6.02 Trigonometric identities

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.

 

 

 

 

 

 

 

 

 

This means that the opposite side of the triangle from $\theta$θ has a length of $\sin\theta$sinθ and the adjacent side of the triangle has a length of $\cos\theta$cosθ. 

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.

 

 

Also since $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}=\frac{\sin\theta}{\cos\theta}$tanθ=oppositeadjacent​=sinθcosθ​ 

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 1

Question 2