6.05 Double 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.

 

Double-angle identities

 

 

Worked Example 1

Simplify the expression: $2\sin45\cos45$2sin45cos45.

This expression looks like the double angle identity for $\sin2\theta$sin2θ.  So we can use this identity to simplify it:
 

$2\sin45\cos45$2sin45cos45	$=$=	$\sin(2\times45)$sin(2×45)
	$=$=	$\sin90$sin90
	$=$=	$1$1

 

Worked Example 2

Simplify the expression: $6\cos^22x-3$6cos22x−3.

This looks like the cosine double identity, we just need to take out a factor of $3$3:

$6\cos^22x-3$6cos22x−3	$=$=	$3(2\cos^22x-1)$3(2cos22x−1)
	$=$=	$3\cos2x$3cos2x

 

Worked Example 3

Expand and simplify: $\left(\sin\theta+\cos\theta\right)^2$(sinθ+cosθ)2.

First we must expand the expression: 

$\left(\sin\theta+\cos\theta\right)^2$(sinθ+cosθ)2

$=$=

$\sin^2\theta+2\sin\theta\cos\theta+\cos^2\theta$sin2θ+2sinθcosθ+cos2θ

 

The first and third term of the expanded expression make up the identity: $\sin^2\theta+\cos^2\theta=1$sin2θ+cos2θ=1. So we can replace these terms with the number $1$1.

 

$\left(\sin\theta+\cos\theta\right)^2$(sinθ+cosθ)2	$=$=	$\sin^2\theta+2\sin\theta\cos\theta+\cos^2\theta$sin2θ+2sinθcosθ+cos2θ
	$=$=	$1+2\sin\theta\cos\theta$1+2sinθcosθ

 

Now we can use the sine double identity to simplify the second term:

$\left(\sin\theta+\cos\theta\right)^2$(sinθ+cosθ)2	$=$=	$\sin^2\theta+2\sin\theta\cos\theta+\cos^2\theta$sin2θ+2sinθcosθ+cos2θ
	$=$=	$1+2\sin\theta\cos\theta$1+2sinθcosθ
	$=$=	$1+\sin2\theta$1+sin2θ

 

 

 

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

 

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​

 

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​)

 

 

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.

QUESTION 1

QUESTION 2

QUESTION 3