6.04 Double and half angle identities
If we take a special case of the sum and difference formulas, we can derive another set of identities known as the double-angle and half-angle identities.
Exploration
Consider the expression $\sin A$sinA. Suppose we want to relate it to the $\sin2A$sin2A. Can we just find the sine of $A$A and double it?
No! (We can find many counterexamples). We can, however, rewrite the expression as $\sin\left(A+A\right)$sin(A+A) and use the sum formula for sine.
| $\sin2A$sin2A | $=$= | $\sin\left(A+A\right)$sin(A+A) |
$2A=A+A$2A=A+A |
| $=$= | $\sin A\cos A+\cos A\sin A$sinAcosA+cosAsinA |
Sum identity for sine |
|
| $=$= | $\sin A\cos A+\sin A\cos A$sinAcosA+sinAcosA |
Commutative property of multiplication |
|
| $=$= | $2\sin A\cos A$2sinAcosA |
Simplify |
This proves the identity $\sin2A=2\sin A\cos A$sin2A=2sinAcosA, which is called the double-angle identitiy for the sine function.
In a similar manner, we can derive the double-angle identities for the cosine and tangent functions.
| Sine Double-Angle Identity | Cosine Double-Angle Identity | Tangent Double-Angle Identity |
|---|---|---|
| $\sin2A=2\sin A\cos A$sin2A=2sinAcosA | $\cos2A=\cos^2\left(A\right)-\sin^2\left(A\right)$cos2A=cos2(A)−sin2(A) | $\tan2A=\frac{2\tan A}{1-\tan^2\left(A\right)}$tan2A=2tanA1−tan2(A) |
By letting $B=2A$B=2A we obtain three corresponding half-angle identities. These identities are sometimes referred to in alternate forms which have been manipulated by applying the Pythagorean identities. The relationships remain the same.
| Sine Half-Angle Identities | Cosine Half-Angle Identities | Tangent Half-Angle Identities |
|---|---|---|
| $\sin B=2\sin\left(\frac{B}{2}\right)\cos\left(\frac{B}{2}\right)$sinB=2sin(B2)cos(B2) | $\cos B=\cos^2\left(\frac{B}{2}\right)-\sin^2\left(\frac{B}{2}\right)$cosB=cos2(B2)−sin2(B2) | $\tan B=\frac{2\tan\left(\frac{B}{2}\right)}{1-\tan^2\left(\frac{B}{2}\right)}$tanB=2tan(B2)1−tan2(B2) |
| $\sin\left(\frac{B}{2}\right)=\pm\sqrt{\frac{1-\cos B}{2}}$sin(B2)=±√1−cosB2 | $\cos\left(\frac{B}{2}\right)=\pm\sqrt{\frac{1+\cos B}{2}}$cos(B2)=±√1+cosB2 | $\tan\left(\frac{B}{2}\right)=\sqrt{\frac{1-\cos B}{1+\cos B}}$tan(B2)=√1−cosB1+cosB |
We can create many variations of these identities by using the Pythagorean or other identities.
For example, we can rearrange $\sin^2\left(A\right)+\cos^2\left(A\right)=1$sin2(A)+cos2(A)=1 to get $\sin^2\left(A\right)=1-\cos^2\left(A\right)$sin2(A)=1−cos2(A) and substitute it in to get:
| $\cos2A$cos2A | $=$= | $\cos^2\left(A\right)-\sin^2\left(A\right)$cos2(A)−sin2(A) |
Double angle formula |
| $\cos2A$cos2A | $=$= | $\cos^2\left(A\right)-\left(1-\cos^2\left(A\right)\right)$cos2(A)−(1−cos2(A)) |
Substitute Pythagorean identity |
| $\cos2A$cos2A | $=$= | $2\cos^2\left(A\right)-1$2cos2(A)−1 |
Simplify |
Similarly, we can use a combination of identities to solve problems involving the other double and half angle formulas.
Practice questions
Question 1
Using a double angle identity, simplify the expression $\sin5x\cos5x$sin5xcos5x.
Express your answer as a single trigonometric function.
Question 2
Rewrite $\sqrt{\frac{1+\cos72^\circ}{2}}$√1+cos72°2 as a single trigonometric value.
Question 3
Find the exact value of $\sin\left(\frac{x}{2}\right)$sin(x2) if $\cos x=-\frac{4}{7}$cosx=−47 and $\frac{\pi}{2}