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
Let $\theta$θ be an acute angle (in degrees).
If $\sin\theta=0.9$sinθ=0.9, find the value of:
$\sin\left(180^\circ-\theta\right)$sin(180°−θ)
$\sin\left(180^\circ+\theta\right)$sin(180°+θ)
$\sin\left(360^\circ-\theta\right)$sin(360°−θ)
$\sin\left(-\theta\right)$sin(−θ)
Question 2
Let $\theta$θ be an acute angle (in degrees).
If $\cos\theta=0.1$cosθ=0.1, find the value of:
$\cos\left(180^\circ-\theta\right)$cos(180°−θ)
$\cos\left(180^\circ+\theta\right)$cos(180°+θ)
$\cos\left(360^\circ-\theta\right)$cos(360°−θ)
$\cos\left(-\theta\right)$cos(−θ)
Question 3
If $\tan x=\tan30^\circ$tanx=tan30° and $180^\circ