Trigonometric ratios of supplementary angles (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

$\sin\left(180^{\circ}-\alpha\right)\equiv\sin\alpha$sin(180∘−α)≡sinα	$\cos\left(180^{\circ}-\alpha\right)\equiv-\cos\alpha$cos(180∘−α)≡−cosα	$\tan\left(180^{\circ}-\alpha\right)\equiv-\tan\alpha$tan(180∘−α)≡−tanα
$\sin\left(180^{\circ}+\alpha\right)\equiv-\sin\alpha$sin(180∘+α)≡−sinα	$\cos\left(180^{\circ}+\alpha\right)\equiv-\cos\alpha$cos(180∘+α)≡−cosα	$\tan\left(180^{\circ}+\alpha\right)\equiv\tan\alpha$tan(180∘+α)≡tanα
$\sin\left(360^{\circ}-\alpha\right)\equiv-\sin\alpha$sin(360∘−α)≡−sinα	$\cos\left(360^{\circ}-\alpha\right)\equiv\cos\alpha$cos(360∘−α)≡cosα	$\tan\left(360^{\circ}-\alpha\right)\equiv-\tan\alpha$tan(360∘−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 $\sin\left(180^{\circ}-\theta\right)$sin(180∘−θ) and $\sin\left(180^{\circ}+\theta\right)$sin(180∘+θ)? Evaluate all three expressions.

The inverse sine of $0.886$0.886 given by calculator, is approximately $62.37^{\circ}$62.37∘. So, the acute angle solution is $\theta\approx62.37$θ≈62.37.

According to the identity $\sin\left(180^{\circ}-\alpha\right)\equiv\sin\alpha$sin(180∘−α)≡sinα, we have also  $\sin\left(180^{\circ}-\theta\right)=0.886$sin(180∘−θ)=0.886. That is, $\sin117.63\approx0.886$sin117.63≈0.886.

Finally, the identity $\sin\left(180^{\circ}+\alpha\right)\equiv-\sin\alpha$sin(180∘+α)≡−sinα  gives $\sin\left(180^{\circ}+\theta\right)=-0.886\approx\sin242.37^{\circ}.$sin(180∘+θ)=−0.886≈sin242.37∘.

Example 2

If $\sin\left(90^{\circ}-\theta\right)=x$sin(90∘−θ)=x, what is $\sin\left(90^{\circ}+\theta\right)$sin(90∘+θ)in terms of $x$x?

From the identities given above, we know that $\sin\left(90^{\circ}-\theta\right)=\sin\left(180^{\circ}-\left(90^{\circ}-\theta\right)\right)=\sin\left(90^{\circ}+\theta\right).$sin(90∘−θ)=sin(180∘−(90∘−θ))=sin(90∘+θ).

So, $\sin\left(90^{\circ}+\theta\right)=x$sin(90∘+θ)=x.

Example 3

If $\sin x=-\cos45^{\circ}$sinx=−cos45∘ (with $x$x in degrees), find all possible values of $x$x between $0^{\circ}$0∘ and $360^{\circ}$360∘.

We have, $\sin x=-\cos45^{\circ}=-\sin45^{\circ}$sinx=−cos45∘=−sin45∘. Since $-\sin45^{\circ}$−sin45∘ is negative, $x$x is in the third or fourth quadrant and, according to the identities, we must have either  $-\sin45^{\circ}=\sin\left(180^{\circ}+45^{\circ}\right)$−sin45∘=sin(180∘+45∘) or $-\sin45^{\circ}=\sin\left(360^{\circ}-45^{\circ}\right)$−sin45∘=sin(360∘−45∘). Therefore, the solutions are $x=225^{\circ}$x=225∘ and $x=315^{\circ}$x=315∘.

 

More Examples

Question 1

Question 2

Question 3