Trigonometric Identities – A Level Maths Year 1

1 The Two Year 1 Identities

An identity is true for every value of the variable, and is written with $\equiv$. In Year 1 you need exactly two trigonometric identities, but you need to use them fluently: to find exact values, to simplify, to prove other identities, and to turn equations into ones you can solve. This page is part of DfE content area E · Trigonometry.

The Year 1 identities

$\tan\theta \equiv \dfrac{\sin\theta}{\cos\theta}$ ($\cos\theta \neq 0$) $\sin^2\theta + \cos^2\theta \equiv 1$

Where they come from. On the unit circle the point at angle $\theta$ is $P(\cos\theta, \sin\theta)$.

The gradient of $OP$ is $\dfrac{\sin\theta}{\cos\theta}$, and that gradient is what $\tan\theta$ means.
$OP = 1$, so by Pythagoras $(\cos\theta)^2 + (\sin\theta)^2 = 1^2$.

Because the unit-circle definition works for every angle, so do the identities.

Notation. $\sin^2\theta$ means $(\sin\theta)^2$. It is not $\sin(\theta^2)$, and it is not $\sin(\sin\theta)$.

Rearrangement	Use it when you want to…
$\sin^2\theta \equiv 1 - \cos^2\theta$	replace $\sin^2$ so an expression is all in $\cos$
$\cos^2\theta \equiv 1 - \sin^2\theta$	replace $\cos^2$ so an expression is all in $\sin$
$\sin\theta \equiv \tan\theta\cos\theta$	get rid of a fraction involving $\tan$

2 Finding One Ratio from Another

If you know one of $\sin\theta$, $\cos\theta$, $\tan\theta$ and which quadrant $\theta$ is in, the identities give the other two exactly.

Use $\sin^2\theta + \cos^2\theta = 1$ to find the size of the missing ratio.
Use the quadrant (CAST, or the graphs on Trigonometric Graphs) to choose the sign.
Use $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$ for the third.

Worked Example 1 — An obtuse angle

Given that $\sin\theta = \tfrac35$ and $\theta$ is obtuse, find the exact values of $\cos\theta$ and $\tan\theta$.

①$\cos^2\theta = 1 - \left(\tfrac35\right)^2 = 1 - \tfrac{9}{25} = \tfrac{16}{25}$, so $\cos\theta = \pm\tfrac45$.

②$\theta$ is obtuse (second quadrant), where cos is negative: $\cos\theta = -\tfrac45$.

③$\tan\theta = \dfrac{3/5}{-4/5} = -\tfrac34$

Worked Example 2 — Starting from tan

Given that $\tan\theta = -2$ and $270° < \theta < 360°$, find the exact values of $\sin\theta$ and $\cos\theta$.

①$\sin\theta = -2\cos\theta$. Substitute into the Pythagorean identity: $4\cos^2\theta + \cos^2\theta = 1$.

②$\cos^2\theta = \tfrac15$, so $\cos\theta = \pm\tfrac{1}{\sqrt5}$.

③Fourth quadrant: cos positive, sin negative. $\cos\theta = \dfrac{1}{\sqrt5} = \dfrac{\sqrt5}{5}$.

④$\sin\theta = -2\cos\theta = -\dfrac{2\sqrt5}{5}$

Alternative: draw a right-angled triangle with opposite $2$ and adjacent $1$. The hypotenuse is $\sqrt5$, giving the sizes; the quadrant gives the signs.

3 Simplifying Expressions

Worked Example 3 — Three quick simplifications

Simplify (a) $4 - 4\sin^2\theta$ (b) $\dfrac{\sin\theta}{\tan\theta}$ (c) $\dfrac{1 - \cos^2\theta}{\sin\theta\cos\theta}$

①(a) $4(1 - \sin^2\theta) = 4\cos^2\theta$

②(b) $\sin\theta \div \dfrac{\sin\theta}{\cos\theta} = \sin\theta \times \dfrac{\cos\theta}{\sin\theta} = \cos\theta$

③(c) $\dfrac{\sin^2\theta}{\sin\theta\cos\theta} = \dfrac{\sin\theta}{\cos\theta} = \tan\theta$

$\sin^2\theta + \cos^2\theta = 1$ needs the same angle. $\sin^2 2x + \cos^2 2x = 1$ is fine, but $\sin^2 x + \cos^2 2x$ does not simplify to $1$.

4 Proving Identities

The rule. Start with one side — usually the more complicated one — and transform it step by step until it becomes the other side. Write "LHS $\equiv$ …" and finish with "$\equiv$ RHS". Do not write the identity down and manipulate both sides as if it were an equation: that assumes what you are trying to prove.

Useful moves

Write $\tan$ as $\dfrac{\sin}{\cos}$ · combine fractions over a common denominator · expand brackets · spot $\sin^2 + \cos^2$ · factorise (especially a difference of two squares)

Worked Example 4 — Expanding a bracket

Prove that $(\sin\theta + \cos\theta)^2 \equiv 1 + 2\sin\theta\cos\theta$.

