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:
| $\cos\left(A+B\right)$cos(A+B) | $=$= | $\cos A\cos B-\sin A\sin B$cosAcosB−sinAsinB |
| $\cos\left(A-B\right)$cos(A−B) | $=$= | $\cos A\cos B+\sin A\sin B$cosAcosB+sinAsinB |
| $\sin\left(A+B\right)$sin(A+B) | $=$= | $\sin A\cos B+\cos A\sin B$sinAcosB+cosAsinB |
| $\sin\left(A-B\right)$sin(A−B) | $=$= | $\sin A\cos B-\cos A\sin B$sinAcosB−cosAsinB |
| $\tan\left(A+B\right)$tan(A+B) | $=$= | $\frac{\tan A+\tan B}{1-\tan A\tan B}$tanA+tanB1−tanAtanB |
| $\tan\left(A-B\right)$tan(A−B) | $=$= | $\frac{\tan A-\tan B}{1+\tan A\tan B}$tanA−tanB1+tanAtanB |
If $B=A$B=A in the above identities, then we obtain the following useful double angle formulae:
| $\sin2A$sin2A | $=$= | $2\sin A\cos A$2sinAcosA |
| $\cos2A$cos2A | $=$= | $\cos^2\left(A\right)-\sin^2\left(A\right)$cos2(A)−sin2(A) |
| $\tan2A$tan2A | $=$= | $\frac{2\tan A}{1-\tan^2\left(A\right)}$2tanA1−tan2(A) |
Worked example
Example 1
Find an 'exact value' expression for $\sin\frac{\pi}{12}$sinπ12.
Think: From lesson 5.04 we found 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 |
| $$ | $=$= | $\sin\frac{\pi}{4}\cos\frac{\pi}{6}-\cos\frac{\pi}{4}\sin\frac{\pi}{6}$sinπ4cosπ6−cosπ4sinπ6 |
| $=$= | $\frac{\sqrt{2}}{2}.\frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2}.\frac{1}{2}$√22.√32−√22.12 | |
| $=$= | $\frac{\sqrt{6}-\sqrt{2}}{4}$√6−√24 |
Practice questions
Question 1
Using the expansion of $\cos\left(A+B\right)$cos(A+B), find the exact value of $\cos\left(\frac{7\pi}{12}\right)$cos(7π12). Express the value in rationalised form.
Question 2
Express $\cos\left(3\theta+x\right)\cos3\theta-\sin\left(3\theta+x\right)\sin3\theta$cos(3θ+x)cos3θ−sin(3θ+x)sin3θ in simplest form.
Question 3
By simplifying the left hand side of the identity , prove that $\tan\left(\theta+\alpha\right)\tan\left(\theta-\alpha\right)=\frac{\tan^2\left(\theta\right)-\tan^2\left(\alpha\right)}{1-\tan^2\left(\theta\right)\tan^2\left(\alpha\right)}$tan(θ+α)tan(θ−α)=tan2(θ)−tan2(α)1−tan2(θ)tan2(α).
Question 4
If $\tan A+\tan B=-5$tanA+tanB=−5 and $\tan A\tan B=6$tanAtanB=6, prove that $\sin\left(A+B\right)=\cos\left(A+B\right)$sin(A+B)=cos(A+B).