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

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




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




The last of these three can be used to express $\sin x$sinx and $\cos x$cosx in terms of $\tan\frac{x}{2}$tanx2. First, we put $\tan\frac{x}{2}=t$tanx2=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 $\sin x$sinx is a multiple of $2t$2t and $\cos x$cosx is the same multiple of $1-t^2$1t2. Say, $\sin x=k.2t$sinx=k.2t and $\cos x=k.\left(1-t^2\right)$cosx=k.(1t2). Then, since we require $\sin^2x+\cos^2x\equiv1$sin2x+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 $\tan x$tanx is $\frac{2t}{1-t^2}$2t1t2, then the hypotenuse must be $1+t^2$1+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,



$\frac{\sin\left(\frac{\pi}{100}\right)}{1+\cos\left(\frac{\pi}{100}\right)}$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 


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



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


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).



Manipulate trigonometric expressions


Apply trigonometric methods in solving problems

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