A bit of trigonometry and algebra

Geometry Level 2

Given that cos ⁡ 2 θ = − 1 3 \cos2\theta=-\frac{1}{3} , find the value of sin ⁡ 6 θ − cos ⁡ 6 θ \sin^6\theta-\cos^6\theta .

− 7 27 -\frac{7}{27} − 1 3 -\frac{1}{3} 1 3 \frac{1}{3} 7 27 \frac{7}{27}

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Philip Lee by Philip Lee , 21, Singapore

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4 solutions

Marco Brezzi
Aug 14, 2017

Identities I'll use:

a 3 − b 3 = ( a − b ) ( a 2 + a b + b 2 ) a^3-b^3=(a-b)(a^2+ab+b^2)

sin ⁡ 2 θ + cos ⁡ 2 θ = 1 \sin^2{\theta}+\cos^2{\theta}=1

sin ⁡ ( 2 θ ) = 2 sin ⁡ θ cos ⁡ θ \sin (2\theta)=2\sin\theta\cos\theta

cos ⁡ ( 2 θ ) = cos ⁡ 2 θ − sin ⁡ 2 θ \cos (2\theta)=\cos^2\theta -\sin^2\theta

Hence

sin ⁡ 6 θ − cos ⁡ 6 θ = ( sin ⁡ 2 θ ) 3 − ( cos ⁡ 2 θ ) 3 = ( sin ⁡ 2 θ − cos ⁡ 2 θ ) ( sin ⁡ 4 θ + sin ⁡ 2 θ cos ⁡ 2 θ + cos ⁡ 4 θ ) = [ − cos ⁡ ( 2 θ ) ] ( sin ⁡ 4 θ + cos ⁡ 4 θ + 2 sin ⁡ 2 θ cos ⁡ 2 θ − sin ⁡ 2 θ cos ⁡ 2 θ ) = 1 3 [ ( sin ⁡ 2 θ + cos ⁡ 2 θ ) 2 − sin ⁡ 2 θ cos ⁡ 2 θ ] = 1 3 [ 1 2 − sin ⁡ 2 ( 2 θ ) 4 ] = 1 3 [ 1 − 1 − cos ⁡ 2 ( 2 θ ) 4 ] = 1 3 ( 1 − 1 − 1 9 4 ) = 1 3 ( 1 − 2 9 ) = 1 3 ⋅ 7 9 = 7 27 \begin{aligned} \sin^6\theta-\cos^6\theta &=(\sin^2\theta)^3-(\cos^2\theta)^3\\ &=(\sin^2\theta -\cos^2\theta)(\sin^4\theta+\sin^2\theta\cos^2\theta+\cos^4\theta)\\ &=[ -\cos(2\theta)](\sin^4\theta +\cos^4\theta +2\sin^2\theta\cos^2\theta-\sin^2\theta\cos^2\theta)\\ &=\dfrac{1}{3}[(\sin^2\theta+\cos^2\theta)^2-\sin^2\theta\cos^2\theta]\\ &=\dfrac{1}{3}\left[1^2-\dfrac{\sin^2(2\theta)}{4}\right]\\ &=\dfrac{1}{3}\left[1-\dfrac{1-\cos^2(2\theta)}{4}\right]\\ &=\dfrac{1}{3}\left(1-\dfrac{1-\frac{1}{9}}{4}\right)=\dfrac{1}{3}\left(1-\dfrac{2}{9}\right)=\dfrac{1}{3}\cdot\dfrac{7}{9}=\boxed{\dfrac{7}{27}} \end{aligned}

6 Helpful 0 Interesting 0 Brilliant 0 Confused

Didn't think of using sin ⁡ ( 2 θ ) = 2 sin ⁡ θ cos ⁡ θ \sin\left(2\theta\right)=2\sin\theta\cos\theta . Nice!

Philip Lee - 3 years, 10 months ago

Dude, l just wanna say your solutions are outstanding. Neat and logical.

