Symmetries
A previous lesson explored the many symmetries in the functions for sine and cosine. Let's start with a recap of a few important relationships that arise from these symmetries.
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:
| $\sin\left(\pi-\theta\right)$sin(π−θ) | $=$= | $\sin\theta$sinθ |
| $\sin\left(\pi+\theta\right)$sin(π+θ) | $=$= | $-\sin\theta$−sinθ |
| $\sin\left(2\pi-\theta\right)$sin(2π−θ) | $=$= | $-\sin\theta$−sinθ |
| $\sin\left(-\theta\right)$sin(−θ) | $=$= | $-\sin\theta$−sinθ |
If we focus on the $x$x-coordinate, we can find:
| $\cos\left(\pi-\theta\right)$cos(π−θ) | $=$= | $-\cos\theta$−cosθ |
| $\cos\left(\pi+\theta\right)$cos(π+θ) | $=$= | $-\cos\theta$−cosθ |
| $\cos\left(2\pi-\theta\right)$cos(2π−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. Try these in the worked example below.
Practice question
Question 1
Let $\theta$θ be an acute angle (in radians).
If $\tan\theta=0.52$tanθ=0.52, find the value of:
$\tan\left(\pi-\theta\right)$tan(π−θ)
$\tan\left(\pi+\theta\right)$tan(π+θ)
$\tan\left(2\pi-\theta\right)$tan(2π−θ)
$\tan\left(-\theta\right)$tan(−θ)
Further symmetries
Let's look at the relationship between trigonometric functions of complementary angles, that is, angles that add to $90^\circ$90° or $\frac{\pi}{2}$π2 radians.
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 $\frac{\pi}{2}-\theta$π2−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:
| $\sin\left(\frac{\pi}{2}-\theta\right)$sin(π2−θ) | $=$= | $\cos\theta$cosθ |
| $\cos\left(\frac{\pi}{2}-\theta\right)$cos(π2−θ) | $=$= | $\sin\theta$sinθ |
Also, look at the relationship between trigonometric functions at an angle $\theta$θ compared to one rotated further by $90^\circ$90° or $\frac{\pi}{2}$π2 radians:
Again, by comparing the $x$x- and $y$y-coordinates of point $P$P compared to point $Q$Q, this leads to the following rules:
| $\sin\left(\frac{\pi}{2}+\theta\right)$sin(π2+θ) | $=$= | $\cos\theta$cosθ |
| $\cos\left(\frac{\pi}{2}+\theta\right)$cos(π2+θ) | $=$= | $-\sin\theta$−sinθ |
Practice questions
Question 2
Given that $\sin x=0.46$sinx=0.46, find the exact value of $\cos\left(90^\circ-x\right)$cos(90°−x).
Question 3
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