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India
Class XI

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 $\sin x$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,

$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 1

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

11.SF.TF.1

Positive and negative angles. Measuring angles in radians and in degrees and conversion from one measure to another. Definition of trigonometric functions with the help of unit circle. Truth of the identity sin^2 x + cos^2 x = 1, for all x. Signs of trigonometric functions and sketch of their graphs. Expressing sin (x + y) and cos (x + y) in terms of sin x, sin y, cos x and cos y. Deducing the identities like following: cot(x + or - y), sin x + sin y, cos x + cos y, sin x - sin y, cos x - cos y (see syllabus document)

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