Hana Wehbi - 3 years, 9 months ago
Chew-Seong Cheong
Aug 14, 2017

cos ⁡ 2 θ = − 1 3 cos ⁡ 2 θ − sin ⁡ 2 θ = − 1 3 sin ⁡ 2 θ − cos ⁡ 2 θ = 1 3 ( sin ⁡ 2 θ − cos ⁡ 2 θ ) 3 = ( 1 3 ) 3 sin ⁡ 6 θ − 3 sin ⁡ 4 θ cos ⁡ 2 θ + 3 sin ⁡ 2 θ cos ⁡ 4 θ − cos ⁡ 6 θ = 1 27 sin ⁡ 6 θ − 3 sin ⁡ 2 θ cos ⁡ 2 θ ( sin ⁡ 2 θ − cos ⁡ 2 θ ) − cos ⁡ 6 θ = 1 27 sin ⁡ 6 θ − 3 sin ⁡ 2 θ cos ⁡ 2 θ ( 1 3 ) − cos ⁡ 6 θ = 1 27 sin ⁡ 6 θ − cos ⁡ 6 θ = 1 27 + sin ⁡ 2 θ cos ⁡ 2 θ = 1 27 + 1 4 sin ⁡ 2 2 θ = 1 27 + 1 4 ( 1 − cos ⁡ 2 2 θ ) = 1 27 + 1 4 ( 1 − 1 9 ) = 7 27 \begin{aligned} \cos 2\theta & = - \frac 13 \\ \cos^2 \theta - \sin^2 \theta & = - \frac 13 \\ \sin^2 \theta - \cos^2 \theta & = \frac 13 \\ \left(\sin^2 \theta - \cos^2 \theta\right)^3 & = \left(\frac 13\right)^3 \\ \sin^6 \theta - 3\sin^4 \theta \cos^2 \theta + 3\sin^2 \theta \cos^4 \theta - \cos^6 \theta & = \frac 1{27} \\ \sin^6 \theta - 3\sin^2 \theta \cos^2 \theta (\sin^2 \theta - \cos^2 \theta) - \cos^6 \theta & = \frac 1{27} \\ \sin^6 \theta - 3\sin^2 \theta \cos^2 \theta \left(\frac 13\right) - \cos^6 \theta & = \frac 1{27} \\ \sin^6 \theta - \cos^6 \theta & = \frac 1{27} + \sin^2 \theta \cos^2 \theta \\ & = \frac 1{27} + \frac 14 \sin^2 2 \theta \\ & = \frac 1{27} + \frac 14 (1 - \cos^2 2 \theta) \\ & = \frac 1{27} + \frac 14 \left(1 - \frac 19 \right) \\ & = \boxed{\dfrac 7{27}} \end{aligned}

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Philip Lee
Aug 14, 2017

Relevant wiki: Factorization of Polynomials

From the double angle identities and pythagorean identities , cos ⁡ 2 θ = cos ⁡ 2 θ − sin ⁡ 2 θ = − 1 3 \cos2\theta=\cos^2\theta-\sin^2\theta=-\frac{1}{3} and sin ⁡ 2 θ + cos ⁡ 2 θ = 1 \sin^2\theta+\cos^2\theta=1 .

Hence, by reversing the perfect cube identity ,

sin ⁡ 6 θ − cos ⁡ 6 θ = − ( cos ⁡ 6 θ − sin ⁡ 6 θ ) = − [ ( cos ⁡ 2 θ − sin ⁡ 2 θ ) 3 + 3 cos ⁡ 4 θ sin ⁡ 2 θ − 3 cos ⁡ 2 θ sin ⁡ 4 θ ] = − [ ( cos ⁡ 2 θ − sin ⁡ 2 θ ) 3 + 3 cos ⁡ 2 θ sin ⁡ 2 θ ( cos ⁡ 2 θ − sin ⁡ 2 θ ) ] = − [ ( cos ⁡ 2 θ − sin ⁡ 2 θ ) 3 + 3 4 ( 4 cos ⁡ 2 θ sin ⁡ 2 θ ) ( cos ⁡ 2 θ − sin ⁡ 2 θ ) ] = − { ( cos ⁡ 2 θ − sin ⁡ 2 θ ) 3 + 3 4 [ ( cos ⁡ 2 θ + sin ⁡ 2 θ ) 2 − ( cos ⁡ 2 θ − sin ⁡ 2 θ ) 2 ] ( cos ⁡ 2 θ − sin ⁡ 2 θ ) } = − { ( − 1 3 ) 3 + 3 4 [ ( 1 ) 2 − ( − 1 3 ) 2 ] ( − 1 3 ) } = 7 27 □ \begin{aligned} \sin^6\theta-\cos^6\theta &= -\left(\cos^6\theta-\sin^6\theta\right) \\ &= -\left[\left(\cos^2\theta-\sin^2\theta\right)^3+3\cos^4\theta\sin^2\theta-3\cos^2\theta\sin^4\theta\right] \\ &= -\left[\left(\cos^2\theta-\sin^2\theta\right)^3+3\cos^2\theta\sin^2\theta\left(\cos^2\theta-\sin^2\theta\right)\right] \\ &= -\left[\left(\cos^2\theta-\sin^2\theta\right)^3+\frac{3}{4}\left(4\cos^2\theta\sin^2\theta\right)\left(\cos^2\theta-\sin^2\theta\right)\right] \\ &= -\left\{\left(\cos^2\theta-\sin^2\theta\right)^3+\frac{3}{4}\left[\left(\cos^2\theta+\sin^2\theta\right)^2-\left(\cos^2\theta-\sin^2\theta\right)^2\right]\left(\cos^2\theta-\sin^2\theta\right)\right\} \\ &= -\left\{\left(-\frac{1}{3}\right)^3+\frac{3}{4}\left[\left(1\right)^2-\left(-\frac{1}{3}\right)^2\right]\left(-\frac{1}{3}\right)\right\} \\ &= \frac{7}{27} \ _\square \end{aligned} or by factoring sin ⁡ 6 θ − cos ⁡ 6 θ \sin^6\theta-\cos^6\theta directly,

