◆LB◆2◆RB◆◆RB◆

tan⁡◆LB◆θ◆RB◆◆LB◆2◆RB◆=◆LB◆sin⁡θ◆RB◆◆LB◆1+cos⁡θ◆RB◆=◆LB◆1−cos⁡θ◆RB◆◆LB◆sin⁡θ◆RB◆\tan\frac◆LB◆\theta◆RB◆◆LB◆2◆RB◆ = \frac◆LB◆\sin\theta◆RB◆◆LB◆1 + \cos\theta◆RB◆ = \frac◆LB◆1 - \cos\theta◆RB◆◆LB◆\sin\theta◆RB◆

7.1 Derivation​

Starting from cos⁡2A=1−2sin⁡2A\cos 2A = 1 - 2\sin^2 A and substituting A=◆LB◆θ◆RB◆◆LB◆2◆RB◆A = \frac◆LB◆\theta◆RB◆◆LB◆2◆RB◆:

cos⁡θ=1−2sin⁡2◆LB◆θ◆RB◆◆LB◆2◆RB◆\cos\theta = 1 - 2\sin^2\frac◆LB◆\theta◆RB◆◆LB◆2◆RB◆

2sin⁡2◆LB◆θ◆RB◆◆LB◆2◆RB◆=1−cos⁡θ2\sin^2\frac◆LB◆\theta◆RB◆◆LB◆2◆RB◆ = 1 - \cos\theta

sin⁡◆LB◆θ◆RB◆◆LB◆2◆RB◆=±◆LB◆◆LB◆1−cos⁡θ◆RB◆◆LB◆2◆RB◆◆RB◆■\sin\frac◆LB◆\theta◆RB◆◆LB◆2◆RB◆ = \pm\sqrt◆LB◆\frac◆LB◆1 - \cos\theta◆RB◆◆LB◆2◆RB◆◆RB◆ \quad \blacksquare

Similarly, from cos⁡2A=2cos⁡2A−1\cos 2A = 2\cos^2 A - 1:

cos⁡θ=2cos⁡2◆LB◆θ◆RB◆◆LB◆2◆RB◆−1\cos\theta = 2\cos^2\frac◆LB◆\theta◆RB◆◆LB◆2◆RB◆ - 1

2cos⁡2◆LB◆θ◆RB◆◆LB◆2◆RB◆=1+cos⁡θ2\cos^2\frac◆LB◆\theta◆RB◆◆LB◆2◆RB◆ = 1 + \cos\theta

cos⁡◆LB◆θ◆RB◆◆LB◆2◆RB◆=±◆LB◆◆LB◆1+cos⁡θ◆RB◆◆LB◆2◆RB◆◆RB◆■\cos\frac◆LB◆\theta◆RB◆◆LB◆2◆RB◆ = \pm\sqrt◆LB◆\frac◆LB◆1 + \cos\theta◆RB◆◆LB◆2◆RB◆◆RB◆ \quad \blacksquare

7.2 Tangent Half-Angle Formulas​

The tangent half-angle formulas avoid the ambiguity of the ±\pm sign:

tan⁡◆LB◆θ◆RB◆◆LB◆2◆RB◆=◆LB◆sin⁡θ◆RB◆◆LB◆1+cos⁡θ◆RB◆\tan\frac◆LB◆\theta◆RB◆◆LB◆2◆RB◆ = \frac◆LB◆\sin\theta◆RB◆◆LB◆1 + \cos\theta◆RB◆

Proof. Using the double angle formulas:

◆LB◆sin⁡θ◆RB◆◆LB◆1+cos⁡θ◆RB◆=◆LB◆2sin⁡◆LB◆θ◆RB◆◆LB◆2◆RB◆cos⁡◆LB◆θ◆RB◆◆LB◆2◆RB◆◆RB◆◆LB◆1+(2cos⁡2◆LB◆θ◆RB◆◆LB◆2◆RB◆−1)◆RB◆=◆LB◆2sin⁡◆LB◆θ◆RB◆◆LB◆2◆RB◆cos⁡◆LB◆θ◆RB◆◆LB◆2◆RB◆◆RB◆◆LB◆2cos⁡2◆LB◆θ◆RB◆◆LB◆2◆RB◆◆RB◆=◆LB◆sin⁡◆LB◆θ◆RB◆◆LB◆2◆RB◆◆RB◆◆LB◆cos⁡◆LB◆θ◆RB◆◆LB◆2◆RB◆◆RB◆=tan⁡◆LB◆θ◆RB◆◆LB◆2◆RB◆■\frac◆LB◆\sin\theta◆RB◆◆LB◆1 + \cos\theta◆RB◆ = \frac◆LB◆2\sin\frac◆LB◆\theta◆RB◆◆LB◆2◆RB◆\cos\frac◆LB◆\theta◆RB◆◆LB◆2◆RB◆◆RB◆◆LB◆1 + (2\cos^2\frac◆LB◆\theta◆RB◆◆LB◆2◆RB◆ - 1)◆RB◆ = \frac◆LB◆2\sin\frac◆LB◆\theta◆RB◆◆LB◆2◆RB◆\cos\frac◆LB◆\theta◆RB◆◆LB◆2◆RB◆◆RB◆◆LB◆2\cos^2\frac◆LB◆\theta◆RB◆◆LB◆2◆RB◆◆RB◆ = \frac◆LB◆\sin\frac◆LB◆\theta◆RB◆◆LB◆2◆RB◆◆RB◆◆LB◆\cos\frac◆LB◆\theta◆RB◆◆LB◆2◆RB◆◆RB◆ = \tan\frac◆LB◆\theta◆RB◆◆LB◆2◆RB◆ \quad \blacksquare

Similarly:

tan⁡◆LB◆θ◆RB◆◆LB◆2◆RB◆=◆LB◆1−cos⁡θ◆RB◆◆LB◆sin⁡θ◆RB◆\tan\frac◆LB◆\theta◆RB◆◆LB◆2◆RB◆ = \frac◆LB◆1 - \cos\theta◆RB◆◆LB◆\sin\theta◆RB◆

Proof. ◆LB◆1−cos⁡θ◆RB◆◆LB◆sin⁡θ◆RB◆=◆LB◆1−(1−2sin⁡2◆LB◆θ◆RB◆◆LB◆2◆RB◆)◆RB◆◆LB◆2sin⁡◆LB◆θ◆RB◆◆LB◆2◆RB◆cos⁡◆LB◆θ◆RB◆◆LB◆2◆RB◆◆RB◆=◆LB◆2sin⁡2◆LB◆θ◆RB◆◆LB◆2◆RB◆◆RB◆◆LB◆2sin⁡◆LB◆θ◆RB◆◆LB◆2◆RB◆cos⁡◆LB◆θ◆RB◆◆LB◆2◆RB◆◆RB◆=◆LB◆sin⁡◆LB◆θ◆RB◆◆LB◆2◆RB◆◆RB◆◆LB◆cos⁡◆LB◆θ◆RB◆◆LB◆2◆RB◆◆RB◆=tan⁡◆LB◆θ◆RB◆◆LB◆2◆RB◆■\frac◆LB◆1 - \cos\theta◆RB◆◆LB◆\sin\theta◆RB◆ = \frac◆LB◆1 - (1 - 2\sin^2\frac◆LB◆\theta◆RB◆◆LB◆2◆RB◆)◆RB◆◆LB◆2\sin\frac◆LB◆\theta◆RB◆◆LB◆2◆RB◆\cos\frac◆LB◆\theta◆RB◆◆LB◆2◆RB◆◆RB◆ = \frac◆LB◆2\sin^2\frac◆LB◆\theta◆RB◆◆LB◆2◆RB◆◆RB◆◆LB◆2\sin\frac◆LB◆\theta◆RB◆◆LB◆2◆RB◆\cos\frac◆LB◆\theta◆RB◆◆LB◆2◆RB◆◆RB◆ = \frac◆LB◆\sin\frac◆LB◆\theta◆RB◆◆LB◆2◆RB◆◆RB◆◆LB◆\cos\frac◆LB◆\theta◆RB◆◆LB◆2◆RB◆◆RB◆ = \tan\frac◆LB◆\theta◆RB◆◆LB◆2◆RB◆ \quad \blacksquare

7.3 Sign Determination​

The ±\pm in the sine and cosine half-angle formulas depends on the quadrant of ◆LB◆θ◆RB◆◆LB◆2◆RB◆\frac◆LB◆\theta◆RB◆◆LB◆2◆RB◆, not the quadrant of θ\theta itself. Always determine which quadrant ◆LB◆θ◆RB◆◆LB◆2◆RB◆\frac◆LB◆\theta◆RB◆◆LB◆2◆RB◆ lies in before choosing the sign.