①LHS $\equiv \sin^2\theta + 2\sin\theta\cos\theta + \cos^2\theta$

②$\equiv (\sin^2\theta + \cos^2\theta) + 2\sin\theta\cos\theta$

③$\equiv 1 + 2\sin\theta\cos\theta \equiv$ RHS ■

Worked Example 5 — Combining fractions

Prove that $\tan\theta + \dfrac{1}{\tan\theta} \equiv \dfrac{1}{\sin\theta\cos\theta}$.

①LHS $\equiv \dfrac{\sin\theta}{\cos\theta} + \dfrac{\cos\theta}{\sin\theta}$

②$\equiv \dfrac{\sin^2\theta + \cos^2\theta}{\sin\theta\cos\theta}$

③$\equiv \dfrac{1}{\sin\theta\cos\theta} \equiv$ RHS ■

Worked Example 6 — Difference of two squares

Prove that $\sin^4\theta - \cos^4\theta \equiv 1 - 2\cos^2\theta$.

①LHS $\equiv (\sin^2\theta - \cos^2\theta)(\sin^2\theta + \cos^2\theta)$

②$\equiv (\sin^2\theta - \cos^2\theta) \times 1$

③$\equiv (1 - \cos^2\theta) - \cos^2\theta \equiv 1 - 2\cos^2\theta \equiv$ RHS ■

Worked Example 7 — Multiplying top and bottom

Prove that $\dfrac{\cos\theta}{1 - \sin\theta} \equiv \dfrac{1 + \sin\theta}{\cos\theta}$.

①LHS $\equiv \dfrac{\cos\theta}{1 - \sin\theta} \times \dfrac{1 + \sin\theta}{1 + \sin\theta} \equiv \dfrac{\cos\theta(1 + \sin\theta)}{1 - \sin^2\theta}$

②$\equiv \dfrac{\cos\theta(1 + \sin\theta)}{\cos^2\theta}$

③$\equiv \dfrac{1 + \sin\theta}{\cos\theta} \equiv$ RHS ■

Multiplying by $1 + \sin\theta$ creates $1 - \sin^2\theta$, which the identity turns into $\cos^2\theta$ — the same trick as rationalising a surd denominator.

Worked Example 8 — Disproving a false "identity"

Show that $\sin(A + B) \equiv \sin A + \sin B$ is false.

①One counter-example is enough. Take $A = B = 90°$.

②LHS $= \sin 180° = 0$, but RHS $= 1 + 1 = 2$.

③$0 \neq 2$, so the statement is not an identity. (The correct formula for $\sin(A+B)$ is a Year 2 result.)

5 Using Identities to Change an Equation

An equation containing both $\sin$ and $\cos$ usually cannot be solved directly. An identity rewrites it in one trig function.

Worked Example 9 — Rewriting in terms of sin

Show that the equation $2\cos^2 x + 3\sin x = 3$ can be written as $2\sin^2 x - 3\sin x + 1 = 0$.

①Replace $\cos^2 x$ by $1 - \sin^2 x$: $2(1 - \sin^2 x) + 3\sin x = 3$

②$2 - 2\sin^2 x + 3\sin x - 3 = 0$

③$-2\sin^2 x + 3\sin x - 1 = 0$, so $2\sin^2 x - 3\sin x + 1 = 0$ ■

This is now a quadratic in $\sin x$. Solving it is covered in Trigonometric Equations.

Worked Example 10 — Turning sin and cos into tan

Write $3\sin x = 4\cos x$ as an equation in $\tan x$.

①Divide both sides by $\cos x$ (allowed, because $\cos x = 0$ would force $\sin x = 0$ too, which is impossible since $\sin^2 x + \cos^2 x = 1$).

②$3\tan x = 4$, so $\tan x = \tfrac43$.

6 Quick Reference

Quotient identity

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

Pythagorean identity

$\sin^2\theta + \cos^2\theta \equiv 1$

Rearranged

$\sin^2\theta \equiv 1 - \cos^2\theta$; $\cos^2\theta \equiv 1 - \sin^2\theta$

Missing ratio

Size from the identity, sign from the quadrant.

Proof layout

LHS $\equiv \ldots \equiv$ RHS. Never work on both sides.

Fractions

Common denominator, then look for $\sin^2 + \cos^2$.

Fourth powers

Difference of two squares.

$1 \pm \sin\theta$

Multiply by the "conjugate" $1 \mp \sin\theta$.

Disproof

One counter-example is enough.

Equations

Use an identity to get one function only.

7 Practice Questions

Question 1

Simplify $5\sin^2\theta + 5\cos^2\theta$.

▶ Show solution

$5(\sin^2\theta + \cos^2\theta) = 5 \times 1 = 5$

Question 2

Simplify $\tan\theta\cos\theta$ and $\dfrac{1 - \sin^2\theta}{\cos\theta}$.

▶ Show solution

$\tan\theta\cos\theta = \dfrac{\sin\theta}{\cos\theta} \times \cos\theta = \sin\theta$

$\dfrac{1 - \sin^2\theta}{\cos\theta} = \dfrac{\cos^2\theta}{\cos\theta} = \cos\theta$

Question 3

Given that $\cos\theta = -\tfrac{5}{13}$ and $90° < \theta < 180°$, find the exact values of $\sin\theta$ and $\tan\theta$.

