1.05PureStretch

Trigonometric Identities

Q: What is the difference between an identity and an equation?

An identity is true for every value of the variable; an equation is true only for particular values, which you solve to find.

Q: Which identities do I need to memorise?

The Pythagorean identity, the tan quotient, the reciprocal definitions and the two identities derived from them. Compound angle formulas are in the formula book, but fluent recall saves time.

Q: Which form of 2A should I use?

The one that leaves a single trigonometric function in the expression — often chosen to cancel a constant or match the rest of the question.

Q: How do I pick between R ( + ) and R ( - )?

Use the form the question specifies. If free to choose, pick one whose expansion matches the signs of your expression, so that comes out acute.

An $\textbf{identity}$ is an equation that holds for every value of the variable, not just particular ones — which is why it is written with $\equiv$ . Trigonometric identities are the tools that turn an awkward expression into a manageable one, and nearly every difficult trigonometry question is really a question about choosing the right identity.

45 min Trigonometry

The big picture

There are only a few identities to learn, and they come in families. The two foundations, $\tan\theta \equiv \dfrac{\sin\theta}{\cos\theta}$ and $\sin^{2}\theta + \cos^{2}\theta \equiv 1$ , follow directly from the unit circle. Dividing the second by $\cos^{2}\theta$ or $\sin^{2}\theta$ gives the versions with secant, cosecant and cotangent. The compound angle formulas for $\sin(A \pm B)$ and $\cos(A \pm B)$ generate the double angle formulas as a special case, and they also let you combine $a\sin\theta + b\cos\theta$ into a single wave $R\sin(\theta + \alpha)$ . Knowing where each identity comes from means you can rebuild any you forget — and recognising which one matches the shape of an expression is the real skill being tested.

What you'll be able to do

Use $\tan\theta \equiv \dfrac{\sin\theta}{\cos\theta}$ and $\sin^{2}\theta + \cos^{2}\theta \equiv 1$
Define $\sec$ , $\operatorname{cosec}$ and $\cot$ and use $1 + \tan^{2}\theta \equiv \sec^{2}\theta$ and $1 + \cot^{2}\theta \equiv \operatorname{cosec}^{2}\theta$
Use compound and double angle formulas
Express $a\sin\theta + b\cos\theta$ in the form $R\sin(\theta \pm \alpha)$ or $R\cos(\theta \pm \alpha)$
Prove identities with clear, one-directional working

The two foundations

On the unit circle a point at angle $\theta$ is $(\cos\theta, \sin\theta)$ and lies on $x^{2} + y^{2} = 1$ . So $\sin^{2}\theta + \cos^{2}\theta \equiv 1$ for every $\theta$ . The gradient of the radius gives $\tan\theta \equiv \dfrac{\sin\theta}{\cos\theta}$ .

Rearranged forms are just as useful: $\sin^{2}\theta \equiv 1 - \cos^{2}\theta$ and $\cos^{2}\theta \equiv 1 - \sin^{2}\theta$ . They let you convert an expression into a single trigonometric function.

\sin^{2}\theta + \cos^{2}\theta \equiv 1

The Pythagorean identity.

\tan\theta \equiv \dfrac{\sin\theta}{\cos\theta}

Tangent as a quotient.

Example 1Given

\sin\theta = \tfrac{5}{13}

and

\theta

is obtuse, find the exact values of

\cos\theta

and

\tan\theta

\cos^{2}\theta = 1 - \tfrac{25}{169} = \tfrac{144}{169}

, so

\cos\theta = \pm\tfrac{12}{13}

2Obtuse means second quadrant, where cosine is negative:

\cos\theta = -\tfrac{12}{13}

\tan\theta = \dfrac{5/13}{-12/13}

Answer

\cos\theta = -\tfrac{12}{13}

\tan\theta = -\tfrac{5}{12}

Why this works

Squaring destroys sign information, so the Pythagorean identity always gives $\pm$ . The quadrant — from the question — is what restores the sign.

Reciprocal functions

The reciprocal functions are $\sec\theta = \dfrac{1}{\cos\theta}$ , $\operatorname{cosec}\theta = \dfrac{1}{\sin\theta}$ and $\cot\theta = \dfrac{1}{\tan\theta} = \dfrac{\cos\theta}{\sin\theta}$ . Each is undefined where its denominator is zero.

Dividing $\sin^{2}\theta + \cos^{2}\theta \equiv 1$ by $\cos^{2}\theta$ gives $\tan^{2}\theta + 1 \equiv \sec^{2}\theta$ . Dividing by $\sin^{2}\theta$ gives $1 + \cot^{2}\theta \equiv \operatorname{cosec}^{2}\theta$ .