Quadrant of ◆LB◆θ◆RB◆◆LB◆2◆RB◆\frac◆LB◆\theta◆RB◆◆LB◆2◆RB◆sin⁡◆LB◆θ◆RB◆◆LB◆2◆RB◆\sin\frac◆LB◆\theta◆RB◆◆LB◆2◆RB◆cos⁡◆LB◆θ◆RB◆◆LB◆2◆RB◆\cos\frac◆LB◆\theta◆RB◆◆LB◆2◆RB◆
I: 0<◆LB◆θ◆RB◆◆LB◆2◆RB◆<◆LB◆π◆RB◆◆LB◆2◆RB◆0 \lt{} \frac◆LB◆\theta◆RB◆◆LB◆2◆RB◆ \lt{} \frac◆LB◆\pi◆RB◆◆LB◆2◆RB◆++++
II: ◆LB◆π◆RB◆◆LB◆2◆RB◆<◆LB◆θ◆RB◆◆LB◆2◆RB◆<π\frac◆LB◆\pi◆RB◆◆LB◆2◆RB◆ \lt{} \frac◆LB◆\theta◆RB◆◆LB◆2◆RB◆ \lt{} \pi++−-
III: π<◆LB◆θ◆RB◆◆LB◆2◆RB◆<◆LB◆3π◆RB◆◆LB◆2◆RB◆\pi \lt{} \frac◆LB◆\theta◆RB◆◆LB◆2◆RB◆ \lt{} \frac◆LB◆3\pi◆RB◆◆LB◆2◆RB◆−-−-
IV: ◆LB◆3π◆RB◆◆LB◆2◆RB◆<◆LB◆θ◆RB◆◆LB◆2◆RB◆<2π\frac◆LB◆3\pi◆RB◆◆LB◆2◆RB◆ \lt{} \frac◆LB◆\theta◆RB◆◆LB◆2◆RB◆ \lt{} 2\pi−-++
Details

Worked example Find the exact value of sin⁡◆LB◆π◆RB◆◆LB◆8◆RB◆\sin\frac◆LB◆\pi◆RB◆◆LB◆8◆RB◆.

Since 0<◆LB◆π◆RB◆◆LB◆8◆RB◆<◆LB◆π◆RB◆◆LB◆2◆RB◆0 \lt{} \frac◆LB◆\pi◆RB◆◆LB◆8◆RB◆ \lt{} \frac◆LB◆\pi◆RB◆◆LB◆2◆RB◆ (first quadrant), sin⁡◆LB◆π◆RB◆◆LB◆8◆RB◆>0\sin\frac◆LB◆\pi◆RB◆◆LB◆8◆RB◆ > 0, so we take the positive root.

sin⁡◆LB◆π◆RB◆◆LB◆8◆RB◆=◆LB◆◆LB◆1−cos⁡π4◆RB◆◆LB◆2◆RB◆◆RB◆=◆LB◆◆LB◆1−22◆RB◆◆LB◆2◆RB◆◆RB◆=◆LB◆◆LB◆2−2◆RB◆◆LB◆4◆RB◆◆RB◆=◆LB◆◆LB◆2−2◆RB◆◆RB◆◆LB◆2◆RB◆\sin\frac◆LB◆\pi◆RB◆◆LB◆8◆RB◆ = \sqrt◆LB◆\frac◆LB◆1 - \cos\frac{\pi}{4}◆RB◆◆LB◆2◆RB◆◆RB◆ = \sqrt◆LB◆\frac◆LB◆1 - \frac{\sqrt{2}}{2}◆RB◆◆LB◆2◆RB◆◆RB◆ = \sqrt◆LB◆\frac◆LB◆2 - \sqrt{2}◆RB◆◆LB◆4◆RB◆◆RB◆ = \frac◆LB◆\sqrt◆LB◆2 - \sqrt{2}◆RB◆◆RB◆◆LB◆2◆RB◆

8. R-Addition Formula (Harmonic Form)​

Theorem. For real numbers aa and bb:

asin⁡θ+bcos⁡θ=Rsin⁡(θ+α)a\sin\theta + b\cos\theta = R\sin(\theta + \alpha)

where R=a2+b2R = \sqrt{a^2 + b^2} and tan⁡α=ba\tan\alpha = \dfrac{b}{a}.

8.1 Cosine Form​

Alternatively, the same expression can be written as:

asin⁡θ+bcos⁡θ=Rcos⁡(θ−β)a\sin\theta + b\cos\theta = R\cos(\theta - \beta)

where R=a2+b2R = \sqrt{a^2 + b^2} and tan⁡β=ab\tan\beta = \dfrac{a}{b}.

Both forms are equivalent; the choice between them is a matter of convenience depending on whether a sine or cosine expansion is more natural for the problem at hand.

8.2 Derivation​

Expand the right-hand side of the sine form:

Rsin⁡(θ+α)=Rsin⁡θcos⁡α+Rcos⁡θsin⁡αR\sin(\theta + \alpha) = R\sin\theta\cos\alpha + R\cos\theta\sin\alpha

Equating coefficients with asin⁡θ+bcos⁡θa\sin\theta + b\cos\theta:

Rcos⁡α=a,Rsin⁡α=bR\cos\alpha = a, \qquad R\sin\alpha = b

Squaring and adding: R2cos⁡2α+R2sin⁡2α=a2+b2R^2\cos^2\alpha + R^2\sin^2\alpha = a^2 + b^2, so R2=a2+b2R^2 = a^2 + b^2 and R=a2+b2R = \sqrt{a^2 + b^2}.

Dividing the second equation by the first: tan⁡α=ba\tan\alpha = \dfrac{b}{a}. ■\blacksquare

The cosine form is derived similarly by expanding Rcos⁡(θ−β)=Rcos⁡θcos⁡β+Rsin⁡θsin⁡βR\cos(\theta - \beta) = R\cos\theta\cos\beta + R\sin\theta\sin\beta and equating coefficients.

8.3 Applications: Maximum and Minimum​

Since −1≤sin⁡(θ+α)≤1-1 \leq \sin(\theta + \alpha) \leq 1:

−R≤asin⁡θ+bcos⁡θ≤R-R \leq a\sin\theta + b\cos\theta \leq R

  • Maximum =R=a2+b2= R = \sqrt{a^2 + b^2}, occurring when θ+α=◆LB◆π◆RB◆◆LB◆2◆RB◆+2nπ\theta + \alpha = \frac◆LB◆\pi◆RB◆◆LB◆2◆RB◆ + 2n\pi.
  • Minimum =−R= -R, occurring when θ+α=◆LB◆3π◆RB◆◆LB◆2◆RB◆+2nπ\theta + \alpha = \frac◆LB◆3\pi◆RB◆◆LB◆2◆RB◆ + 2n\pi.
Details

Example: Finding maximum and minimum Find the maximum and minimum values of 3sin⁡θ−4cos⁡θ3\sin\theta - 4\cos\theta and the values of θ\theta at which they occur, for 0≤θ<2π0 \leq \theta \lt{} 2\pi.

