NZ Level 7 (NZC) Level 2 (NCEA)
Trigonometric ratios of supplementary angles (rad)
Lesson

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(πθ) and $\sin\left(\pi+\theta\right)$sin(π+θ)

According to the identity $\sin\left(\pi-\alpha\right)\equiv\sin\alpha$sin(πα)sinα, we have also  $\sin\left(\pi-\theta\right)=0.886$sin(πθ)=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(π2θ)=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(π2θ)=sin(π(π2θ))=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(2ππ4). Therefore, the solutions are $x=\frac{5\pi}{4}$x=5π4 and $x=\frac{7\pi}{4}$x=7π4.

#### More Examples

##### Question 1

Let $\theta$θ be an acute angle (in radians).

If $\tan\theta=0.52$tanθ=0.52, find the value of:

1. $\tan\left(\pi-\theta\right)$tan(πθ)

2. $\tan\left(\pi+\theta\right)$tan(π+θ)

3. $\tan\left(2\pi-\theta\right)$tan(2πθ)

4. $\tan\left(-\theta\right)$tan(θ)

1. A

B

C

A

B

C

### Outcomes

#### M7-4

Apply trigonometric relationships, including the sine and cosine rules, in two and three dimensions

#### 91259

Apply trigonometric relationships in solving problems