5.09 Angle sum identities

Angle sum identities

The angle sum and difference identities allow us to expand 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.

We can write two different expressions for the length of the green line $PQ$PQ. First, we use the cosine rule.

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

$c^2$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, we write the distance $PQ$PQ using the coordinates of the points and the distance formula. We have: 

$d^2$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 we can generate several others by using substitution and properties we have already established. 

To establish a rule for $\cos\left(A+B\right)$cos(A+B) we can 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, we have:

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

$=$=

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

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

If we write $\sin\left(A-B\right)$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

We can obtain the rule for $\sin\left(A+B\right)$sin(A+B) 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, we have

$\tan\left(A+B\right)=\frac{\tan A+\tan B}{1-\tan A\tan B}$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, we have:

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

Worked example

Example 1

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

Think: We know that we can find exact values for the trig ratios of  integer multiples of $\frac{\pi}{4}$π4​ and $\frac{\pi}{6}$π6​ . $\frac{\pi}{12}$π12​ is not an integer multiple of either of these but can we write it as a sum or difference of angles that we know exact values for? 

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
$\sin\left(\frac{\pi}{4}-\frac{\pi}{6}\right)$sin(π4​−π6​)	$=$=	$\sin\frac{\pi}{4}\cos\frac{\pi}{6}-\cos\frac{\pi}{4}\sin\frac{\pi}{6}$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 1

Question 2

Question 3

Question 4

Question 5