Evaluate trig expressions using double and half angle identities

Just as certain trigonometric expressions involving a variable $$θ or $$x can be simplified with the help of identities, expressions with known values of  $$θ or $$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

$$sin2θ≡2sinθcosθ

$$cos2θ≡cos2θ−sin2θ

$$tan2θ≡2tanθ1−tan2θ​

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

$$sinα≡2sinα2​cosα2​

$$cosα≡cos2α2​−sin2α2​

$$

The last of these three can be used to express $$sinx and $$cosx in terms of $$tanx2​. First, we put $$tanx2​=t. Then, we have

$$tanx≡2t1−t2​

Now, because $$tanx≡sinxcosx​, it must be that $$sinx is a multiple of $$2t and $$cosx is the same multiple of $$1−t2. Say, $$sinx=k.2t and $$cosx=k.(1−t2). Then, since we require $$sin2x+cos2x≡1, we have  $$(k.2t)2+k2(1−t2)2=1 and in a few steps we deduce that $$k=11+t2​. Hence,

$$sinx≡2t1+t2​ and

$$cosx≡1−t21+t2​

These relationships can be verified geometrically, using Pythagoras's theorem in the right-angled triangle below. If $$tanx is $$2t1−t2​, then the hypotenuse must be $$1+t2 because $$ simplifies to $$1+t2. The expressions for sine and cosine follow.

Example 1

Simplify and find an approximate value for $$sin1.8°1+cos1.8°​.

The numerator is $$2sin0.9°cos0.9°

and the denominator is $$1+cos20.9°−sin20.9° or equivalently,

$$1+cos20.9°−(1−cos20.9°)=2cos20.9°.

So, 

$$sin1.8°1+cos1.8°​	$$=	$$2sin0.9°cos0.9°2cos20.9°​
	$$=	$$tan0.9°

This simplification could have been done whatever the angle had been. But for a small angle like $$0.9° measured in radians, we can use the fact that $$tanx is close to $$x itself. So, $$sin1.8°1+cos1.8°​ is approximately $$0.9°.

This should be verified by calculator.

 

Example 2

Evaluate $$sin67.5°.

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

We can use the identity $$cosα≡cos2α2​−sin2α2​=1−2sin2α2​. This can be rearranged to give 

$$sinα2​=√12​(1−cosα)

So, $$sin67.5°=√12​(1−cos135°)=√12​(1+1√2​)=12​√2+√2.

More Worked Examples

QUESTION 1

QUESTION 2

QUESTION 3