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6.04 Double and half angle identities

Lesson

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.

Double-Angle Identities
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=2tanA1tan2(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.

Half-Angle Identities

 

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)1tan2(B2)
$\sin\left(\frac{B}{2}\right)=\pm\sqrt{\frac{1-\cos B}{2}}$sin(B2)=±1cosB2 $\cos\left(\frac{B}{2}\right)=\pm\sqrt{\frac{1+\cos B}{2}}$cos(B2)=±1+cosB2 $\tan\left(\frac{B}{2}\right)=\sqrt{\frac{1-\cos B}{1+\cos B}}$tan(B2)=1cosB1+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)=1cos2(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)(1cos2(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 $\frac{\pi}{2}π2<x$<$<$\pi$π.

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