These are the identities to reach for when an expression mixes $\tan$ with $\sec$ , or $\cot$ with $\operatorname{cosec}$ .

1 + \tan^{2}\theta \equiv \sec^{2}\theta

Divide the Pythagorean identity by

\cos^{2}\theta

1 + \cot^{2}\theta \equiv \operatorname{cosec}^{2}\theta

Divide by

\sin^{2}\theta

Tip — Match the first letters: sec goes with cos, cosec goes with sin. Students swap these constantly.

Compound and double angles

The compound angle formulas expand a trigonometric function of a sum or difference. They are given in the OCR formula book, but you must be able to use them fluently.

Setting $B = A$ gives the double angle formulas. The cosine version has three forms, obtained using $\sin^{2}A + \cos^{2}A \equiv 1$ ; choose the one that leaves a single function.

\sin(A \pm B) \equiv \sin A\cos B \pm \cos A\sin B

Compound angle, sine.

\cos(A \pm B) \equiv \cos A\cos B \mp \sin A\sin B

Compound angle, cosine (note the sign flip).

\sin 2A \equiv 2\sin A\cos A

Double angle, sine.

\cos 2A \equiv \cos^{2}A - \sin^{2}A \equiv 2\cos^{2}A - 1 \equiv 1 - 2\sin^{2}A

Double angle, cosine.

Example 1Find the exact value of

\cos 75^{\circ}

\cos 75^{\circ} = \cos(45^{\circ} + 30^{\circ}) = \cos 45^{\circ}\cos 30^{\circ} - \sin 45^{\circ}\sin 30^{\circ}

= \tfrac{\sqrt2}{2} \cdot \tfrac{\sqrt3}{2} - \tfrac{\sqrt2}{2} \cdot \tfrac12

Answer

\dfrac{\sqrt6 - \sqrt2}{4}

The harmonic form

Any expression $a\sin\theta + b\cos\theta$ is a single sine wave. Write it as $R\sin(\theta + \alpha)$ , expand, and compare coefficients: $R\cos\alpha = a$ and $R\sin\alpha = b$ .

Then $R = \sqrt{a^{2} + b^{2}}$ and $\tan\alpha = \dfrac{b}{a}$ . With $R > 0$ and $\alpha$ acute, the signs of $a$ and $b$ decide which of the four forms ( $R\sin(\theta \pm \alpha)$ , $R\cos(\theta \pm \alpha)$ ) to use.

The harmonic form immediately gives the maximum value $R$ , the minimum $-R$ , and where they occur — which is why it appears in modelling questions about tides, temperatures and alternating current.

Example 1Express

3\sin\theta + 4\cos\theta

in the form

R\sin(\theta + \alpha)

with

R > 0

and

0 < \alpha < 90^{\circ}

. Hence find its maximum value and the smallest positive

\theta

at which it occurs.

R\sin(\theta + \alpha) = R\sin\theta\cos\alpha + R\cos\theta\sin\alpha

, so

R\cos\alpha = 3

and

R\sin\alpha = 4

R = \sqrt{9 + 16} = 5

and

\tan\alpha = \tfrac43

, so

\alpha = 53.13^{\circ}

3The maximum is

5

, when

\theta + 53.13^{\circ} = 90^{\circ}

Answer

5\sin(\theta + 53.13^{\circ})

; maximum

5

\theta = 36.87^{\circ}

Tip — Write the two comparison equations $R\cos\alpha = \dots$ and $R\sin\alpha = \dots$ before calculating. Getting $\tan\alpha$ upside down is the common error, and these lines prevent it.

Why the addition formulae follow from geometry

Take unit perpendicular directions $\mathbf u=(\cos A,\sin A)$ and $\mathbf v=(-\sin A,\cos A)$ . They are the original horizontal and vertical directions rotated anticlockwise by angle A. The coordinates of v follow by a further quarter-turn.

A unit segment making angle B with u has projections $\cos B$ along u and $\sin B$ along v. It therefore has vector $\cos B\,\mathbf u+\sin B\,\mathbf v$ . It also makes angle A+B with the original horizontal axis. Comparing horizontal and vertical projections proves the two addition identities.

Example 1Use these projections to prove the sine and cosine addition formulae, then obtain a double-angle identity.