R=9+16=5R = \sqrt{9 + 16} = 5.

Here a=3a = 3 and b=−4b = -4. Writing 3sin⁡θ−4cos⁡θ=5sin⁡(θ+α)3\sin\theta - 4\cos\theta = 5\sin(\theta + \alpha) where tan⁡α=−43\tan\alpha = \dfrac{-4}{3}, so α=−arctan⁡43\alpha = -\arctan\dfrac{4}{3}.

Maximum =5= 5 when θ+α=◆LB◆π◆RB◆◆LB◆2◆RB◆\theta + \alpha = \frac◆LB◆\pi◆RB◆◆LB◆2◆RB◆:

θ=◆LB◆π◆RB◆◆LB◆2◆RB◆−α=◆LB◆π◆RB◆◆LB◆2◆RB◆+arctan⁡43≈2.214rad\theta = \frac◆LB◆\pi◆RB◆◆LB◆2◆RB◆ - \alpha = \frac◆LB◆\pi◆RB◆◆LB◆2◆RB◆ + \arctan\frac{4}{3} \approx 2.214 \mathrm{ rad}

Minimum =−5= -5 when θ+α=◆LB◆3π◆RB◆◆LB◆2◆RB◆\theta + \alpha = \frac◆LB◆3\pi◆RB◆◆LB◆2◆RB◆:

θ=◆LB◆3π◆RB◆◆LB◆2◆RB◆−α=◆LB◆3π◆RB◆◆LB◆2◆RB◆+arctan⁡43≈5.356rad\theta = \frac◆LB◆3\pi◆RB◆◆LB◆2◆RB◆ - \alpha = \frac◆LB◆3\pi◆RB◆◆LB◆2◆RB◆ + \arctan\frac{4}{3} \approx 5.356 \mathrm{ rad}

8.4 Solving Equations​

The R-addition formula converts asin⁡θ+bcos⁡θ=ka\sin\theta + b\cos\theta = k into Rsin⁡(θ+α)=kR\sin(\theta + \alpha) = k, a standard trigonometric equation.

Details

Example: Solving an equation Solve sin⁡θ+cos⁡θ=1\sin\theta + \cos\theta = 1 for 0≤θ<2π0 \leq \theta \lt{} 2\pi.

R=1+1=2R = \sqrt{1 + 1} = \sqrt{2}, α=arctan⁡1=◆LB◆π◆RB◆◆LB◆4◆RB◆\alpha = \arctan 1 = \frac◆LB◆\pi◆RB◆◆LB◆4◆RB◆.

2sin⁡ ⁣(θ+◆LB◆π◆RB◆◆LB◆4◆RB◆)=1\sqrt{2}\sin\!\left(\theta + \frac◆LB◆\pi◆RB◆◆LB◆4◆RB◆\right) = 1

sin⁡ ⁣(θ+◆LB◆π◆RB◆◆LB◆4◆RB◆)=◆LB◆1◆RB◆◆LB◆2◆RB◆=sin⁡◆LB◆π◆RB◆◆LB◆4◆RB◆\sin\!\left(\theta + \frac◆LB◆\pi◆RB◆◆LB◆4◆RB◆\right) = \frac◆LB◆1◆RB◆◆LB◆\sqrt{2}◆RB◆ = \sin\frac◆LB◆\pi◆RB◆◆LB◆4◆RB◆

θ+◆LB◆π◆RB◆◆LB◆4◆RB◆=◆LB◆π◆RB◆◆LB◆4◆RB◆+2nπorθ+◆LB◆π◆RB◆◆LB◆4◆RB◆=◆LB◆3π◆RB◆◆LB◆4◆RB◆+2nπ\theta + \frac◆LB◆\pi◆RB◆◆LB◆4◆RB◆ = \frac◆LB◆\pi◆RB◆◆LB◆4◆RB◆ + 2n\pi \quad \mathrm{or} \quad \theta + \frac◆LB◆\pi◆RB◆◆LB◆4◆RB◆ = \frac◆LB◆3\pi◆RB◆◆LB◆4◆RB◆ + 2n\pi

θ=2nπorθ=◆LB◆π◆RB◆◆LB◆2◆RB◆+2nπ\theta = 2n\pi \quad \mathrm{or} \quad \theta = \frac◆LB◆\pi◆RB◆◆LB◆2◆RB◆ + 2n\pi

For 0≤θ<2π0 \leq \theta \lt{} 2\pi: θ=0\theta = 0 or θ=◆LB◆π◆RB◆◆LB◆2◆RB◆\theta = \frac◆LB◆\pi◆RB◆◆LB◆2◆RB◆.

9. Trigonometric Identities: Proof Strategies​

Proving trigonometric identities is a core exam skill. The following strategies cover the most common approaches.

9.1 Strategy 1: Work with One Side​

Start from the more complicated side and simplify it until it matches the simpler side. This avoids the logical error of assuming what you are trying to prove.

9.2 Strategy 2: Express Everything in Sine and Cosine​

Replace tan⁡\tan, sec⁡\sec, csc⁡\csc, cot⁡\cot with their definitions in terms of sin⁡\sin and cos⁡\cos: tan⁡θ=◆LB◆sin⁡θ◆RB◆◆LB◆cos⁡θ◆RB◆\tan\theta = \frac◆LB◆\sin\theta◆RB◆◆LB◆\cos\theta◆RB◆, sec⁡θ=◆LB◆1◆RB◆◆LB◆cos⁡θ◆RB◆\sec\theta = \frac◆LB◆1◆RB◆◆LB◆\cos\theta◆RB◆, etc.

9.3 Strategy 3: Use Known Identities​

Apply the Pythagorean identity, compound angle, or double angle formulas to create simplifications.

9.4 Strategy 4: Multiply by the Conjugate​

When you see expressions of the form a±ba \pm b in a denominator or numerator, multiply top and bottom by the conjugate a∓ba \mp b to produce a difference of squares.

Details

Example 1: Strategy 2 (express in sin and cos) Prove that cot⁡A+tan⁡A=◆LB◆2◆RB◆◆LB◆sin⁡2A◆RB◆\cot A + \tan A = \dfrac◆LB◆2◆RB◆◆LB◆\sin 2A◆RB◆.

cot⁡A+tan⁡A=◆LB◆cos⁡A◆RB◆◆LB◆sin⁡A◆RB◆+◆LB◆sin⁡A◆RB◆◆LB◆cos⁡A◆RB◆=◆LB◆cos⁡2A+sin⁡2A◆RB◆◆LB◆sin⁡Acos⁡A◆RB◆=◆LB◆1◆RB◆◆LB◆sin⁡Acos⁡A◆RB◆=◆LB◆2◆RB◆◆LB◆2sin⁡Acos⁡A◆RB◆=◆LB◆2◆RB◆◆LB◆sin⁡2A◆RB◆■\begin{aligned} \cot A + \tan A &= \frac◆LB◆\cos A◆RB◆◆LB◆\sin A◆RB◆ + \frac◆LB◆\sin A◆RB◆◆LB◆\cos A◆RB◆ \\ &= \frac◆LB◆\cos^2 A + \sin^2 A◆RB◆◆LB◆\sin A \cos A◆RB◆ \\ &= \frac◆LB◆1◆RB◆◆LB◆\sin A \cos A◆RB◆ \\ &= \frac◆LB◆2◆RB◆◆LB◆2\sin A \cos A◆RB◆ = \frac◆LB◆2◆RB◆◆LB◆\sin 2A◆RB◆ \quad \blacksquare \end{aligned}
Details

Example 2: Strategy 3 (use known identities) Prove that sin⁡3A=3sin⁡A−4sin⁡3A\sin 3A = 3\sin A - 4\sin^3 A.

