NZ Level 8 (NZC) Level 3 (NCEA) [In development]
Evaluate trig expressions using double and half angle identities
Lesson

Just as certain trigonometric expressions involving a variable $\theta$θ or $x$x can be simplified with the help of identities, expressions with known values of  $\theta$θ or $x$x can sometimes be evaluated using the same identities.

Here, we are making a distinction between 'simplification', meaning writing an expression in an equivalent but less complex form, and 'evaluation', meaning giving a numerical value to an expression.

Half-angle identities

We  derive some further identities from the double-angle identities, as follows:

We have the double angle identities

$\sin2\theta\equiv2\sin\theta\cos\theta$sin2θ2sinθcosθ

$\cos2\theta\equiv\cos^2\theta-\sin^2\theta$cos2θcos2θsin2θ

$\tan2\theta\equiv\frac{2\tan\theta}{1-\tan^2\theta}$tan2θ2tanθ1tan2θ

By putting $\alpha=2\theta$α=2θ we obtain the corresponding half-angle formulae:

$\sin\alpha\equiv2\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}$sinα2sinα2cosα2

$\cos\alpha\equiv\cos^2\frac{\alpha}{2}-\sin^2\frac{\alpha}{2}$cosαcos2α2sin2α2

$$The last of these three can be used to express \sin xsinx and \cos xcosx in terms of \tan\frac{x}{2}tanx2. First, we put \tan\frac{x}{2}=ttanx2=t. Then, we have \tan x\equiv\frac{2t}{1-t^2}tanx2t1t2 Now, because \tan x\equiv\frac{\sin x}{\cos x}tanxsinxcosx, it must be that \sin xsinx is a multiple of 2t2t and \cos xcosx is the same multiple of 1-t^21t2. Say, \sin x=k.2tsinx=k.2t and \cos x=k.\left(1-t^2\right)cosx=k.(1t2). Then, since we require \sin^2x+\cos^2x\equiv1sin2x+cos2x1, we have \left(k.2t\right)^2+k^2\left(1-t^2\right)^2=1(k.2t)2+k2(1t2)2=1 and in a few steps we deduce that k=\frac{1}{1+t^2}k=11+t2. Hence, \sin x\equiv\frac{2t}{1+t^2}sinx2t1+t2 and \cos x\equiv\frac{1-t^2}{1+t^2}cosx1t21+t2 These relationships can be verified geometrically, using Pythagoras's theorem in the right-angled triangle below. If \tan xtanx is \frac{2t}{1-t^2}2t1t2, then the hypotenuse must be 1+t^21+t2 because$$ simplifies to $1+t^2.$1+t2. The expressions for sine and cosine follow.

Example 1

Simplify and find an approximate value for $\frac{\sin\left(\frac{\pi}{100}\right)}{1+\cos\left(\frac{\pi}{100}\right)}$sin(π100)1+cos(π100).

The numerator is $2\sin\left(\frac{\pi}{200}\right)\cos\left(\frac{\pi}{200}\right)$2sin(π200)cos(π200)

and the denominator is $1+\cos^2\left(\frac{\pi}{200}\right)-\sin^2\left(\frac{\pi}{200}\right)$1+cos2(π200)sin2(π200) or equivalently,

$1+\cos^2\left(\frac{\pi}{200}\right)-\left(1-\cos^2\left(\frac{\pi}{200}\right)\right)=2\cos^2\left(\frac{\pi}{200}\right)$1+cos2(π200)(1cos2(π200))=2cos2(π200).

So,

 $\frac{\sin\left(\frac{\pi}{100}\right)}{1+\cos\left(\frac{\pi}{100}\right)}$sin(π100​)1+cos(π100​)​ $=$= $\frac{2\sin\left(\frac{\pi}{200}\right)\cos\left(\frac{\pi}{200}\right)}{2\cos^2\left(\frac{\pi}{200}\right)}$2sin(π200​)cos(π200​)2cos2(π200​)​ $=$= $\tan\left(\frac{\pi}{200}\right)$tan(π200​)

This simplification could have been done whatever the angle had been. But for a small angle like $\frac{\pi}{200}$π200 measured in radians, we can use the fact that $\tan x$tanx is close to $x$x itself. So, $\frac{\sin\left(\frac{\pi}{100}\right)}{1+\cos\left(\frac{\pi}{100}\right)}$sin(π100)1+cos(π100) is approximately $\frac{\pi}{200}.$π200.

This should be verified by calculator.

Example 2

Evaluate $\sin\frac{3\pi}{8}$sin3π8.

We observe that $\frac{3\pi}{8}$3π8 is half of $\frac{3\pi}{4}$3π4, a second quadrant angle that is related to the first quadrant angle $\frac{\pi}{4}$π4 for which we have exact values of the trigonometric functions.

We can use the identity $\cos\alpha\equiv\cos^2\frac{\alpha}{2}-\sin^2\frac{\alpha}{2}=1-2\sin^2\frac{\alpha}{2}$cosαcos2α2sin2α2=12sin2α2. This can be rearranged to give

$\sin\frac{\alpha}{2}=\sqrt{\frac{1}{2}\left(1-\cos\alpha\right)}$sinα2=12(1cosα)

So, $\sin\frac{3\pi}{8}=\sqrt{\frac{1}{2}\left(1-\cos\frac{3\pi}{4}\right)}=\sqrt{\frac{1}{2}\left(1+\frac{1}{\sqrt{2}}\right)}=\frac{1}{2}\sqrt{2+\sqrt{2}}.$sin3π8=12(1cos3π4)=12(1+12)=122+2.

More Worked Examples

QUESTION 2

Given $\cos\theta=\frac{4}{5}$cosθ=45 and $\sin\theta$sinθ$<$<$0$0, find:

1. $\sin\theta$sinθ

2. $\sin2\theta$sin2θ

3. $\cos2\theta$cos2θ

4. $\tan2\theta$tan2θ

QUESTION 3

Consider the following.

1. Simplify $\frac{1+\cos x}{2}$1+cosx2.

2. We want to find the exact value of $\cos\left(\frac{x}{2}\right)$cos(x2) if $\sin x=\frac{5}{13}$sinx=513 and $\frac{\pi}{2}π2<x<π. First find$\cos x$cosx. 3. Hence or otherwise, find$\cos\left(\frac{x}{2}\right)\$cos(x2).

Outcomes

M8-6

Manipulate trigonometric expressions

91575

Apply trigonometric methods in solving problems