There are many symmetries in the functions for sine and cosine from the unit circle.
Comparing the coordinates of the points around the circle leads to the following rules:
If we focus on the $y$y-coordinate, we can find:
$\sin\left(180^\circ-\theta\right)$sin(180°−θ) | $=$= | $\sin\theta$sinθ |
$\sin\left(180^\circ+\theta\right)$sin(180°+θ) | $=$= | $-\sin\theta$−sinθ |
$\sin\left(-\theta\right)$sin(−θ) | $=$= | $-\sin\theta$−sinθ |
If we focus on the $x$x-coordinate, we can find:
$\cos\left(180^\circ-\theta\right)$cos(180°−θ) | $=$= | $-\cos\theta$−cosθ |
$\cos\left(180^\circ+\theta\right)$cos(180°+θ) | $=$= | $-\cos\theta$−cosθ |
$\cos\left(-\theta\right)$cos(−θ) | $=$= | $\cos\theta$cosθ |
By using the definition $\tan\theta=\frac{\sin\theta}{\cos\theta}$tanθ=sinθcosθ similar rules can be created for $\tan\theta$tanθ.
Let's look at the relationship between trigonometric functions of complementary angles, that is, angles that add to $90^\circ$90°.
From the diagram, notice that the triangle formed by the angle $\theta$θ to the point $P$P has its sides flipped to form the triangle at angle $90^\circ-\theta$90°−θ to point $Q$Q. Since the $x$x- and $y$y-coordinates have swapped when comparing the two points, the following rules can be extracted:
$\sin\left(90^\circ-\theta\right)$sin(90°−θ) | $=$= | $\cos\theta$cosθ |
$\cos\left(90^\circ-\theta\right)$cos(90°−θ) | $=$= | $\sin\theta$sinθ |
Given that $\sin x=0.46$sinx=0.46, find the exact value of $\cos\left(90^\circ-x\right)$cos(90°−x).
Prove that $\frac{\sin x\sin\left(90^\circ-x\right)}{\cos x\cos\left(90^\circ-x\right)}=1$sinxsin(90°−x)cosxcos(90°−x)=1