sin⁡3A=sin⁡(2A+A)=sin⁡2Acos⁡A+cos⁡2Asin⁡A=2sin⁡Acos⁡2A+(1−2sin⁡2A)sin⁡A=2sin⁡A(1−sin⁡2A)+sin⁡A−2sin⁡3A=2sin⁡A−2sin⁡3A+sin⁡A−2sin⁡3A=3sin⁡A−4sin⁡3A■\begin{aligned} \sin 3A &= \sin(2A + A) \\ &= \sin 2A \cos A + \cos 2A \sin A \\ &= 2\sin A \cos^2 A + (1 - 2\sin^2 A)\sin A \\ &= 2\sin A(1 - \sin^2 A) + \sin A - 2\sin^3 A \\ &= 2\sin A - 2\sin^3 A + \sin A - 2\sin^3 A \\ &= 3\sin A - 4\sin^3 A \quad \blacksquare \end{aligned}
Details

Example 3: Strategy 4 (multiply by conjugate) Prove that ◆LB◆1◆RB◆◆LB◆sec⁡A+tan⁡A◆RB◆=sec⁡A−tan⁡A\dfrac◆LB◆1◆RB◆◆LB◆\sec A + \tan A◆RB◆ = \sec A - \tan A.

Multiply numerator and denominator by sec⁡A−tan⁡A\sec A - \tan A:

◆LB◆1◆RB◆◆LB◆sec⁡A+tan⁡A◆RB◆⋅◆LB◆sec⁡A−tan⁡A◆RB◆◆LB◆sec⁡A−tan⁡A◆RB◆=◆LB◆sec⁡A−tan⁡A◆RB◆◆LB◆sec⁡2A−tan⁡2A◆RB◆=◆LB◆sec⁡A−tan⁡A◆RB◆◆LB◆1+tan⁡2A−tan⁡2A◆RB◆=◆LB◆sec⁡A−tan⁡A◆RB◆◆LB◆1◆RB◆=sec⁡A−tan⁡A■\begin{aligned} \frac◆LB◆1◆RB◆◆LB◆\sec A + \tan A◆RB◆ \cdot \frac◆LB◆\sec A - \tan A◆RB◆◆LB◆\sec A - \tan A◆RB◆ &= \frac◆LB◆\sec A - \tan A◆RB◆◆LB◆\sec^2 A - \tan^2 A◆RB◆ \\ &= \frac◆LB◆\sec A - \tan A◆RB◆◆LB◆1 + \tan^2 A - \tan^2 A◆RB◆ \\ &= \frac◆LB◆\sec A - \tan A◆RB◆◆LB◆1◆RB◆ \\ &= \sec A - \tan A \quad \blacksquare \end{aligned}

where we used sec⁡2A=1+tan⁡2A\sec^2 A = 1 + \tan^2 A so that sec⁡2A−tan⁡2A=1\sec^2 A - \tan^2 A = 1.

Details

Example 4: Strategy 1 (work with one side) Prove that ◆LB◆cos⁡2A◆RB◆◆LB◆1+sin⁡2A◆RB◆=◆LB◆cos⁡A−sin⁡A◆RB◆◆LB◆cos⁡A+sin⁡A◆RB◆\dfrac◆LB◆\cos 2A◆RB◆◆LB◆1 + \sin 2A◆RB◆ = \dfrac◆LB◆\cos A - \sin A◆RB◆◆LB◆\cos A + \sin A◆RB◆.

Working from the LHS:

◆LB◆cos⁡2A◆RB◆◆LB◆1+sin⁡2A◆RB◆=◆LB◆cos⁡2A−sin⁡2A◆RB◆◆LB◆1+2sin⁡Acos⁡A◆RB◆=◆LB◆(cos⁡A−sin⁡A)(cos⁡A+sin⁡A)◆RB◆◆LB◆cos⁡2A+2sin⁡Acos⁡A+sin⁡2A◆RB◆=◆LB◆(cos⁡A−sin⁡A)(cos⁡A+sin⁡A)◆RB◆◆LB◆(cos⁡A+sin⁡A)2◆RB◆=◆LB◆cos⁡A−sin⁡A◆RB◆◆LB◆cos⁡A+sin⁡A◆RB◆■\begin{aligned} \frac◆LB◆\cos 2A◆RB◆◆LB◆1 + \sin 2A◆RB◆ &= \frac◆LB◆\cos^2 A - \sin^2 A◆RB◆◆LB◆1 + 2\sin A \cos A◆RB◆ \\ &= \frac◆LB◆(\cos A - \sin A)(\cos A + \sin A)◆RB◆◆LB◆\cos^2 A + 2\sin A\cos A + \sin^2 A◆RB◆ \\ &= \frac◆LB◆(\cos A - \sin A)(\cos A + \sin A)◆RB◆◆LB◆(\cos A + \sin A)^2◆RB◆ \\ &= \frac◆LB◆\cos A - \sin A◆RB◆◆LB◆\cos A + \sin A◆RB◆ \quad \blacksquare \end{aligned}

10. Trigonometric Graphs​

10.1 Key Features​

FunctionPeriodAmplitudeDomainRange
sin⁡x\sin x2π2\pi11R\mathbb{R}[−1,1][-1, 1]
cos⁡x\cos x2π2\pi11R\mathbb{R}[−1,1][-1, 1]
tan⁡x\tan xπ\piundefinedx≠◆LB◆π◆RB◆◆LB◆2◆RB◆+nπx \neq \frac◆LB◆\pi◆RB◆◆LB◆2◆RB◆ + n\piR\mathbb{R}

Key values of sin⁡\sin and cos⁡\cos:

Angle00◆LB◆π◆RB◆◆LB◆6◆RB◆\frac◆LB◆\pi◆RB◆◆LB◆6◆RB◆◆LB◆π◆RB◆◆LB◆4◆RB◆\frac◆LB◆\pi◆RB◆◆LB◆4◆RB◆◆LB◆π◆RB◆◆LB◆3◆RB◆\frac◆LB◆\pi◆RB◆◆LB◆3◆RB◆◆LB◆π◆RB◆◆LB◆2◆RB◆\frac◆LB◆\pi◆RB◆◆LB◆2◆RB◆π\pi◆LB◆3π◆RB◆◆LB◆2◆RB◆\frac◆LB◆3\pi◆RB◆◆LB◆2◆RB◆2π2\pi
sin⁡\sin0012\frac{1}{2}◆LB◆2◆RB◆◆LB◆2◆RB◆\frac◆LB◆\sqrt{2}◆RB◆◆LB◆2◆RB◆◆LB◆3◆RB◆◆LB◆2◆RB◆\frac◆LB◆\sqrt{3}◆RB◆◆LB◆2◆RB◆1100−1-100
cos⁡\cos11◆LB◆3◆RB◆◆LB◆2◆RB◆\frac◆LB◆\sqrt{3}◆RB◆◆LB◆2◆RB◆◆LB◆2◆RB◆◆LB◆2◆RB◆\frac◆LB◆\sqrt{2}◆RB◆◆LB◆2◆RB◆12\frac{1}{2}00−1-10011

10.2 Transformations​

For the general form y=Asin⁡(Bx+C)+Dy = A\sin(Bx + C) + D (and similarly for cos⁡\cos):

  • ∣A∣|A| is the amplitude (vertical stretch from the midline)
  • The period is ◆LB◆2π◆RB◆◆LB◆∣B∣◆RB◆\dfrac◆LB◆2\pi◆RB◆◆LB◆|B|◆RB◆
  • The phase shift is −CB-\dfrac{C}{B} (horizontal shift)
  • DD is the vertical shift (midline is y=Dy = D)

For tan⁡\tan, the period is ◆LB◆π◆RB◆◆LB◆∣B∣◆RB◆\dfrac◆LB◆\pi◆RB◆◆LB◆|B|◆RB◆ and amplitude is not defined.

y = A sin(Bx + C) + D

Use the sliders to adjust the amplitude, period, phase shift, and vertical translation of the trigonometric functions. Observe how each parameter affects the graph of y=Asin⁡(Bx+C)+Dy = A\sin(Bx + C) + D.

