Symmetries
From previous lessons we discovered 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.
By comparing the coordinates of the points around the circle we can create 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 we can create similar rules for $\tan\theta$tanθ. Try these for yourself in the practice question 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, we can see 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, we can extract the 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 |
We can 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, we can extract the 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
Fill in the blank with the acute angle that makes the statement true.
$\sin$sin$\editable{}$°$=$=$\cos25^\circ$cos25°
Question 3
Simplify $\sin\left(90^\circ-y\right)\times\tan y$sin(90°−y)×tany.
Question 4
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
Pythagorean identity
A very useful expression can be found by applying Pythagoras' theorem to the right-angled triangle formed in a unit circle. Consider the picture below.
Applying Pythagoras' theorem:
| $a^2+b^2$a2+b2 | $=$= | $c^2$c2 |
| $\left(\cos\theta\right)^2+\left(\sin\theta\right)^2$(cosθ)2+(sinθ)2 | $=$= | $1$1 |
Which can also be written as:
| $\cos^2\theta+\sin^2\theta$cos2θ+sin2θ | $=$= | $1$1 |
This relationship holds for any angle $\theta$θ.
| $\cos^2\theta+\sin^2\theta$cos2θ+sin2θ | $=$= | $1$1 |
Worked examples
Example 1
Evaluate the expression $5\cos^2\left(80^\circ\right)+5\sin^2\left(80^\circ\right)$5cos2(80°)+5sin2(80°).
Although neither $\cos\left(80^\circ\right)$cos(80°) or $\sin\left(80^\circ\right)$sin(80°) can be written in exact form using decimals or surds, we can simplify the expression by recognising the Pythagorean identity.
| $5\cos^2\left(80^\circ\right)+5\sin^2\left(80^\circ\right)$5cos2(80°)+5sin2(80°) | $=$= | $5\left(\cos^2\left(80^\circ\right)+\sin^2\left(80^\circ\right)\right)$5(cos2(80°)+sin2(80°)) |
| $=$= | $5\times1$5×1 | |
| $=$= | $5$5 |
Example 2
Find the exact value of $\cos\theta$cosθ and $\tan\theta$tanθ given that $\sin\theta=\frac{12}{13}$sinθ=1213 and $\theta$θ is in the second quadrant.
Think: We can find $\cos\theta$cosθ using the Pythagorean identity and the additional information about the quadrant will tell us if the ratio is positive or negative. One we know both $\sin\theta$sinθ and $\cos\theta$cosθ we can find $\tan\theta$tanθ using the definition $\tan\theta=\frac{\sin\theta}{\cos\theta}$tanθ=sinθcosθ
Do: Write out the Pythagorean identity and substitute the known value in.
| $\cos^2\theta+\sin^2\theta$cos2θ+sin2θ | $=$= | $1$1 |
| $\cos^2\theta+\left(\frac{12}{13}\right)^2$cos2θ+(1213)2 | $=$= | $1$1 |
| $\cos^2\theta+\frac{144}{169}$cos2θ+144169 | $=$= | $1$1 |
| $\cos^2\theta$cos2θ | $=$= | $1-\frac{144}{169}$1−144169 |
| $\cos\theta$cosθ | $=$= | $\pm\sqrt{\frac{25}{169}}$±√25169 |
| $=$= | $\pm\frac{5}{13}$±513 |
We have been given the additional information that $\theta$θ is in the second quadrant. In the second quadrant sine is positive but cosine and tangent will be negative. Hence, $\cos\theta=\frac{-5}{13}$cosθ=−513.
Using $\tan\theta=\frac{\sin\theta}{\cos\theta}$tanθ=sinθcosθ, we find:
| $\tan\theta$tanθ | $=$= | $\frac{\sin\theta}{\cos\theta}$sinθcosθ |
| $=$= | $\frac{12}{13}\div\frac{-5}{13}$1213÷−513 | |
| $=$= | $\frac{-12}{5}$−125 |
Practice questions
Question 5
Answer the following questions given that $\cos y=-\frac{5}{13}$cosy=−513, where $180^\circ
In which quadrant does angle $y$y lie?
Quadrant $I$I
AQuadrant $II$II
BQuadrant $III$III
CQuadrant $IV$IV
DUse a Pythagorean identity to find the value of $\tan y$tany.
Question 6
Simplify $\frac{1-\cos^2\left(\theta\right)}{1-\sin^2\left(\theta\right)}$1−cos2(θ)1−sin2(θ).
Question 7
Simplify $\left(\cos\theta-1\right)\left(\cos\theta+1\right)$(cosθ−1)(cosθ+1).