6
$\begingroup$

Prove that: $$\cos^2\theta\sin^4\theta=\frac{1}{32}(\cos6\theta-\cos2\theta+2-2\cos4\theta)$$

Attempt: \begin{align*} L.H.S & = \cos^2\theta\sin^4\theta\\ & = \cos^2\theta\sin^2\theta\sin^2\theta\\ & = \frac{1+\cos2\theta}{2}.\frac{1-\cos2\theta}{2}.\frac{1-\cos2\theta}{2}\\ & = \frac{1}{8} (1-\cos^22\theta)(1-\cos2\theta) \end{align*}

Now, what should I do?

$\endgroup$
2
  • 1
    $\begingroup$ I suggest starting off with $\cos^2 \theta \sin^2 \theta = \frac{1}{4}\sin^2 2\theta$ and remember $\cos^2 x+ \sin^2 x = 1$ etc etc $\endgroup$
    – Chinny84
    Feb 11 '16 at 16:30
  • $\begingroup$ starting from RHS to reduec it to the form of LHS $\endgroup$
    – runaround
    Feb 11 '16 at 16:30
2
$\begingroup$

Use the linearisation and the duplication formulae: \begin{align*} \cos^2\theta\sin^4\theta & = \cos^2\theta\sin^2\theta\sin^2\theta = \frac14 \sin^2 2\theta\sin^2\theta\\ &=\frac14\frac{1-\cos4\theta}{2}\frac{1-\cos2\theta}{2}= \frac{1}{32} (2\cos4\theta\cos2\theta-2\cos4\theta-2\cos2\theta+2)\\ &=\frac1{32}(\cos 6\theta+\cos2\theta-2\cos4\theta-2\cos2\theta+2)\\ &=\frac1{32}(\cos 6\theta-2\cos4\theta- \cos2\theta+2)\ \end{align*}

If you are allowed to use complex numbers, this is much easier: set $u=\mathrm e^{\mathrm i\theta}$. Then we have, by Euler's formulae: $$\bar u=\mathrm e^{-\mathrm i\theta},\quad \cos\theta=\frac{u+\bar u}2,\quad \sin\theta =\frac{u-\bar u}{2\mathrm i},$$ whence (note $u\bar u=1$) \begin{align*} \cos^2\theta\sin^4\theta & = \frac{(u+\bar u)^2}4 \frac{(u-\bar u)^4}{16}=\frac1{64}(u^2-\bar u^2)^2(u-\bar u)^2\\ &=\frac1{64}(u^4-2+\bar u^4)(u^2-2+\bar u^2)\\ &=\frac1{64}(u^6-2u^4+u^2-2u^2+4-2\bar u^2+\bar u^2-2\bar u^4+\bar u^6)\\ &=\frac1{64}(u^6+\bar u^6-2(u^4+\bar u^4) -(u^2+\bar u^2)+4)\\ &=\frac1{32}(\cos6\theta-2\cos4\theta-\cos2\theta+2). \end{align*}

$\endgroup$
0
$\begingroup$

HINT: use that $$\cos(6\theta)=32\, \left( \cos \left( \theta \right) \right) ^{6}-48\, \left( \cos \left( \theta \right) \right) ^{4}+18\, \left( \cos \left( \theta \right) \right) ^{2}-1 $$ $$\cos(2\theta)=2\, \left( \cos \left( \theta \right) \right) ^{2}-1$$ $$\cos(4\theta)=8\, \left( \cos \left( \theta \right) \right) ^{4}-8\, \left( \cos \left( \theta \right) \right) ^{2}+1 $$

$\endgroup$
0
$\begingroup$

With $z=e^{i\theta}$,

$$\left(\frac{z+z^{-1}}2\right)^2\left(\frac{z-z^{-1}}{2i}\right)^4=\frac1{32}\left(\frac{z^6+z^{-6}-z^2-z^{-2}+4-2z^4-2z^{-4}}2\right).$$

For convenience, multiply by $64z^6$, set $t:=z^2$ and rewrite

$$(t^2-1)^2(t-1)^2=t^6+1-t^4-t^2+4t^3-2t^5-2t.$$

By direct expansion of the LHS you get the equality.