Details

Worked example Describe the key features of y=2sin⁡ ⁣(2x−◆LB◆π◆RB◆◆LB◆3◆RB◆)y = 2\sin\!\left(2x - \frac◆LB◆\pi◆RB◆◆LB◆3◆RB◆\right) for 0≤x≤2π0 \leq x \leq 2\pi.

Amplitude: ∣A∣=2|A| = 2, so the range is [−2,2][-2, 2].

Period: ◆LB◆2π◆RB◆◆LB◆∣B∣◆RB◆=◆LB◆2π◆RB◆◆LB◆2◆RB◆=π\dfrac◆LB◆2\pi◆RB◆◆LB◆|B|◆RB◆ = \dfrac◆LB◆2\pi◆RB◆◆LB◆2◆RB◆ = \pi.

Phase shift: −CB=−◆LB◆−π/3◆RB◆◆LB◆2◆RB◆=◆LB◆π◆RB◆◆LB◆6◆RB◆-\dfrac{C}{B} = -\dfrac◆LB◆-\pi/3◆RB◆◆LB◆2◆RB◆ = \dfrac◆LB◆\pi◆RB◆◆LB◆6◆RB◆ (shift right by ◆LB◆π◆RB◆◆LB◆6◆RB◆\frac◆LB◆\pi◆RB◆◆LB◆6◆RB◆).

Key points. The first cycle begins at x=◆LB◆π◆RB◆◆LB◆6◆RB◆x = \frac◆LB◆\pi◆RB◆◆LB◆6◆RB◆ (where the curve crosses the midline upward). Subsequent key points within [0,2π][0, 2\pi]:

  • Maximum at x=◆LB◆5π◆RB◆◆LB◆12◆RB◆x = \frac◆LB◆5\pi◆RB◆◆LB◆12◆RB◆ (value 22)
  • Midline crossing (down) at x=◆LB◆2π◆RB◆◆LB◆3◆RB◆x = \frac◆LB◆2\pi◆RB◆◆LB◆3◆RB◆
  • Minimum at x=◆LB◆11π◆RB◆◆LB◆12◆RB◆x = \frac◆LB◆11\pi◆RB◆◆LB◆12◆RB◆ (value −2-2)
  • Midline crossing (up) at x=◆LB◆7π◆RB◆◆LB◆6◆RB◆x = \frac◆LB◆7\pi◆RB◆◆LB◆6◆RB◆

Since the period is π\pi, the second cycle repeats with all xx-values shifted by π\pi:

  • Maximum at x=◆LB◆17π◆RB◆◆LB◆12◆RB◆x = \frac◆LB◆17\pi◆RB◆◆LB◆12◆RB◆ (value 22)
  • Minimum at x=◆LB◆23π◆RB◆◆LB◆12◆RB◆x = \frac◆LB◆23\pi◆RB◆◆LB◆12◆RB◆ (value −2-2)

yy-intercept: y=2sin⁡ ⁣(−◆LB◆π◆RB◆◆LB◆3◆RB◆)=−3≈−1.73y = 2\sin\!\left(-\frac◆LB◆\pi◆RB◆◆LB◆3◆RB◆\right) = -\sqrt{3} \approx -1.73.

11. Problem Set​

Problem 1. Prove that ◆LB◆sin⁡2θ◆RB◆◆LB◆1+cos⁡2θ◆RB◆=tan⁡θ\frac◆LB◆\sin 2\theta◆RB◆◆LB◆1 + \cos 2\theta◆RB◆ = \tan\theta.

Solution
◆LB◆sin⁡2θ◆RB◆◆LB◆1+cos⁡2θ◆RB◆=◆LB◆2sin⁡θcos⁡θ◆RB◆◆LB◆1+(2cos⁡2θ−1)◆RB◆=◆LB◆2sin⁡θcos⁡θ◆RB◆◆LB◆2cos⁡2θ◆RB◆=◆LB◆sin⁡θ◆RB◆◆LB◆cos⁡θ◆RB◆=tan⁡θ■\begin{aligned} \frac◆LB◆\sin 2\theta◆RB◆◆LB◆1 + \cos 2\theta◆RB◆ &= \frac◆LB◆2\sin\theta\cos\theta◆RB◆◆LB◆1 + (2\cos^2\theta - 1)◆RB◆ \\ &= \frac◆LB◆2\sin\theta\cos\theta◆RB◆◆LB◆2\cos^2\theta◆RB◆ \\ &= \frac◆LB◆\sin\theta◆RB◆◆LB◆\cos\theta◆RB◆ = \tan\theta \quad \blacksquare \end{aligned}

If you get this wrong, revise: Double angle formulas

Problem 2. Solve sin⁡2x=sin⁡x\sin 2x = \sin x for 0≤x<2π0 \leq x < 2\pi.

Details

Solution 2sin⁡xcos⁡x=sin⁡x2\sin x\cos x = \sin x

sin⁡x(2cos⁡x−1)=0\sin x(2\cos x - 1) = 0

sin⁡x=0  ⟹  x=0,π\sin x = 0 \implies x = 0, \pi.

2cos⁡x−1=0  ⟹  cos⁡x=12  ⟹  x=◆LB◆π◆RB◆◆LB◆3◆RB◆,◆LB◆5π◆RB◆◆LB◆3◆RB◆2\cos x - 1 = 0 \implies \cos x = \frac{1}{2} \implies x = \frac◆LB◆\pi◆RB◆◆LB◆3◆RB◆, \frac◆LB◆5\pi◆RB◆◆LB◆3◆RB◆.

Solutions: x=0,◆LB◆π◆RB◆◆LB◆3◆RB◆,π,◆LB◆5π◆RB◆◆LB◆3◆RB◆x = 0, \frac◆LB◆\pi◆RB◆◆LB◆3◆RB◆, \pi, \frac◆LB◆5\pi◆RB◆◆LB◆3◆RB◆.

If you get this wrong, revise: Solving trig equations

Problem 3. Express 4sin⁡θ−3cos⁡θ4\sin\theta - 3\cos\theta in the form Rsin⁡(θ−α)R\sin(\theta - \alpha), where R>0R > 0 and 0<α<◆LB◆π◆RB◆◆LB◆2◆RB◆0 < \alpha < \frac◆LB◆\pi◆RB◆◆LB◆2◆RB◆.

Details

Solution Rsin⁡(θ−α)=R(sin⁡θcos⁡α−cos⁡θsin⁡α)=Rcos⁡αsin⁡θ−Rsin⁡αcos⁡θR\sin(\theta - \alpha) = R(\sin\theta\cos\alpha - \cos\theta\sin\alpha) = R\cos\alpha\sin\theta - R\sin\alpha\cos\theta.

Comparing coefficients: Rcos⁡α=4R\cos\alpha = 4 and Rsin⁡α=3R\sin\alpha = 3.

R2=42+32=25  ⟹  R=5R^2 = 4^2 + 3^2 = 25 \implies R = 5

tan⁡α=34  ⟹  α=arctan⁡(34)≈0.6435rad\tan\alpha = \frac{3}{4} \implies \alpha = \arctan\left(\frac{3}{4}\right) \approx 0.6435 \mathrm{ rad}

4sin⁡θ−3cos⁡θ=5sin⁡(θ−0.6435)4\sin\theta - 3\cos\theta = 5\sin(\theta - 0.6435)

If you get this wrong, revise: Compound angle formulas

Problem 4. Find the exact value of sin⁡75∘\sin 75^\circ.

