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Wize High School Algebra II Textbook (Common Core) > Trigonometric Identities & Proofs

Double & Half Angle Identities

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Double Angle Identities

The following identities are known as Double Angle Identities.

sin⁡2θ=2sin⁡θcos⁡θcos⁡2θ=cos⁡2θ−sin⁡2θ=2cos⁡2θ−1=1−2sin⁡2θtan⁡2θ=2tan⁡θ1−tan⁡2θ\boxed{\begin{array} {} \sin2\theta&=&2\sin\theta\cos\theta\\\\\\ \cos2\theta&=&\cos^2\theta-\sin^2\theta\\\\ &=&2\cos^2\theta-1\\\\ &=&1-2\sin^2\theta\\\\\\ \tan2\theta&=&\dfrac{2\tan\theta}{1-\tan^2\theta}\\\\ \end{array}}

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Example

Simplify cos⁡2(4θ)−sin⁡2(4θ)\cos^2(4\theta)-\sin^2(4\theta) into one trigonometric function

Let a=4θa=4\theta be an angle measured in radians. Then,

cos⁡2(4θ)−sin⁡2(4θ)=cos⁡2(a)−sin⁡2(a)\begin{array}{rcl} \cos^2(4\theta)-\sin^2(4\theta)&=&\cos^2(a)-\sin^2(a)\\\\ \end{array}
Therefore,

cos⁡2(a)−sin⁡2(a)=cos⁡(2a)cos⁡2(4θ)−sin⁡2(4θ)=cos⁡(8θ)\begin{array}{rcl} \cos^2(a)-\sin^2(a)&=&\cos(2a)\\\\ \cos^2(4\theta)-\sin^2(4\theta)&=&\cos(8\theta) \end{array}


So, cos⁡2(4θ)−sin⁡2(4θ)=cos⁡(8θ)\boxed{\cos^2(4\theta)-\sin^2(4\theta)=\cos(8\theta)}
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Double Angle Identities


Let sin⁡θ=23\sin\theta=\dfrac{2}{3} and 0≤θ≤π20\leq{\theta}\leq{}\dfrac{\pi}{2}. Determine sin⁡2θ, cos⁡2θ, tan⁡2θ.\sin2\theta,~\cos2\theta, ~\tan2\theta.


Since sin⁡θ=23=yr\sin\theta=\dfrac{2}{3}=\dfrac{y}{r}, then using Pythagorean Theorem, we can find 'x' and the other trigonometric ratios

x2+y2=r2x2+22=32x=9−4x=5\begin{array}{rcl} x^2+y^2&=&r^2\\\\ x^2+2^2&=&3^2\\\\ x&=&\sqrt{9-4}\\\\ x&=&\sqrt{5} \end{array} ∴ cos⁡θ=53sin⁡θ=23tan⁡θ=25\begin{array}{rcl} \therefore~\cos\theta&=&\dfrac{\sqrt{5}}{3}\\\\ \sin\theta&=&\dfrac{2}{3}\\\\ \tan\theta&=&\dfrac{2}{\sqrt{5}} \end{array}


sin⁡2θ‾:\underline{\sin2\theta}: cos⁡2θ‾:\underline{\cos2\theta}: tan⁡2θ‾:\underline{\tan2\theta}:
sin⁡2θ=2sin⁡θcos⁡θ=2(23)(53)=459\begin{array}{rcl} \sin2\theta&=&2\sin\theta\cos\theta\\\\ &=&2\Bigg(\dfrac{2}{3}\Bigg)\Bigg(\dfrac{\sqrt{5}}{3}\Bigg)\\\\ &=&\boxed{\dfrac{4\sqrt{5}}{9}} \end{array} cos⁡2θ=cos⁡2θ−sin⁡2θ=(53)2−(23)2=(59)−(49)=19\begin{array}{rcl} \cos2\theta&=&\cos^2\theta-\sin^2\theta\\\\ &=&\Bigg(\dfrac{\sqrt{5}}{3}\Bigg)^2-\Bigg(\dfrac{2}{3}\Bigg)^2\\\\ &=&\Bigg(\dfrac{5}{9}\Bigg)-\Bigg(\dfrac{4}{9}\Bigg)\\\\ &=&\boxed{\dfrac{1}{9}} \end{array} tan⁡2θ=2tan⁡θ1−tan⁡2θ=2⋅(25)(1−(25)2)=(45)(1−45)=(45)(15)=205\begin{array}{rcl} \tan2\theta&=&\dfrac{2\tan\theta}{1-\tan^2\theta}\\\\ &=&\dfrac{2\cdot\Bigg(\dfrac{2}{\sqrt{5}}\Bigg)}{\Bigg(1-\Bigg(\dfrac{2}{\sqrt{5}}\Bigg)^2\Bigg)}\\\\ &=&\dfrac{\Bigg(\dfrac{4}{\sqrt{5}}\Bigg)}{\Bigg(1-\dfrac{4}{5}\Bigg)}\\\\ &=&\dfrac{\Bigg(\dfrac{4}{\sqrt{5}}\Bigg)}{\Bigg(\dfrac{1}{5}\Bigg)}\\\\ &=&\dfrac{20}{\sqrt{5}} \end{array}
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Half-Angle Identities

The following identities are known as Half-Angle identities.

