Evaluate trig expressions using double and half angle identities (deg/rad)

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

$\sin2\theta\equiv2\sin\theta\cos\theta$sin2θ≡2sinθcosθ

$\cos2\theta\equiv\cos^2\theta-\sin^2\theta$cos2θ≡cos2θ−sin2θ

$\tan2\theta\equiv\frac{2\tan\theta}{1-\tan^2\theta}$tan2θ≡2tanθ1−tan2θ​

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

$\sin\alpha\equiv2\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}$sinα≡2sinα2​cosα2​

$\cos\alpha\equiv\cos^2\frac{\alpha}{2}-\sin^2\frac{\alpha}{2}$cosα≡cos2α2​−sin2α2​

$\tan\alpha\equiv\frac{2\tan\frac{\alpha}{2}}{1-\tan^2\frac{\alpha}{2}}$tanα≡2tanα2​1−tan2α2​​

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

$\tan x\equiv\frac{2t}{1-t^2}$tanx≡2t1−t2​

Now, because $\tan x\equiv\frac{\sin x}{\cos x}$tanx≡sinxcosx​, 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$1−t2. Say, $\sin x=k.2t$sinx=k.2t and $\cos x=k.\left(1-t^2\right)$cosx=k.(1−t2). Then, since we require $\sin^2x+\cos^2x\equiv1$sin2x+cos2x≡1, we have  $\left(k.2t\right)^2+k^2\left(1-t^2\right)^2=1$(k.2t)2+k2(1−t2)2=1 and in a few steps we deduce that $k=\frac{1}{1+t^2}$k=11+t2​. Hence,

$\sin x\equiv\frac{2t}{1+t^2}$sinx≡2t1+t2​ and

$\cos x\equiv\frac{1-t^2}{1+t^2}$cosx≡1−t21+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}$2t1−t2​, then the hypotenuse must be $1+t^2$1+t2 because $\sqrt{\left(2t\right)^2+\left(1-t^2\right)^2}$√(2t)2+(1−t2)2 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​)−(1−cos2(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 $\sin67.5^\circ$sin67.5°.

We observe that $67.5^\circ$67.5° is half of $135^\circ$135°, a second quadrant angle that is related to the first quadrant angle $45^\circ$45° for which we have exact values of the trigonometric functions.

We can use the identity $\cos\alpha\equiv\cos^2\frac{\alpha}{2}-\sin^2\frac{\alpha}{2}=1-2\sin^2\frac{\alpha}{2}$cosα≡cos2α2​−sin2α2​=1−2sin2α2​. This can be rearranged to give 

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

So, $\sin67.5^\circ=\sqrt{\frac{1}{2}\left(1-\cos135^\circ\right)}=\sqrt{\frac{1}{2}\left(1+\frac{1}{\sqrt{2}}\right)}=\frac{1}{2}\sqrt{2+\sqrt{2}}.$sin67.5°=√12​(1−cos135°)=√12​(1+1√2​)=12​√2+√2.

More Worked Examples

QUESTION 1

QUESTION 2

QUESTION 3