Details

Solution sin⁡75°=sin⁡(45°+30°)=sin⁡45°cos⁡30°+cos⁡45°sin⁡30∘\sin 75° = \sin(45° + 30°) = \sin 45°\cos 30° + \cos 45°\sin 30^\circ

=◆LB◆2◆RB◆◆LB◆2◆RB◆⋅◆LB◆3◆RB◆◆LB◆2◆RB◆+◆LB◆2◆RB◆◆LB◆2◆RB◆⋅12=◆LB◆6+2◆RB◆◆LB◆4◆RB◆= \frac◆LB◆\sqrt{2}◆RB◆◆LB◆2◆RB◆ \cdot \frac◆LB◆\sqrt{3}◆RB◆◆LB◆2◆RB◆ + \frac◆LB◆\sqrt{2}◆RB◆◆LB◆2◆RB◆ \cdot \frac{1}{2} = \frac◆LB◆\sqrt{6} + \sqrt{2}◆RB◆◆LB◆4◆RB◆

If you get this wrong, revise: Compound angle formulas

Problem 5. Prove that ◆LB◆1−cos⁡2θ◆RB◆◆LB◆sin⁡2θ◆RB◆=tan⁡θ\frac◆LB◆1 - \cos 2\theta◆RB◆◆LB◆\sin 2\theta◆RB◆ = \tan\theta.

Solution
◆LB◆1−cos⁡2θ◆RB◆◆LB◆sin⁡2θ◆RB◆=◆LB◆1−(1−2sin⁡2θ)◆RB◆◆LB◆2sin⁡θcos⁡θ◆RB◆=◆LB◆2sin⁡2θ◆RB◆◆LB◆2sin⁡θcos⁡θ◆RB◆=◆LB◆sin⁡θ◆RB◆◆LB◆cos⁡θ◆RB◆=tan⁡θ■\begin{aligned} \frac◆LB◆1 - \cos 2\theta◆RB◆◆LB◆\sin 2\theta◆RB◆ &= \frac◆LB◆1 - (1 - 2\sin^2\theta)◆RB◆◆LB◆2\sin\theta\cos\theta◆RB◆ \\ &= \frac◆LB◆2\sin^2\theta◆RB◆◆LB◆2\sin\theta\cos\theta◆RB◆ \\ &= \frac◆LB◆\sin\theta◆RB◆◆LB◆\cos\theta◆RB◆ = \tan\theta \quad \blacksquare \end{aligned}

If you get this wrong, revise: Double angle formulas

Problem 6. A sector of a circle of radius 8 cm has an angle of 1.2 radians. Find the arc length and the area of the sector.

Details

Solution Arc length: s=rθ=8×1.2=9.6s = r\theta = 8 \times 1.2 = 9.6 cm.

Area: A=12r2θ=12(64)(1.2)=38.4A = \frac{1}{2}r^2\theta = \frac{1}{2}(64)(1.2) = 38.4 cm².

If you get this wrong, revise: Radian measure

Problem 7. Solve 3cos⁡2x+2sin⁡x−2=03\cos^2 x + 2\sin x - 2 = 0 for −π≤x≤π-\pi \leq x \leq \pi.

Details

Solution 3(1−sin⁡2x)+2sin⁡x−2=03(1 - \sin^2 x) + 2\sin x - 2 = 0

3−3sin⁡2x+2sin⁡x−2=03 - 3\sin^2 x + 2\sin x - 2 = 0

−3sin⁡2x+2sin⁡x+1=0-3\sin^2 x + 2\sin x + 1 = 0

3sin⁡2x−2sin⁡x−1=03\sin^2 x - 2\sin x - 1 = 0

(3sin⁡x+1)(sin⁡x−1)=0(3\sin x + 1)(\sin x - 1) = 0

sin⁡x=−13  ⟹  x=arcsin⁡(−13)≈−0.3398\sin x = -\frac{1}{3} \implies x = \arcsin(-\frac{1}{3}) \approx -0.3398 or x=−π−arcsin⁡(−13)=−π+0.3398≈−2.802x = -\pi - \arcsin(-\frac{1}{3}) = -\pi + 0.3398 \approx -2.802 (since 3.481>π3.481 > \pi, subtract 2π2\pi to stay in [−π,π][-\pi, \pi]).

sin⁡x=1  ⟹  x=◆LB◆π◆RB◆◆LB◆2◆RB◆\sin x = 1 \implies x = \frac◆LB◆\pi◆RB◆◆LB◆2◆RB◆.

Solutions: x≈−2.802,−0.340,◆LB◆π◆RB◆◆LB◆2◆RB◆x \approx -2.802, -0.340, \frac◆LB◆\pi◆RB◆◆LB◆2◆RB◆.

If you get this wrong, revise: Solving trig equations

Problem 8. Use small angle approximations to estimate ◆LB◆sin⁡0.05◆RB◆◆LB◆cos⁡0.05◆RB◆\frac◆LB◆\sin 0.05◆RB◆◆LB◆\cos 0.05◆RB◆.

Details

Solution sin⁡0.05≈0.05\sin 0.05 \approx 0.05

cos⁡0.05≈1−0.0522=1−0.00125=0.99875\cos 0.05 \approx 1 - \frac{0.05^2}{2} = 1 - 0.00125 = 0.99875

◆LB◆sin⁡0.05◆RB◆◆LB◆cos⁡0.05◆RB◆≈0.050.99875≈0.05006\frac◆LB◆\sin 0.05◆RB◆◆LB◆\cos 0.05◆RB◆ \approx \frac{0.05}{0.99875} \approx 0.05006

(Alternatively, tan⁡0.05≈0.05\tan 0.05 \approx 0.05 directly.)

If you get this wrong, revise: Small angle approximations

Problem 9. Prove that ◆LB◆1+sin⁡2A◆RB◆◆LB◆cos⁡2A◆RB◆=tan⁡ ⁣(A+◆LB◆π◆RB◆◆LB◆4◆RB◆)\dfrac◆LB◆1 + \sin 2A◆RB◆◆LB◆\cos 2A◆RB◆ = \tan\!\left(A + \dfrac◆LB◆\pi◆RB◆◆LB◆4◆RB◆\right).

Details

Solution Working from the LHS:

◆LB◆1+sin⁡2A◆RB◆◆LB◆cos⁡2A◆RB◆=◆LB◆1+2sin⁡Acos⁡A◆RB◆◆LB◆cos⁡2A−sin⁡2A◆RB◆=◆LB◆sin⁡2A+2sin⁡Acos⁡A+cos⁡2A◆RB◆◆LB◆(cos⁡A−sin⁡A)(cos⁡A+sin⁡A)◆RB◆=◆LB◆(sin⁡A+cos⁡A)2◆RB◆◆LB◆(cos⁡A−sin⁡A)(cos⁡A+sin⁡A)◆RB◆=◆LB◆sin⁡A+cos⁡A◆RB◆◆LB◆cos⁡A−sin⁡A◆RB◆\begin{aligned} \frac◆LB◆1 + \sin 2A◆RB◆◆LB◆\cos 2A◆RB◆ &= \frac◆LB◆1 + 2\sin A \cos A◆RB◆◆LB◆\cos^2 A - \sin^2 A◆RB◆ \\ &= \frac◆LB◆\sin^2 A + 2\sin A \cos A + \cos^2 A◆RB◆◆LB◆(\cos A - \sin A)(\cos A + \sin A)◆RB◆ \\ &= \frac◆LB◆(\sin A + \cos A)^2◆RB◆◆LB◆(\cos A - \sin A)(\cos A + \sin A)◆RB◆ \\ &= \frac◆LB◆\sin A + \cos A◆RB◆◆LB◆\cos A - \sin A◆RB◆ \end{aligned}

