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≡2sinα2cosα2
$\cos\alpha\equiv\cos^2\frac{\alpha}{2}-\sin^2\frac{\alpha}{2}$cosα≡cos2α2−sin2α2
$$
The last of these three can be used to express $\sin x$sinx and $\cos x$cosx in terms of $\tan\frac{x}{2}$tanx2. First, we put $\tan\frac{x}{2}=t$tanx2=t. Then, we have
$\tan x\equiv\frac{2t}{1-t^2}$tanx≡2t1−t2
Now, because $\tan x\equiv\frac{\sin x}{\cos x}$tanx≡sinxcosx, it must be that $\sin x$sinx is a multiple of $2t$2t and $\cos x$cosx is the same multiple of $1-t^2$1−t2. Say, $\sin x=k.2t$sinx=k.2t and $\cos x=k.\left(1-t^2\right)$cosx=k.(1−t2). Then, since we require $\sin^2x+\cos^2x\equiv1$sin2x+cos2x≡1, we have $\left(k.2t\right)^2+k^2\left(1-t^2\right)^2=1$(k.2t)2+k2(1−t2)2=1 and in a few steps we deduce that $k=\frac{1}{1+t^2}$k=11+t2. Hence,
$\sin x\equiv\frac{2t}{1+t^2}$sinx≡2t1+t2 and
$\cos x\equiv\frac{1-t^2}{1+t^2}$cosx≡1−t21+t2
These relationships can be verified geometrically, using Pythagoras's theorem in the right-angled triangle below. If $\tan x$tanx is $\frac{2t}{1-t^2}$2t1−t2, then the hypotenuse must be $1+t^2$1+t2 because $$ simplifies to $1+t^2.$1+t2. The expressions for sine and cosine follow.
Simplify and find an approximate value for $\frac{\sin1.8^\circ}{1+\cos1.8^\circ}$sin1.8°1+cos1.8°.
The numerator is $2\sin0.9^\circ\cos0.9^\circ$2sin0.9°cos0.9°
and the denominator is $1+\cos^20.9^\circ-\sin^20.9^\circ$1+cos20.9°−sin20.9° or equivalently,
$1+\cos^20.9^\circ-\left(1-\cos^20.9^\circ\right)=2\cos^20.9^\circ$1+cos20.9°−(1−cos20.9°)=2cos20.9°.
So,
$\frac{\sin1.8^\circ}{1+\cos1.8^\circ}$sin1.8°1+cos1.8° | $=$= | $\frac{2\sin0.9^\circ\cos0.9^\circ}{2\cos^20.9^\circ}$2sin0.9°cos0.9°2cos20.9° |
$=$= | $\tan0.9^\circ$tan0.9° |
This simplification could have been done whatever the angle had been. But for a small angle like $0.9^\circ$0.9° measured in radians, we can use the fact that $\tan x$tanx is close to $x$x itself. So, $\frac{\sin1.8^\circ}{1+\cos1.8^\circ}$sin1.8°1+cos1.8° is approximately $0.9^\circ.$0.9°.
This should be verified by calculator.
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.
Find the exact value of $\cos157.5^\circ$cos157.5°.
Express your answer with a rational denominator.
Given $\cos\theta=\frac{4}{5}$cosθ=45 and $\sin\theta$sinθ$<$<$0$0, find:
$\sin\theta$sinθ
$\sin2\theta$sin2θ
$\cos2\theta$cos2θ
$\tan2\theta$tan2θ
Use the double angle identity for the sine ratio to simplify the expression $\frac{1}{6}\sin157.5^\circ\cos157.5^\circ$16sin157.5°cos157.5°.