Trigonometric ratios of supplementary angles (rad)

In this chapter we present ideas in the context of radian measure that were previously discussed for angles measured in degrees.  In other chapters, it was shown how the trigonometric functions of angles of any magnitude are found using the unit circle definitions of the functions and by relating angles to a relative acute angle. The following diagram revisits these ideas and helps to explain the identities that follow.

SOME USEFUL IDENTITIES

$\sin\left(\pi-\alpha\right)\equiv\sin\alpha$sin(π−α)≡sinα	$\cos\left(\pi-\alpha\right)\equiv-\cos\alpha$cos(π−α)≡−cosα	$\tan\left(\pi-\alpha\right)\equiv-\tan\alpha$tan(π−α)≡−tanα
$\sin\left(\pi+\alpha\right)\equiv-\sin\alpha$sin(π+α)≡−sinα	$\cos\left(\pi+\alpha\right)\equiv-\cos\alpha$cos(π+α)≡−cosα	$\tan\left(\pi+\alpha\right)\equiv\tan\alpha$tan(π+α)≡tanα
$\sin\left(2\pi-\alpha\right)\equiv-\sin\alpha$sin(2π−α)≡−sinα	$\cos\left(2\pi-\alpha\right)\equiv\cos\alpha$cos(2π−α)≡cosα	$\tan\left(2\pi-\alpha\right)\equiv-\tan\alpha$tan(2π−α)≡−tanα

 

 

 

It is unnecessary to memorise these identities since they can easily be recovered from a mental or physical diagram like the one above.

Examples

Example 1

If $\sin\theta=0.886$sinθ=0.886, what are $\sin\left(\pi-\theta\right)$sin(π−θ) and $\sin\left(\pi+\theta\right)$sin(π+θ)? 

According to the identity $\sin\left(\pi-\alpha\right)\equiv\sin\alpha$sin(π−α)≡sinα, we have also  $\sin\left(\pi-\theta\right)=0.886$sin(π−θ)=0.886. 

Finally, the identity $\sin\left(\pi+\alpha\right)\equiv-\sin\alpha gives\sin\left(\pi+\theta\right)=-0.886$sin(π+α)≡−sinαgivessin(π+θ)=−0.886

The inverse sine of $0.886$0.886 is approximately $1.08865$1.08865 which is about $\frac{\pi}{2.88578}$π2.88578​. 

Example 2

If $\sin\left(\frac{\pi}{2}-\theta\right)=x$sin(π2​−θ)=x, what is $\sin\left(\frac{\pi}{2}+\theta\right)$sin(π2​+θ) in terms of $x$x?

From the identities given above, we know that $\sin\left(\frac{\pi}{2}-\theta\right)=\sin\left(\pi-\left(\frac{\pi}{2}-\theta\right)\right)=\sin\left(\frac{\pi}{2}+\theta\right)$sin(π2​−θ)=sin(π−(π2​−θ))=sin(π2​+θ).

So, $\sin\left(\frac{\pi}{2}+\theta\right)=x$sin(π2​+θ)=x.

Example 3

If $\sin x=-\cos\frac{\pi}{4}$sinx=−cosπ4​, find all possible values of $x$x between $0$0 and $2\pi$2π.

We have, $\sin x=-\cos\frac{\pi}{4}=-\sin\frac{\pi}{4}.$sinx=−cosπ4​=−sinπ4​. Since $-\sin\frac{\pi}{4}$−sinπ4​ is negative, $x$x is in the third or fourth quadrant and, according to the identities, we must have either  $-\sin\frac{\pi}{4}=\sin\left(\pi+\frac{\pi}{4}\right)$−sinπ4​=sin(π+π4​) or $-\sin\frac{\pi}{4}=\sin\left(2\pi-\frac{\pi}{4}\right)$−sinπ4​=sin(2π−π4​). Therefore, the solutions are $x=\frac{5\pi}{4}$x=5π4​ and $x=\frac{7\pi}{4}$x=7π4​.

 

More Examples

Question 1

Question 2

Question 3