▶ Show solution

$\sin^2\theta = 1 - \tfrac{25}{169} = \tfrac{144}{169}$, so $\sin\theta = \pm\tfrac{12}{13}$.

Second quadrant: sin positive, so $\sin\theta = \tfrac{12}{13}$.

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

Question 4

Given that $\tan\theta = 2$ and $180° < \theta < 270°$, find the exact values of $\sin\theta$ and $\cos\theta$.

▶ Show solution

$\sin\theta = 2\cos\theta$, so $4\cos^2\theta + \cos^2\theta = 1$ and $\cos^2\theta = \tfrac15$.

Third quadrant: both sin and cos negative (tan positive ✓).

$\cos\theta = -\dfrac{1}{\sqrt5} = -\dfrac{\sqrt5}{5}$, $\sin\theta = 2\cos\theta = -\dfrac{2\sqrt5}{5}$.

Question 5

Prove that $(1 - \sin^2\theta)\tan\theta \equiv \sin\theta\cos\theta$.

▶ Show solution

LHS $\equiv \cos^2\theta \times \dfrac{\sin\theta}{\cos\theta}$

$\equiv \sin\theta\cos\theta \equiv$ RHS ■

Question 6

Prove that $(\cos\theta + \sin\theta)^2 + (\cos\theta - \sin\theta)^2 \equiv 2$.

▶ Show solution

LHS $\equiv (\cos^2\theta + 2\sin\theta\cos\theta + \sin^2\theta) + (\cos^2\theta - 2\sin\theta\cos\theta + \sin^2\theta)$

$\equiv 2\cos^2\theta + 2\sin^2\theta$

$\equiv 2(\cos^2\theta + \sin^2\theta) \equiv 2 \equiv$ RHS ■

Question 7

Prove that $1 + \tan^2\theta \equiv \dfrac{1}{\cos^2\theta}$.

▶ Show solution

LHS $\equiv 1 + \dfrac{\sin^2\theta}{\cos^2\theta} \equiv \dfrac{\cos^2\theta + \sin^2\theta}{\cos^2\theta}$

$\equiv \dfrac{1}{\cos^2\theta} \equiv$ RHS ■

(In Year 2 this becomes $1 + \tan^2\theta \equiv \sec^2\theta$.)

Question 8

Prove that $\dfrac{\sin\theta}{1 + \cos\theta} + \dfrac{1 + \cos\theta}{\sin\theta} \equiv \dfrac{2}{\sin\theta}$.

▶ Show solution

LHS $\equiv \dfrac{\sin^2\theta + (1 + \cos\theta)^2}{\sin\theta(1 + \cos\theta)}$

Numerator: $\sin^2\theta + 1 + 2\cos\theta + \cos^2\theta = 2 + 2\cos\theta = 2(1 + \cos\theta)$

LHS $\equiv \dfrac{2(1 + \cos\theta)}{\sin\theta(1 + \cos\theta)} \equiv \dfrac{2}{\sin\theta} \equiv$ RHS ■

Question 9

A curve is given by $x = 2 + 3\cos\theta$, $y = 3\sin\theta - 1$. Use an identity to find an equation linking $x$ and $y$ only, and describe the curve.

▶ Show solution

$\cos\theta = \dfrac{x - 2}{3}$ and $\sin\theta = \dfrac{y + 1}{3}$.

$\sin^2\theta + \cos^2\theta = 1$: $\dfrac{(y+1)^2}{9} + \dfrac{(x-2)^2}{9} = 1$

$(x - 2)^2 + (y + 1)^2 = 9$: a circle with centre $(2, -1)$ and radius $3$ (see Circles).

Question 10

(a) Show that $2\sin^2\theta + 3\cos\theta = 0$ can be written as $2\cos^2\theta - 3\cos\theta - 2 = 0$. (b) Hence solve $2\sin^2\theta + 3\cos\theta = 0$ for $0° \leqslant \theta \leqslant 360°$. (c) Explain why one of the factors gives no solutions. (d) Hence solve $2\sin^2 2x + 3\cos 2x = 0$ for $0° \leqslant x \leqslant 180°$.

▶ Show solution

(a) $2(1 - \cos^2\theta) + 3\cos\theta = 0 \;\Rightarrow\; 2 - 2\cos^2\theta + 3\cos\theta = 0$

Multiply by $-1$: $2\cos^2\theta - 3\cos\theta - 2 = 0$ ■

(b) $(2\cos\theta + 1)(\cos\theta - 2) = 0$, so $\cos\theta = -\tfrac12$ or $\cos\theta = 2$.

$\cos\theta = -\tfrac12$: $\theta = 120°$ or $\theta = 360° - 120° = 240°$.

(c) $-1 \leqslant \cos\theta \leqslant 1$, so $\cos\theta = 2$ is impossible.

(d) The same working with $\theta = 2x$, where $0° \leqslant 2x \leqslant 360°$: $2x = 120°$ or $240°$.

$x = 60°$ or $x = 120°$.

Trigonometric Identities · A Level Maths Year 1 · Created with MathJax