$\endgroup$
0
$\begingroup$

$$\cos^2\theta\sin^4\theta=\frac{1}{32}(\cos6\theta-\cos2\theta+2-2\cos4\theta)$$

Use $\sin^2 \theta + \cos^2 \theta = 1$ to put everything in terms of cosine:

$$\begin{align} \cos^2\theta (1 - \cos^2 \theta )^2 &= \frac{1}{32}(\cos6\theta-\cos2\theta+2-2\cos4\theta) \\ \cos^2\theta (1 - 2\cos^2\theta + \cos^4 \theta ) &= \frac{1}{32}(\cos6\theta-\cos2\theta+2-2\cos4\theta) \\ \cos^2\theta - 2\cos^4\theta + \cos^6 \theta &= \frac{1}{32}(\cos6\theta-\cos2\theta+2-2\cos4\theta) \end{align}$$

Use $\cos(\theta) = \frac{e^{i\theta} + e^{-i\theta}}{2}$ :

$$ \left(\frac{e^{i\theta} + e^{-i\theta}}{2}\right)^2 - 2\left(\frac{e^{i\theta} + e^{-i\theta}}{2}\right)^4 + \left(\frac{e^{i\theta} + e^{-i\theta}}{2}\right)^6 = \frac{1}{32}\left( \left(\frac{e^{6i\theta} + e^{-6i\theta}}{2}\right) - \left(\frac{e^{2i\theta} + e^{-2i\theta}}{2}\right) + 2 - 2\left(\frac{e^{4i\theta} + e^{-4i\theta}}{2}\right) \right)$$

Multiply it out:

$$ \begin{align} \frac{1}{4} & \left( e^{2i\theta} + 2 + e^{-2i\theta} \right) \\ - \frac{1}{8} & \left( e^{4i\theta} + 4 e^{2i\theta} + 6 + 4 e^{-2i \theta} + e^{-4i\theta} \right) \\ + \frac{1}{64} & \left( e^{6i\theta} + 6 e^{4i\theta} + 15 e^{2i\theta} + 20 + 15 e^{-2i\theta} + 6 e^{-4i\theta} + e^{6i\theta} \right) \\ = \frac{1}{32} & \left( \frac{ \left( e^{6i\theta} + e^{-6i\theta} \right) - \left( e^{2i\theta} + e^{-2i\theta} \right) + 4 - \left(2e^{4i\theta} + 2e^{-4i\theta}\right) }{2} \right) \end{align}$$

Combine like terms:

$$\begin{array} {c} 16 e^{2i\theta} + 32 + 16 e^{-2i\theta} \\ -~ 8 e^{4i\theta} - 32 e^{2i\theta} - 48 - 32 e^{-2i \theta} - 8 e^{-4i\theta} \\ +~ e^{6i\theta} + 6 e^{4i\theta} + 15 e^{2i\theta} + 20 + 15 e^{-2i\theta} + 6 e^{-4i\theta} + e^{6i\theta} \\ =~ e^{6i\theta} - 2e^{4i\theta} - e^{2i\theta} + 4 - e^{-2i\theta} - 2e^{-4i\theta} + e^{-6i\theta} \end{array}$$

So

$$\begin{array} {c} e^{6i\theta} + 6 e^{4i\theta} + 15 e^{2i\theta} + 20 + 15 e^{-2i\theta} + 6 e^{-4i\theta} + e^{6i\theta} \\ = \\ e^{6i\theta} + 6 e^{4i\theta} + 15 e^{2i\theta} + 20 + 15 e^{-2i\theta} + 6 e^{-4i\theta} + e^{6i\theta} \end{array}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.