sin⁡(θ2)=±1−cos⁡θ2cos⁡(θ2)=±1+cos⁡θ2tan⁡(θ2)=1−cos⁡θsin⁡θ=sin⁡θ1+cos⁡θ\boxed{\begin{array} {} \sin\Bigg(\dfrac{\theta}{2}\Bigg)&=&\pm\sqrt{\dfrac{1-\cos\theta}{2}}\\\\\\ \cos\Bigg(\dfrac{\theta}{2}\Bigg)&=&\pm\sqrt{\dfrac{1+\cos\theta}{2}}\\\\ \tan\Bigg(\dfrac{\theta}{2}\Bigg)&=&\dfrac{1-\cos\theta}{\sin\theta}=\dfrac{\sin\theta}{1+\cos\theta}\\\\ \end{array}}

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Example

Let cos⁡θ=47\cos\theta=\dfrac{4}{7} and 0≤θ≤π20\leq{}\theta\leq{}\dfrac{\pi}{2}. Find tan⁡(θ2)\tan\Bigg(\dfrac{\theta}{2}\Bigg).


Since cos⁡θ=47=xr\cos\theta=\dfrac{4}{7}=\dfrac{x}{r}, then, using Pythagorean Theorem, we can find yy and find sin⁡θ\sin\theta.
x2+y2=r242+y2=72y=72−42y=33\begin{array}{rcl} x^2+y^2&=&r^2\\\\ 4^2+y^2&=&7^2\\\\ y&=&\sqrt{7^2-4^2}\\\\ y&=&\sqrt{33} \end{array} ∴ sin⁡θ=337\therefore~\sin\theta=\dfrac{\sqrt{33}}{7}


Therefore,

tan⁡(θ2)=1−cos⁡θsin⁡θ=(1−47)(337)=37337=333\begin{array}{rcl} \tan\Bigg(\dfrac{\theta}{2}\Bigg)&=&\dfrac{1-\cos\theta}{\sin\theta}\\\\ &=&\dfrac{\Bigg(1-\dfrac{4}{7}\Bigg)}{\Bigg(\dfrac{\sqrt{33}}{7}\Bigg)}\\\\ &=&\dfrac{\dfrac{3}{7}}{\dfrac{\sqrt{33}}{7}}\\\\ &=&\boxed{\dfrac{3}{\sqrt{33}}} \end{array}
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Example: Half Angle Identities

What is the exact value of sin⁡(π8)\sin\Bigg(\dfrac{\pi}{8}\Bigg)?

Let θ2=π8\dfrac{\theta}{2}=\dfrac{\pi}{8}

So, θ=π4\theta=\dfrac{\pi}{4}

Using the half-angle identity, we get:

sin⁡(π8)=sin⁡(π42)=±1−cos⁡π42=±1−222=±2−222=±2−24=±2−22\begin{array}{rcl} \sin\Bigg(\dfrac{\pi}{8}\Bigg)=\sin\Bigg(\dfrac{\frac{\pi}{4}}{2}\Bigg)&=&\pm\sqrt{\dfrac{1-\cos\frac{\pi}{4}}{2}}\\\\ &=&\pm\sqrt{\dfrac{1-\frac{\sqrt{2}}{2}}{2}}\\\\ &=&\pm\sqrt{\dfrac{\frac{2-\sqrt{2}}{2}}{2}}\\\\ &=&\pm\sqrt{\dfrac{2-\sqrt{2}}{4}}\\\\ &=&\pm\dfrac{\sqrt{2-\sqrt{2}}}{2} \end{array}

Practice: Double & Half Angle Identities


Let tan⁡θ=83\tan\theta=\dfrac{8}{3} and 0≤θ≤π20\leq{}\theta\leq{}\dfrac{\pi}{2}.


Practice: Double & Half Angle Identities


Determine the exact value for each of the following:

Practice: Double & Half Angle Identities

True or False? 1−tan⁡2θ1+tan⁡2θ=cos⁡2θ\dfrac{1-\tan^2\theta}{1+\tan^2\theta}=\cos2\theta

Extra Practice
Double & Half Angle Identities
Practice: Double & Half Angle Identities
True or False? csc⁡2θ−2csc⁡2θ=sin⁡2θ\dfrac{\csc^2\theta-2}{\csc^2\theta}=\sin2\theta
Double & Half Angle Identities
Practice: Double & Half Angle Identities
True or False? sin⁡2θ1−cos⁡2θ=2csc⁡2θ−tan⁡θ\dfrac{\sin2\theta}{1-\cos2\theta}=2\csc2\theta-\tan\theta
Double & Half Angle Identities
Practice: Double & Half Angle Identities
Determine the exact value for each of the following:
Double & Half Angle Identities
Practice: Double & Half Angle Identities
Determine the exact value for each of the following:
Double & Half Angle Identities
Practice: Double & Half Angle Identities
Let cos⁡θ=16\cos\theta=\dfrac{1}{6} and 0≤θ≤π20\leq{}\theta\leq{}\dfrac{\pi}{2}.
Double & Half Angle Identities
Practice: Double & Half Angle Identities
Let sin⁡θ=25\sin\theta=\dfrac{2}{5} and 0≤θ≤π20\leq{}\theta\leq{}\dfrac{\pi}{2}.