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If $$u = (1+\cos\theta)(1+\cos2\theta) - \sin\theta \sin 2\theta \qquad v = \sin\theta (1+\cos2\theta) + \sin2\theta(1+\cos\theta)$$ then show that $$u^2 + v^2 = 4(1+\cos\theta)(1+\cos2\theta)$$

I have simplified the values of $u$ and $v$ and got:

$$u = 2\cos\theta (1+\cos2\theta) \qquad v=\sin2\theta(1+\cos\theta)$$

Then I tried to square both $u$ and $v$ individually before doing summation. Still could not prove the statement.

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Assigning $a$, $b$, $c$, $d$ as $$a := 1+\cos\theta \qquad b := 1+\cos2\theta \qquad c := \sin\theta \qquad d := \sin 2\theta$$

we can write $$u = ab - c d \qquad v = bc + ad$$ When squaring and adding, the $-2abcd$ will cancel with $2abcd$, leaving $$u^2 + v^2 = a^2 b^2 + c^2 d^2 + b^2 c^2 + a^2 d^2 = \left(a^2+c^2\right)\left(b^2+d^2\right)$$

Now, $$\begin{align} a^2 + c^2 &= \left(1 + 2 \cos\theta + \cos^2\theta\right) + \sin^2\theta = 2\left(1 + \cos\theta\right) \\[4pt] b^2 + d^2 &= \left(1 + 2 \cos 2\theta + \cos^22\theta\right) + \sin^22\theta = 2\left(1 + \cos 2\theta\right) \end{align}$$ and the result follows. $\square$


Note. The sum reduces even further to $$16\cos^2\frac{\theta}{2}\cos^2\theta$$

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    $\begingroup$ Thanks @Blue.. It's a really a good approach to solve it. $\endgroup$ – Desi boys Jun 23 at 10:54
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Let $f(\theta)=1+\cos\theta+i\sin\theta$.

Then $|f(\theta)|^2=(1+\cos\theta)^2+\sin^2\theta=2(1+\cos\theta)$.

Note that $f(\theta)f(2\theta)=u+iv$.

So, $u^2+v^2=|f(\theta)f(2\theta)|^2=|f(\theta)|^2|f(2\theta)|^2=4(1+\cos\theta)(1+\cos2\theta)$.

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$$u^2+v^2=(1+\cos\theta+\cos2\theta+\cos3\theta)^2+(\sin\theta+\sin2\theta+\sin3\theta)^2=$$ $$=4+2(\cos\theta+\cos2\theta+\cos3\theta)+2(\cos\theta\cos2\theta+\sin\theta\sin2\theta)+$$ $$+2(\cos\theta\cos3\theta+\sin\theta\sin3\theta)+2(\cos3\theta\cos2\theta+\sin3\theta\sin2\theta)=$$ $$=4+2(\cos\theta+\cos2\theta+\cos3\theta)+2(\cos\theta+\cos2\theta+\cos\theta)=$$ $$=4+6\cos\theta+4\cos2\theta+2\cos3\theta=4+4\cos\theta+4\cos2\theta+4\cos\theta\cos2\theta=$$ $$=4(1+\cos\theta)(1+\cos2\theta).$$

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Simplifying $u$ we get $$u=2\,\sin \left( x \right) \cos \left( x \right) \left( 2\,\cos \left( x \right) +1 \right) $$ and $v$ $$v=2\,\sin \left( x \right) \cos \left( x \right) \left( 2\,\cos \left( x \right) +1 \right) $$ so $$u^2+v^2=8\, \left( \cos \left( x \right) \right) ^{2} \left( 1+\cos \left( x \right) \right) $$ and the right-hand side:$$8\, \left( \cos \left( x \right) \right) ^{2} \left( 1+\cos \left( x \right) \right) $$ which is clearly the same.

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  • $\begingroup$ I think you have $u$ and $v$ switched. Please check. $\endgroup$ – Somos Jun 23 at 14:41
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You simplified the values incorrectly. See below.

Alternatively: $$\begin{align}\color{blue}u&=(1+\cos\theta)(1+\cos2\theta) - \sin\theta \sin 2\theta=\\ &=1+\cos \theta +\cos 2\theta +\cos \theta \cos 2\theta-\sin\theta \sin2\theta=\\ &=1+\cos\theta+\cos2\theta+\cos3\theta=\\ &=1+\cos2\theta+2\cos\theta\cos2\theta= \ \ \ \ [\color{red}{\ne 2\cos\theta (1+\cos2\theta)}]\\ &=\color{blue}{1+\cos2\theta(1+2\cos\theta)};\\ \color{green}v&=\sin\theta (1+\cos2\theta) + \sin2\theta(1+\cos\theta)=\\ &=\sin\theta+\sin2\theta+\sin\theta\cos2\theta+\cos\theta\sin2\theta=\\ &=\sin\theta+\sin2\theta+\sin3\theta=\\ &=\sin2\theta+2\sin2\theta\cos\theta=\\ &=\color{green}{\sin2\theta(1+2\cos\theta)}; \ \ \ \ \ \ \ \ [\color{red}{\ne \sin2\theta(1+\cos\theta)}]\\ \color{blue}{u^2}+\color{green}{v^2}&=\color{blue}{1+2\cos2\theta(1+2\cos\theta)+\cos^22\theta(1+2\cos\theta)^2}+\color{green}{\sin^22\theta(1+2\cos\theta)^2}=\\ &=1+2\cos2\theta(1+2\cos\theta)+1\cdot(1+2\cos\theta)^2=\\ &=1+2\cos2\theta+4\cos\theta\cos2\theta+1+4\cos\theta+4\cos^2\theta=\\ &=2+2\cos2\theta+4\cos\theta\cos2\theta+4\cos\theta+2(\cos2\theta+1)=\\ &= 4(1+\cos\theta)(1+\cos2\theta).\end{align}$$

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