1The horizontal coordinate is

\cos B\cos A-\sin B\sin A

, and is also

\cos(A+B)

2The vertical coordinate is

\cos B\sin A+\sin B\cos A

, and is also

\sin(A+B)

3Set B = A to obtain

\sin2A=2\sin A\cos A

and

\cos2A=\cos^2A-\sin^2A

4Replace B by -B, using the evenness of cosine and oddness of sine, to obtain the difference formulae. The tangent addition formula follows by dividing the sine result by the cosine result where the denominator is nonzero.

AnswerThe formulae express the horizontal and vertical projections of the same rotated unit segment in two coordinate systems.

Think like an examiner

Prove identities by working from one side to the other; do not treat the identity as an equation to be solved.
Choose the form of $\cos 2A$ that leaves a single trigonometric function.
State $R\cos\alpha$ and $R\sin\alpha$ equations explicitly when finding the harmonic form.
Use the quadrant to choose signs after using the Pythagorean identity.

Common misconceptions

Writing

\sec\theta = \dfrac{1}{\sin\theta}

\sec\theta = \dfrac{1}{\cos\theta}

; cosec goes with sine.

Expanding

\cos(A + B)

\cos A\cos B + \sin A\sin B

The sign flips:

\cos(A+B) = \cos A\cos B - \sin A\sin B

Writing

\sin 2A = 2\sin A

\sin 2A = 2\sin A\cos A

Starting a proof by writing the identity and manipulating both sides.

Start from one side and transform it until it becomes the other.

Trigonometric identities

\sin^{2}\theta + \cos^{2}\theta \equiv 1

Pythagorean.

1 + \tan^{2}\theta \equiv \sec^{2}\theta

Secant form.

\sin 2A \equiv 2\sin A\cos A

Double angle.

a\sin\theta + b\cos\theta \equiv R\sin(\theta + \alpha)

Harmonic form,

R = \sqrt{a^{2}+b^{2}}

Stretch yourself

Prove that $\dfrac{1 - \cos 2\theta}{\sin 2\theta} \equiv \tan\theta$ , and state the values of $\theta$ in $0 \le \theta < 2\pi$ for which the identity is not defined.

Hint — Use $\cos 2\theta \equiv 1 - 2\sin^{2}\theta$ in the numerator so the 1s cancel, and $\sin 2\theta \equiv 2\sin\theta\cos\theta$ in the denominator.

Reveal solution

SolutionStart from the left-hand side. Using

\cos 2\theta \equiv 1 - 2\sin^{2}\theta

, the numerator is

1 - (1 - 2\sin^{2}\theta) = 2\sin^{2}\theta

. Using

\sin 2\theta \equiv 2\sin\theta\cos\theta

, the expression becomes

\dfrac{2\sin^{2}\theta}{2\sin\theta\cos\theta} = \dfrac{\sin\theta}{\cos\theta} \equiv \tan\theta

, which is the right-hand side. The left side is undefined when

\sin 2\theta = 0

, that is

\theta = 0, \tfrac{\pi}{2}, \pi, \tfrac{3\pi}{2}

; the cancellation of

\sin\theta

also requires

\sin\theta \neq 0

. So the identity holds for all

\theta

in the interval except

0, \tfrac{\pi}{2}, \pi

and

\tfrac{3\pi}{2}

. Check at

\theta = \tfrac{\pi}{4}

: left

= \dfrac{1 - 0}{1} = 1 = \tan\tfrac{\pi}{4}

✓.

\blacksquare

Questions students ask

What is the difference between an identity and an equation?

An identity is true for every value of the variable; an equation is true only for particular values, which you solve to find.

Which identities do I need to memorise?

The Pythagorean identity, the tan quotient, the reciprocal definitions and the two identities derived from them. Compound angle formulas are in the formula book, but fluent recall saves time.

Which form of

\cos 2A

should I use?

The one that leaves a single trigonometric function in the expression — often chosen to cancel a constant or match the rest of the question.

How do I pick between

R\sin(\theta + \alpha)

and

R\cos(\theta - \alpha)

Use the form the question specifies. If free to choose, pick one whose expansion matches the signs of your expression, so that

\alpha

comes out acute.

Key takeaways

Foundations: $\sin^{2}\theta + \cos^{2}\theta \equiv 1$ and $\tan\theta \equiv \sin\theta/\cos\theta$ .
Reciprocal forms: $1 + \tan^{2}\theta \equiv \sec^{2}\theta$ , $1 + \cot^{2}\theta \equiv \operatorname{cosec}^{2}\theta$ .
Compound angle formulas give the double angle formulas when $B = A$ .
$a\sin\theta + b\cos\theta \equiv R\sin(\theta + \alpha)$ with $R = \sqrt{a^{2}+b^{2}}$ .
Prove identities by transforming one side into the other.