Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$$\frac{\cos{2x}-\sin{4x}-\cos{6x}}{\cos{2x}+\sin{4x}-\cos{6x}}=\tan{(x-15^{\circ})}cot{(x+15^{\circ})}$$ So, here's what I've done so far, but don't know what do do next: $$\frac{\cos{2x}-2\sin{2x}\cos{2x}-\cos{6x}}{\cos{2x}+2\sin{2x}\cos{2x}-\cos{6x}}=$$ $$\frac{\cos{2x}-4\sin{x}\cos{x}\cos{2x}-\cos{(4x+2x)}}{\cos{2x}+4\sin{x}\cos{x}\cos{2x}-\cos{(4x+2x)}}=$$ $$\frac{\cos{2x}-4\sin{x}\cos{x}\cos{2x}-\cos{4x}\cos{2x}+\sin{4x}\sin{2x}}{\cos{2x}+4\sin{x}\cos{x}\cos{2x}-\cos{4x}\cos{2x}+\sin{4x}\sin{2x}}=...$$ I have no idea what to do next. I have a big feeling that I'm going in a totally wrong direction. Is there anything I can do with the expression in the beginning?

share|improve this question
1  
How does the "write everything in terms of $\sin x$ and $\cos x$ approach work? –  DanielV Jun 17 at 21:37
    
Have you tried testing any values for $x$? Like $x=1$? –  DanielV Jun 17 at 22:18

4 Answers 4

Well, don't expand to $\sin(x)$ or $\cos(x)$...

Consider complex numbers...

Let $z = \exp( 2 \textbf{i} x )$

Then

$ \begin{eqnarray} 2 \cos(2x) &=& z + \bar{z}\\ 2 \sin(2x) &=& - \textbf{i} \Big( z - \bar{z} \Big)\\ 2 \sin(4x) &=& - \textbf{i} \Big( z^2 - \bar{z}^2 \Big)\\ &=& - \Big( z + \bar{z} \Big) \textbf{i} \Big( z - \bar{z} \Big)\\ 2 \cos(6x) &=& z^3 + \bar{z}^3\\ &=& \Big( z + \bar{z} \Big) \Big( z^2 - z\bar{z} + \bar{z}^2 \Big) \end{eqnarray} $

Then we can write

$ \begin{eqnarray} 2\cos(2x) + 2\sin(4x) - 2\cos(6x) &=& \Big(z + \bar{z}\Big) \left[ 1 - \textbf{i} \Big( z - \bar{z} \Big) - z^2 + z\bar{z} - \bar{z}^2 \right]\\ &=& \Big(z + \bar{z}\Big) \left[ - \textbf{i} \Big( z - \bar{z} \Big) - z^2 + 2 z\bar{z} - \bar{z}^2 \right]\\ &=& \Big(z + \bar{z}\Big) \left[ \textbf{i}^2 \Big( z - \bar{z} \Big)^2 - \textbf{i} \Big( z - \bar{z} \Big) \right]\\ &=& - \textbf{i} \Big( z - \bar{z} \Big) \Big(z + \bar{z}\Big) \left[ - \textbf{i} \Big( z - \bar{z} \Big) + 1 \right]\\ \end{eqnarray} $

and

$ \begin{eqnarray} 2\cos(2x) - 2\sin(4x) - 2\cos(6x) &=& - \textbf{i} \Big( z - \bar{z} \Big) \Big(z + \bar{z}\Big) \left[ - \textbf{i} \Big( z - \bar{z} \Big) - 1 \right]\\ \end{eqnarray} $

So

$ \begin{eqnarray} \frac{\cos(2x) - \sin(4x) - \cos(6x)}{\cos(2x) + \sin(4x) - \cos(6x)} &=& \frac{ - \textbf{i} \Big( z - \bar{z} \Big) - 1 }{ - \textbf{i} \Big( z - \bar{z} \Big) + 1 }\\ \end{eqnarray} $

But $\sin(2x) = - \textbf{i} \Big( z - \bar{z} \Big)$, whence

$ \begin{eqnarray} \frac{\cos(2x) - \sin(4x) - \cos(6x)}{\cos(2x) + \sin(4x) - \cos(6x)} &=& \frac{ 2 \sin(2x) - 1 }{ 2 \sin(2x) + 1 }\\ \end{eqnarray} $


We can write

$ \begin{eqnarray} \frac{ 2 \sin(2x) - 1 }{ 2 \sin(2x) + 1 } &=& \frac{ 2 \sin(2x) - 2 \sin(2y) }{ 2 \sin(2x) + 2 \sin(2y) }\\ \end{eqnarray} $

where $2 \sin(2y) = 1$, thus $\sin(2 y) = \frac{1}{2}$, so $2 y = 30^o$, whence $y = 15^o$.

But

$ \begin{eqnarray} \sin(2x) \pm 2 \sin(2y) &=& 2 \sin\big( x \pm y \big) \cos\big( x \mp y \big) \end{eqnarray} $

so

$ \begin{eqnarray} \frac{\cos(2x) - \sin(4x) - \cos(6x)}{\cos(2x) + \sin(4x) - \cos(6x)} &=& \frac{ \sin\big( x - 30^o \big) \cos\big( x + 30^o \big) }{ \sin\big( x - 30^o \big) \cos\big( x - 30^o \big) }\\ \end{eqnarray} $

So we obtain

$ \begin{eqnarray} \frac{\cos(2x) - \sin(4x) - \cos(6x)}{\cos(2x) + \sin(4x) - \cos(6x)} &=& \tan\big( x - 30^o \big) \cot\big( x + 30^o\big) \end{eqnarray} $

share|improve this answer

I'd start by using $2x=4x-2x$ and $6x=4x+2x$ to get to $$ \cos(2x)-\cos(6x)=2\sin(2x)\sin(4x) $$

share|improve this answer

This is an exercise in the sum-to-product and product-to-sum identities. On the LHS, use $$ \cos a - \cos b = -2\sin(\tfrac12(a+b))\sin(\tfrac12(a-b)) $$ on the top and bottom (as suggested by Lutzl). For the RHS, write $$ \tan(x-15^\circ)\cot(x+15^\circ) = \frac{\sin(x-15^\circ)\cos(x+15^\circ)}{\sin(x+15^\circ)\cos(x-15^\circ)} $$ and then use $$ \sin a \cos b = \tfrac12(\sin(a+b) + \sin(a-b)) $$ on the top and bottom.

share|improve this answer

Using Prosthaphaeresis Formula, $$\frac{\cos2x-\cos6x}{\sin4x}=\frac{2\sin4x\sin2x}{2\sin4x}=\frac{\sin2x}{\dfrac12}$$

$$\implies \frac{\cos2x-\cos6x}{\sin4x}=\frac{\sin2x}{\sin30^\circ}$$

Now apply Componendo and dividendo and again apply Prosthaphaeresis formulae $$\sin C\pm\sin D$$

share|improve this answer
    
@A6Tech, How about this? –  lab bhattacharjee Jun 19 at 3:18

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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