Trigonometric Identities | CAIE 9709 Pure Mathematics 1 Notes

Pure Mathematics 1

Mathematics9709

Two identities, two reasons

An is true for every input where its expressions are defined. The symbol $\equiv$ emphasises this. Unlike an equation to solve, an identity does not ask for selected angles.

The notation $\sin^2x$ means $(\sin x)^2$ : find sine, then square its value. It does not mean $\sin(x^2)$ or $\sin(2x)$ .

Pythagoras: (cos θ)² + (sin θ)² = 1².

\sin^2x+\cos^2x=1

This is Pythagoras on a circle of radius 1. Squaring the signed coordinates makes it true in every quadrant. The second identity follows by dividing the two coordinate ratios:

\tan x=\frac{\sin x}{\cos x}\quad(\cos x\ne0)

Both are supplied in MF19. To leave only cosine, replace $\sin^2x$ with $1-\cos^2x$ . To leave only sine, replace $\cos^2x$ instead. For example:

3\sin^2x+5\cos^2x

=3(1-\cos^2x)+5\cos^2x

=3+2\cos^2x

Show how one side becomes the other

Start from one side, usually the more complicated one. Keep each line equal to the previous line; do not begin by assuming the identity is true.

Worked example

Simplify $\dfrac1{1-\sin x}+\dfrac1{1+\sin x}$ .

Both denominators must be non-zero. Give the fractions a common denominator by multiplying each top and bottom by the other denominator:

\frac{(1+\sin x)+(1-\sin x)} {(1-\sin x)(1+\sin x)}

The numerator is 2. The denominator uses difference of squares, $(1-a)(1+a)=1-a^2$ :

=\frac2{1-\sin^2x}=\frac2{\cos^2x}

Only common factors can be cancelled. For instance $\cos^2x/\cos x=\cos x$ when $\cos x\ne0$ ; you cannot cancel a term inside a sum. Original excluded angles stay excluded after simplification.

Hence asks you to use the result just proved. If a denominator is doubled, the whole fraction is halved; you do not need to repeat the proof.

Your turn

Your turn9709/11 May/Jun 2024 Q5 [6 marks]

(a) Prove the identity.

\frac{\sin^2x-\cos x-1}{1+\cos x}\equiv-\cos x

(b) Hence solve, for $0^\circ\le x\le360^\circ$ :

\frac{\sin^2x-\cos x-1}{2+2\cos x}=\frac14

Show worked answer

(a) The original denominator excludes $\cos x=-1$ . Replace sine squared, then factor the numerator:

\frac{1-\cos^2x-\cos x-1}{1+\cos x}

=\frac{-\cos x(1+\cos x)}{1+\cos x}

=-\cos x

(b) The denominator is $2(1+\cos x)$ , so the left side is half the proven expression:

-\frac{\cos x}{2}=\frac14

\cos x=-\frac12

Answer

x=120^\circ,\ 240^\circ

Neither angle makes the original denominator zero.