Dividing numerator and denominator by cos⁡A\cos A:

=◆LB◆tan⁡A+1◆RB◆◆LB◆1−tan⁡A◆RB◆= \frac◆LB◆\tan A + 1◆RB◆◆LB◆1 - \tan A◆RB◆

From the tangent compound angle formula:

tan⁡ ⁣(A+◆LB◆π◆RB◆◆LB◆4◆RB◆)=◆LB◆tan⁡A+tan⁡◆LB◆π◆RB◆◆LB◆4◆RB◆◆RB◆◆LB◆1−tan⁡Atan⁡◆LB◆π◆RB◆◆LB◆4◆RB◆◆RB◆=◆LB◆tan⁡A+1◆RB◆◆LB◆1−tan⁡A◆RB◆■\tan\!\left(A + \frac◆LB◆\pi◆RB◆◆LB◆4◆RB◆\right) = \frac◆LB◆\tan A + \tan\frac◆LB◆\pi◆RB◆◆LB◆4◆RB◆◆RB◆◆LB◆1 - \tan A \tan\frac◆LB◆\pi◆RB◆◆LB◆4◆RB◆◆RB◆ = \frac◆LB◆\tan A + 1◆RB◆◆LB◆1 - \tan A◆RB◆ \quad \blacksquare

If you get this wrong, revise: Compound angle formulas

Problem 10. The area of a sector is 20cm220\mathrm{ cm}^2 and the arc length is 10cm10\mathrm{ cm}. Find the radius and the angle.

Details

Solution s=rθ=10  ⟹  θ=10rs = r\theta = 10 \implies \theta = \frac{10}{r}

A=12r2θ=12r2⋅10r=5r=20  ⟹  r=4A = \frac{1}{2}r^2\theta = \frac{1}{2}r^2 \cdot \frac{10}{r} = 5r = 20 \implies r = 4

θ=104=2.5radians\theta = \frac{10}{4} = 2.5 \mathrm{ radians}

If you get this wrong, revise: Radian measure

Problem 11. Find the exact value of cos⁡◆LB◆π◆RB◆◆LB◆12◆RB◆\cos\dfrac◆LB◆\pi◆RB◆◆LB◆12◆RB◆ using a half-angle formula.

Details

Solution cos⁡◆LB◆π◆RB◆◆LB◆12◆RB◆=cos⁡◆LB◆π/6◆RB◆◆LB◆2◆RB◆\cos\frac◆LB◆\pi◆RB◆◆LB◆12◆RB◆ = \cos\frac◆LB◆\pi/6◆RB◆◆LB◆2◆RB◆, so we apply the half-angle formula with θ=◆LB◆π◆RB◆◆LB◆6◆RB◆\theta = \frac◆LB◆\pi◆RB◆◆LB◆6◆RB◆.

Since 0<◆LB◆π◆RB◆◆LB◆12◆RB◆<◆LB◆π◆RB◆◆LB◆2◆RB◆0 \lt{} \frac◆LB◆\pi◆RB◆◆LB◆12◆RB◆ \lt{} \frac◆LB◆\pi◆RB◆◆LB◆2◆RB◆ (first quadrant), cos⁡◆LB◆π◆RB◆◆LB◆12◆RB◆>0\cos\frac◆LB◆\pi◆RB◆◆LB◆12◆RB◆ > 0.

cos⁡◆LB◆π◆RB◆◆LB◆12◆RB◆=◆LB◆◆LB◆1+cos⁡π6◆RB◆◆LB◆2◆RB◆◆RB◆=◆LB◆◆LB◆1+32◆RB◆◆LB◆2◆RB◆◆RB◆=◆LB◆◆LB◆2+3◆RB◆◆LB◆4◆RB◆◆RB◆=◆LB◆◆LB◆2+3◆RB◆◆RB◆◆LB◆2◆RB◆\begin{aligned} \cos\frac◆LB◆\pi◆RB◆◆LB◆12◆RB◆ &= \sqrt◆LB◆\frac◆LB◆1 + \cos\frac{\pi}{6}◆RB◆◆LB◆2◆RB◆◆RB◆ \\ &= \sqrt◆LB◆\frac◆LB◆1 + \frac{\sqrt{3}}{2}◆RB◆◆LB◆2◆RB◆◆RB◆ \\ &= \sqrt◆LB◆\frac◆LB◆2 + \sqrt{3}◆RB◆◆LB◆4◆RB◆◆RB◆ \\ &= \frac◆LB◆\sqrt◆LB◆2 + \sqrt{3}◆RB◆◆RB◆◆LB◆2◆RB◆ \end{aligned}

This can also be verified using the compound angle formula: cos⁡◆LB◆π◆RB◆◆LB◆12◆RB◆=cos⁡ ⁣(◆LB◆π◆RB◆◆LB◆3◆RB◆−◆LB◆π◆RB◆◆LB◆4◆RB◆)=cos⁡◆LB◆π◆RB◆◆LB◆3◆RB◆cos⁡◆LB◆π◆RB◆◆LB◆4◆RB◆+sin⁡◆LB◆π◆RB◆◆LB◆3◆RB◆sin⁡◆LB◆π◆RB◆◆LB◆4◆RB◆=◆LB◆6+2◆RB◆◆LB◆4◆RB◆\cos\frac◆LB◆\pi◆RB◆◆LB◆12◆RB◆ = \cos\!\left(\frac◆LB◆\pi◆RB◆◆LB◆3◆RB◆ - \frac◆LB◆\pi◆RB◆◆LB◆4◆RB◆\right) = \cos\frac◆LB◆\pi◆RB◆◆LB◆3◆RB◆\cos\frac◆LB◆\pi◆RB◆◆LB◆4◆RB◆ + \sin\frac◆LB◆\pi◆RB◆◆LB◆3◆RB◆\sin\frac◆LB◆\pi◆RB◆◆LB◆4◆RB◆ = \frac◆LB◆\sqrt{6} + \sqrt{2}◆RB◆◆LB◆4◆RB◆, and one can check that ◆LB◆◆LB◆2+3◆RB◆◆RB◆◆LB◆2◆RB◆=◆LB◆6+2◆RB◆◆LB◆4◆RB◆\dfrac◆LB◆\sqrt◆LB◆2+\sqrt{3}◆RB◆◆RB◆◆LB◆2◆RB◆ = \dfrac◆LB◆\sqrt{6}+\sqrt{2}◆RB◆◆LB◆4◆RB◆.

If you get this wrong, revise: Half-angle formulas

Problem 12. Find the maximum value of 2sin⁡θ+5cos⁡θ2\sin\theta + 5\cos\theta and the smallest positive value of θ\theta at which it occurs.

Details

Solution R=4+25=29R = \sqrt{4 + 25} = \sqrt{29}.

The maximum value is R=29R = \sqrt{29}.

Writing 2sin⁡θ+5cos⁡θ=29sin⁡(θ+α)2\sin\theta + 5\cos\theta = \sqrt{29}\sin(\theta + \alpha) where tan⁡α=52\tan\alpha = \dfrac{5}{2}.

The maximum occurs when sin⁡(θ+α)=1\sin(\theta + \alpha) = 1, i.e., θ+α=◆LB◆π◆RB◆◆LB◆2◆RB◆\theta + \alpha = \frac◆LB◆\pi◆RB◆◆LB◆2◆RB◆.

θ=◆LB◆π◆RB◆◆LB◆2◆RB◆−α=◆LB◆π◆RB◆◆LB◆2◆RB◆−arctan⁡52≈0.3805rad\theta = \frac◆LB◆\pi◆RB◆◆LB◆2◆RB◆ - \alpha = \frac◆LB◆\pi◆RB◆◆LB◆2◆RB◆ - \arctan\frac{5}{2} \approx 0.3805 \mathrm{ rad}

Since arctan⁡52≈1.1903<◆LB◆π◆RB◆◆LB◆2◆RB◆\arctan\frac{5}{2} \approx 1.1903 < \frac◆LB◆\pi◆RB◆◆LB◆2◆RB◆, this θ\theta is positive and is the smallest positive value.

