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A question on submultiple angles states...

Prove that:$$\sin^{6}{\frac{\theta}{2}}+\cos^{6}{\frac{\theta}{2}}=\frac{1}{4}(1+3\cos^2\theta)$$

My efforts

I tried using the formula $$a^3+b^3=(a+b)^3-3ab(a+b)$$ and $$\cos^2{\frac{\theta}{2}=\frac{\cos\theta + 1}{2}}$$ Then I tried simplifying it: $$\require{cancel} \begin{align} \sin^{6}{\frac{\theta}{2}}+\cos^{6}{\frac{\theta}{2}} & = (\sin^{2}{\frac{\theta}{2}}+\cos^{2}{\frac{\theta}{2}})^3 - 3\sin^2{\frac{\theta}{2}}\cos^2{\frac{\theta}{2}}(\sin^{2}{\frac{\theta}{2}}+\cos^{2}{\frac{\theta}{2}})\\ & = 1 - 3\sin^2{\frac{\theta}{2}}\cos^2{\frac{\theta}{2}}\\ & = 1 - 3(1 - \cos^2{\frac{\theta}{2}})\cos^2{\frac{\theta}{2}}\\ & = 1 - \frac{3}{2}(\cos\theta+1) + \frac{3}{4}(\cos\theta+1)^2\\ & = \frac{4\cancel{-6\cos\theta}-2+3\cos^2+3+\cancel{6\cos\theta}}{4}\\ & = \frac{1}{4}(5+3\cos^2\theta)\end{align} $$

I suspect I must have messed up with some sign somewhere. The trouble is, I can't seem to find where. Please help me in this regard.


Update: I am not accepting an answer because all the answers are equally good. It would be an injustice to the other answers.


Note to the editors: I also suspect that my post is a little short on appropriate tags. If required, please do the needful.

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    $\begingroup$ -2 is wrong. It's -3*2=-6 $\endgroup$ – Exodd Oct 6 '15 at 19:22
  • $\begingroup$ @Exodd Oh... that's where I slipped up. I ought to be more careful. Thank you very much! $\endgroup$ – Hungry Blue Dev Oct 6 '15 at 19:25
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$$\sin^6\frac{\theta}{2}+\cos^6\frac{\theta}{2}=\cdots=1-3\left(1-\cos^2\frac{\theta}{2}\right)\cos^2\frac{\theta}{2}$$ is correct. From here, note that

$$\begin{align}\sin^6\frac{\theta}{2}+\cos^6\frac{\theta}{2}&=1-3\left(1-\color{red}{\cos^2\frac{\theta}{2}}\right)\color{red}{\cos^2\frac{\theta}{2}}\\&=1-3\left(1-\color{red}{\frac{\cos\theta+1}{2}}\right)\cdot\color{red}{\frac{\cos\theta+1}{2}}\\&=1-3\cdot\frac{1-\cos\theta}{2}\cdot\frac{1+\cos\theta}{2}\\&=\frac{4-3(1-\cos^2\theta)}{4}\\&=\frac{1+3\cos^2\theta}{4}\end{align}$$

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You may begin factoring: \begin{align*} \sin^6\frac\theta2+\cos^6\frac\theta2 &=\Bigl(\sin^2\frac\theta2+\cos^2\frac\theta2\Bigr)\Bigl(\sin^4\frac\theta2-\sin^2\frac\theta2\cos^2\frac\theta2+\cos^4\frac\theta2\Bigr)\\ &=\Bigl(\cos^2\frac\theta2-\sin^2\frac\theta2\Bigr)^2+\sin^2\frac\theta2\cos^2\frac\theta2=\cos^2\theta+\frac14\sin^2\theta\\ &=\frac14(3\cos^2\theta+1). \end{align*} You also may linearise: $$\cos^2\theta+\frac14\sin^2\theta=\frac12(1+\cos2\theta)+\frac18(1-\cos2\theta)=\frac18(3\cos2\theta+5).$$

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You may use $$ 2\sin x \cos x= \sin (2x) \tag1 $$ giving $$ \begin{align} \sin^{6}{\frac{\theta}{2}}+\cos^{6}{\frac{\theta}{2}} & = (\sin^{2}{\frac{\theta}{2}}+\cos^{2}{\frac{\theta}{2}})^3 - 3\sin^2{\frac{\theta}{2}}\cos^2{\frac{\theta}{2}}(\sin^{2}{\frac{\theta}{2}}+\cos^{2}{\frac{\theta}{2}})\\ & = 1 - 3\sin^2{\frac{\theta}{2}}\cos^2{\frac{\theta}{2}} \\ & = 1 - \frac34\sin^2{\theta}\quad (\text{using} \,(1))\\ & = 1 - \frac34(1-\cos^2{\theta})\\ & = \frac14(1+3\cos^2{\theta}) \end{align} $$ as desired.

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Hint: $$\sin^2\theta/2\cos^2\theta/2\\=\frac{1}{4}(1-\cos\theta)(1+\cos\theta)=\frac{1}{4}(1-\cos^2\theta)$$

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A little life-saving trick: $$ \require{cancel} \begin{align} \sin^{6}{\frac{\theta}{2}}+\cos^{6}{\frac{\theta}{2}} & = (\sin^{2}{\frac{\theta}{2}}+\cos^{2}{\frac{\theta}{2}})^3 - 3\sin^2{\frac{\theta}{2}}\cos^2{\frac{\theta}{2}}(\sin^{2}{\frac{\theta}{2}}+\cos^{2}{\frac{\theta}{2}})\\ & = 1 - 3\sin^2{\frac{\theta}{2}}\cos^2{\frac{\theta}{2}}\\ & = 1 - 3(1 - \cos^2{\frac{\theta}{2}})\cos^2{\frac{\theta}{2}}\\ & = 1 - \frac{3}{2}(\cos\theta+1) + \frac{3}{4}(\cos\theta+1)^2\\ & = \frac{4\cancel{-6\cos\theta}-2+3\cos^2+3+\cancel{6\cos\theta}}{4}\\ & = \frac{1}{4}(5+3\cos^2\theta)\end{align} $$ is wrong because, if you evaluate at $\theta=0$ you get $$ \require{cancel} \begin{align} 1 & = 1\\ & = 1 \\ & = 1 \\ & = 1 \\ & = 2\\ & = 2\end{align} $$ and this also shows the passage that is wrong.

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  • $\begingroup$ Life-saver indeed. I believe that this is similar to testing the formula $$\cos {3\theta}=4\cos^3\theta - 3\cos\theta$$ by putting $\theta=0$. $\endgroup$ – Hungry Blue Dev Oct 6 '15 at 19:32
  • $\begingroup$ It isn't a test for validity, but a way to find your errors $\endgroup$ – Exodd Oct 6 '15 at 19:57
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Let $\theta/2=x$. Then the LHS is:

$$\sin^6 x+\cos^6 x=\underbrace{\left(\sin^2 x+\cos^2 x\right)}_{=1}\left(\sin^4 x-\sin^2 x\cos^2 x+\cos^4 x\right)$$

$$=\left(1-\cos^2 x\right)^2-\left(1-\cos^2 x\right)\cos^2 x+\cos^4 x\tag{1}$$

The RHS is:

$$\frac{1}{4}\left(1+3\cos^2 2x\right)=\frac{1}{4}\left(1+3\left(2\cos^2 x-1\right)^2\right)\tag{2}$$

Now prove $(1)=(2)$.

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