8.10 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.

 

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.

 

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

Angle sum identities

The Angle Sum and Difference identities allow for the expansion of expressions like $\sin\left(A+B\right)$sin(A+B). One way to establish these identities involves using the cosine rule and the distance formula together with an arbitrary triangle with vertices at the centre and circumference of the unit circle.

Two different expressions can be written for the length of the green line $PQ$PQ. First, using the cosine rule.

The arms of the triangle have unit length because they are radii of the unit circle. Therefore:

$c^2$c2	$=$=	$a^2+b^2-2ab\cos C$a2+b2−2abcosC
$PQ^2$PQ2	$=$=	$1^2+1^2-2\times1\times1\times\cos\left(A-B\right)$12+12−2×1×1×cos(A−B)
	$=$=	$2-2\cos\left(A-B\right)$2−2cos(A−B) .......... [1]

Next, write the distance $PQ$PQ using the coordinates of the points and the distance formula. This gives: 

$d^2$d2	$=$=	$\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2$(x2​−x1​)2+(y2​−y1​)2
$PQ^2$PQ2	$=$=	$\left(\cos A-\cos B\right)^2+\left(\sin A-\sin B\right)^2$(cosA−cosB)2+(sinA−sinB)2
	$=$=	$\cos^2A-2\cos A\cos B+\cos^2B+\sin^2A-2\sin A\sin B+\sin^2B$cos2A−2cosAcosB+cos2B+sin2A−2sinAsinB+sin2B
	$=$=	$\cos^2A+\sin^2A+\cos^2B+\sin^2B-2\cos A\cos B-2\sin A\sin B$cos2A+sin2A+cos2B+sin2B−2cosAcosB−2sinAsinB
	$=$=	$2-2\left(\cos A\cos B+\sin A\sin B\right)$2−2(cosAcosB+sinAsinB) .......... [2]

Equating the two expressions, [1] and [2], gives:

$2-2\cos\left(A-B\right)$2−2cos(A−B)	$=$=	$2-2\left(\cos A\cos B+\sin A\sin B\right)$2−2(cosAcosB+sinAsinB)
Hence, $\cos\left(A-B\right)$cos(A−B)	$=$=	$\cos A\cos B+\sin A\sin B$cosAcosB+sinAsinB

 

From this identity, it is possible to generate several others by using substitution and other properties already established. 

To establish a rule for $\cos\left(A+B\right)$cos(A+B), make the substitution of $-B$−B for angle $B$B:

$\cos\left(A-\left(-B\right)\right)$cos(A−(−B))

$=$=

$\cos A\cos\left(-B\right)+\sin A\sin\left(-B\right)$cosAcos(−B)+sinAsin(−B)

Using the fact that $\cos\left(-\theta\right)=\cos\theta$cos(−θ)=cosθ and $\sin\left(-\theta\right)=-\sin\theta$sin(−θ)=−sinθ, leads to:

$\cos\left(A+B\right)$cos(A+B)

$=$=

$\cos A\cos B-\sin A\sin B$cosAcosB−sinAsinB

The corresponding formulas for sine are obtained by using the complementary angle relationship: $\sin\theta=\cos\left(\frac{\pi}{2}-\theta\right)$sinθ=cos(π2​−θ).

Writing $\sin\left(A-B\right)$sin(A−B) as $\cos\left(\frac{\pi}{2}-\left(A-B\right)\right)$cos(π2​−(A−B)), it follows that: 

$\sin\left(A-B\right)$sin(A−B)	$=$=	$\cos\left(\left(\frac{\pi}{2}-A\right)+B\right)$cos((π2​−A)+B)
	$=$=	$\cos\left(\frac{\pi}{2}-A\right)\cos B-\sin\left(\frac{\pi}{2}-A\right)\sin B$cos(π2​−A)cosB−sin(π2​−A)sinB
	$=$=	$\sin A\cos B-\cos A\sin B$sinAcosB−cosAsinB

The rule for $\sin\left(A+B\right)$sin(A+B) can be obtained by again making the substitution $-B$−B for the angle $B$B. Hence:

$\sin\left(A+B\right)\equiv\sin A\cos B+\cos A\sin B$sin(A+B)≡sinAcosB+cosAsinB

The corresponding formulas for the tangent function are obtained by expanding $\tan\left(A+B\right)=\frac{\sin\left(A+B\right)}{\cos\left(A+B\right)}$tan(A+B)=sin(A+B)cos(A+B)​

After some simplification:

$\tan\left(A+B\right)=\frac{\tan A+\tan B}{1-\tan A\tan B}$tan(A+B)=tanA+tanB1−tanAtanB​

and

$\tan\left(A-B\right)=\frac{\tan A-\tan B}{1+\tan A\tan B}$tan(A−B)=tanA−tanB1+tanAtanB​

Summarising all these rules together:

If $B=A$B=A in the above identities, then the following useful double angle formulae are obtained:

Worked example

Example 3

Find an 'exact value' expression for $\sin\frac{\pi}{12}$sinπ12​.

Think: Recall that integer multiples of $\frac{\pi}{4}$π4​ and $\frac{\pi}{6}$π6​ have exact values when evaluated using a trigonometric function. $\frac{\pi}{12}$π12​ is not an integer multiple of either of these but can we write this as a sum or difference of angles with familiar values? 

Do: It may be easier to think in multiples of $\frac{1}{12}$112​. We notice that $\frac{1}{12}=\frac{3}{12}-\frac{2}{12}$112​=312​−212​ and hence,$\frac{\pi}{12}=\frac{\pi}{4}-\frac{\pi}{6}$π12​=π4​−π6​. Using our rule for $\sin\left(A-B\right)$sin(A−B):

$\sin\left(A-B\right)$sin(A−B)	$=$=	$\sin A\cos B-\cos A\sin B$sinAcosB−cosAsinB
$$	$=$=	$\sin\frac{\pi}{4}\cos\frac{\pi}{6}-\cos\frac{\pi}{4}\sin\frac{\pi}{6}$sinπ4​cosπ6​−cosπ4​sinπ6​
	$=$=	$\frac{\sqrt{2}}{2}.\frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2}.\frac{1}{2}$√22​.√32​−√22​.12​
	$=$=	$\frac{\sqrt{6}-\sqrt{2}}{4}$√6−√24​

Practice questions

Question 3

Question 4

Question 5

Question 6