sin ⁡ 6 θ − cos ⁡ 6 θ = − ( cos ⁡ 2 θ − sin ⁡ 2 θ ) ( cos ⁡ 4 θ + cos ⁡ 2 θ sin ⁡ 2 θ + sin ⁡ 4 θ ) = − ( cos ⁡ 2 θ − sin ⁡ 2 θ ) [ ( cos ⁡ 2 θ + sin ⁡ 2 θ ) 2 − cos ⁡ 2 θ sin ⁡ 2 θ ] = − ( cos ⁡ 2 θ − sin ⁡ 2 θ ) [ ( cos ⁡ 2 θ + sin ⁡ 2 θ ) 2 − 1 4 ( 4 cos ⁡ 2 θ sin ⁡ 2 θ ) ] = − ( cos ⁡ 2 θ − sin ⁡ 2 θ ) { ( cos ⁡ 2 θ + sin ⁡ 2 θ ) 2 − 1 4 [ ( cos ⁡ 2 θ + sin ⁡ 2 θ ) 2 − ( cos ⁡ 2 θ − sin ⁡ 2 θ ) 2 ] } = − ( − 1 3 ) { ( 1 ) 2 − 1 4 [ ( 1 ) 2 − ( − 1 3 ) 2 ] } = 7 27 □ \begin{aligned} \sin^6\theta-\cos^6\theta &= -\left(\cos^2\theta-\sin^2\theta\right)\left(\cos^4\theta+\cos^2\theta\sin^2\theta+\sin^4\theta\right) \\ &= -\left(\cos^2\theta-\sin^2\theta\right)\left[\left(\cos^2\theta+\sin^2\theta\right)^2-\cos^2\theta\sin^2\theta\right] \\ &= -\left(\cos^2\theta-\sin^2\theta\right)\left[\left(\cos^2\theta+\sin^2\theta\right)^2-\frac{1}{4}\left(4\cos^2\theta\sin^2\theta\right)\right] \\ &= -\left(\cos^2\theta-\sin^2\theta\right)\left\{\left(\cos^2\theta+\sin^2\theta\right)^2-\frac{1}{4}\left[\left(\cos^2\theta+\sin^2\theta\right)^2-\left(\cos^2\theta-\sin^2\theta\right)^2\right]\right\} \\ &= -\left(-\frac{1}{3}\right)\left\{\left(1\right)^2-\frac{1}{4}\left[\left(1\right)^2-\left(-\frac{1}{3}\right)^2\right]\right\} \\ &= \frac{7}{27} \ _\square \end{aligned}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Please post a solution if you have any easier ones :)

Philip Lee - 3 years, 10 months ago
Daniel Branscombe
Aug 16, 2017

using

S i n 2 ( θ ) = 1 − C o s ( 2 θ ) 2 = 1 − − 1 3 2 = 2 3 Sin^2(\theta)=\frac{1-Cos(2\theta)}{2}=\frac{1-\frac{-1}{3}}{2}=\frac{2}{3}

and

C o s 2 ( θ ) = 1 + C o s ( 2 θ ) 2 = 1 + − 1 3 2 = 1 3 Cos^2(\theta)=\frac{1+Cos(2\theta)}{2}=\frac{1+\frac{-1}{3}}{2}=\frac{1}{3}

we get

S i n 6 ( θ ) = ( S i n 2 ( θ ) ) 3 = ( 2 3 ) 3 = 8 27 Sin^6(\theta)=(Sin^2(\theta))^3=(\frac{2}{3})^3=\frac{8}{27}

C o s 6 ( θ ) = ( C o s 2 ( θ ) ) 3 = ( 1 3 ) 3 = 1 27 Cos^6(\theta)=(Cos^2(\theta))^3=(\frac{1}{3})^3=\frac{1}{27}

Thus

S i n 6 ( θ ) − C o s 6 ( θ ) = 8 27 − 1 27 = 1 27 Sin^6(\theta)-Cos^6(\theta)=\frac{8}{27}-\frac{1}{27}=\frac{1}{27}

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