If you get this wrong, revise: R-addition formula

Problem 13. Prove that ◆LB◆sin⁡3A◆RB◆◆LB◆sin⁡A◆RB◆−◆LB◆cos⁡3A◆RB◆◆LB◆cos⁡A◆RB◆=2\dfrac◆LB◆\sin 3A◆RB◆◆LB◆\sin A◆RB◆ - \dfrac◆LB◆\cos 3A◆RB◆◆LB◆\cos A◆RB◆ = 2.

Solution
◆LB◆sin⁡3A◆RB◆◆LB◆sin⁡A◆RB◆−◆LB◆cos⁡3A◆RB◆◆LB◆cos⁡A◆RB◆=◆LB◆sin⁡3Acos⁡A−cos⁡3Asin⁡A◆RB◆◆LB◆sin⁡Acos⁡A◆RB◆=◆LB◆sin⁡(3A−A)◆RB◆◆LB◆sin⁡Acos⁡A◆RB◆=◆LB◆sin⁡2A◆RB◆◆LB◆sin⁡Acos⁡A◆RB◆=◆LB◆2sin⁡Acos⁡A◆RB◆◆LB◆sin⁡Acos⁡A◆RB◆=2■\begin{aligned} \frac◆LB◆\sin 3A◆RB◆◆LB◆\sin A◆RB◆ - \frac◆LB◆\cos 3A◆RB◆◆LB◆\cos A◆RB◆ &= \frac◆LB◆\sin 3A \cos A - \cos 3A \sin A◆RB◆◆LB◆\sin A \cos A◆RB◆ \\ &= \frac◆LB◆\sin(3A - A)◆RB◆◆LB◆\sin A \cos A◆RB◆ \\ &= \frac◆LB◆\sin 2A◆RB◆◆LB◆\sin A \cos A◆RB◆ \\ &= \frac◆LB◆2\sin A \cos A◆RB◆◆LB◆\sin A \cos A◆RB◆ \\ &= 2 \quad \blacksquare \end{aligned}

If you get this wrong, revise: Proof strategies

Problem 14. Solve sin⁡θ+3cos⁡θ=1\sin\theta + \sqrt{3}\cos\theta = 1 for 0≤θ<2π0 \leq \theta \lt{} 2\pi.

Details

Solution R=1+3=2R = \sqrt{1 + 3} = 2, α=arctan⁡3=◆LB◆π◆RB◆◆LB◆3◆RB◆\alpha = \arctan\sqrt{3} = \frac◆LB◆\pi◆RB◆◆LB◆3◆RB◆.

2sin⁡ ⁣(θ+◆LB◆π◆RB◆◆LB◆3◆RB◆)=12\sin\!\left(\theta + \frac◆LB◆\pi◆RB◆◆LB◆3◆RB◆\right) = 1

sin⁡ ⁣(θ+◆LB◆π◆RB◆◆LB◆3◆RB◆)=12\sin\!\left(\theta + \frac◆LB◆\pi◆RB◆◆LB◆3◆RB◆\right) = \frac{1}{2}

θ+◆LB◆π◆RB◆◆LB◆3◆RB◆=◆LB◆π◆RB◆◆LB◆6◆RB◆+2nπorθ+◆LB◆π◆RB◆◆LB◆3◆RB◆=◆LB◆5π◆RB◆◆LB◆6◆RB◆+2nπ\theta + \frac◆LB◆\pi◆RB◆◆LB◆3◆RB◆ = \frac◆LB◆\pi◆RB◆◆LB◆6◆RB◆ + 2n\pi \quad \mathrm{or} \quad \theta + \frac◆LB◆\pi◆RB◆◆LB◆3◆RB◆ = \frac◆LB◆5\pi◆RB◆◆LB◆6◆RB◆ + 2n\pi

Case 1: θ=◆LB◆π◆RB◆◆LB◆6◆RB◆−◆LB◆π◆RB◆◆LB◆3◆RB◆+2nπ=−◆LB◆π◆RB◆◆LB◆6◆RB◆+2nπ\theta = \frac◆LB◆\pi◆RB◆◆LB◆6◆RB◆ - \frac◆LB◆\pi◆RB◆◆LB◆3◆RB◆ + 2n\pi = -\frac◆LB◆\pi◆RB◆◆LB◆6◆RB◆ + 2n\pi.

For n=1n = 1: θ=−◆LB◆π◆RB◆◆LB◆6◆RB◆+2π=◆LB◆11π◆RB◆◆LB◆6◆RB◆\theta = -\frac◆LB◆\pi◆RB◆◆LB◆6◆RB◆ + 2\pi = \frac◆LB◆11\pi◆RB◆◆LB◆6◆RB◆.

Case 2: θ=◆LB◆5π◆RB◆◆LB◆6◆RB◆−◆LB◆π◆RB◆◆LB◆3◆RB◆+2nπ=◆LB◆π◆RB◆◆LB◆2◆RB◆+2nπ\theta = \frac◆LB◆5\pi◆RB◆◆LB◆6◆RB◆ - \frac◆LB◆\pi◆RB◆◆LB◆3◆RB◆ + 2n\pi = \frac◆LB◆\pi◆RB◆◆LB◆2◆RB◆ + 2n\pi.

For n=0n = 0: θ=◆LB◆π◆RB◆◆LB◆2◆RB◆\theta = \frac◆LB◆\pi◆RB◆◆LB◆2◆RB◆.

Solutions: θ=◆LB◆π◆RB◆◆LB◆2◆RB◆\theta = \frac◆LB◆\pi◆RB◆◆LB◆2◆RB◆ or θ=◆LB◆11π◆RB◆◆LB◆6◆RB◆\theta = \frac◆LB◆11\pi◆RB◆◆LB◆6◆RB◆.

If you get this wrong, revise: R-addition formula

Problem 15. Prove that ◆LB◆1−tan⁡2A◆RB◆◆LB◆1+tan⁡2A◆RB◆=cos⁡2A\dfrac◆LB◆1 - \tan^2 A◆RB◆◆LB◆1 + \tan^2 A◆RB◆ = \cos 2A.

Details

Solution Using sec⁡2A=1+tan⁡2A\sec^2 A = 1 + \tan^2 A:

◆LB◆1−tan⁡2A◆RB◆◆LB◆1+tan⁡2A◆RB◆=◆LB◆1−tan⁡2A◆RB◆◆LB◆sec⁡2A◆RB◆=(1−tan⁡2A)cos⁡2A=cos⁡2A−tan⁡2Acos⁡2A=cos⁡2A−◆LB◆sin⁡2A◆RB◆◆LB◆cos⁡2A◆RB◆⋅cos⁡2A=cos⁡2A−sin⁡2A=cos⁡2A■\begin{aligned} \frac◆LB◆1 - \tan^2 A◆RB◆◆LB◆1 + \tan^2 A◆RB◆ &= \frac◆LB◆1 - \tan^2 A◆RB◆◆LB◆\sec^2 A◆RB◆ \\ &= (1 - \tan^2 A)\cos^2 A \\ &= \cos^2 A - \tan^2 A \cos^2 A \\ &= \cos^2 A - \frac◆LB◆\sin^2 A◆RB◆◆LB◆\cos^2 A◆RB◆ \cdot \cos^2 A \\ &= \cos^2 A - \sin^2 A \\ &= \cos 2A \quad \blacksquare \end{aligned}

If you get this wrong, revise: Proof strategies

tip

tip Ready to test your understanding of Trigonometry? The diagnostic test contains the hardest questions within the A-Level specification for this topic, each with a full worked solution.

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