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The question is to express $\cos 4x$ in terms of only $\sin 2x$ and hence solve the trigonometric equation with the restriction of $\theta \in (0^{\circ}, 135^{\circ})$

$$\frac{\cos 5 \theta}{\sin \theta} + \frac{\sin 5 \theta}{\cos \theta} = 2.$$

I have come up to this point:

$$\cos 4x = \cos (2x + 2x) = \cos 2x \cos 2x - \sin 2x \sin 2x = (\cos(2x))^2 - (\sin(2x))^2.$$

Now I only need to express $\cos 2x$ in terms of $\sin 2x$, but for all quadrants. Regarding the trigonometric equation, I have done these steps:

\begin{align} \frac{\cos(5x)\cos(x) + \sin(5x)\sin(x)}{\sin(x)\cos(x)}=2 &\implies \cos(5x-x)=2\sin(x)\cos(x) \\ &\implies \cos(4x)=\sin(2x). \end{align}

So I believe that when I express $\cos(4x)$ in terms of $\sin 2x$ I continue with solving the equation.

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    $\begingroup$ $$\sin^2{(2x)}+\cos^2{(2x)}=1$$ $\endgroup$ – Peter Foreman Aug 31 '19 at 14:43
  • $\begingroup$ $sin(x) = cos(x)$ if and only if $x = 45°$ or $x = 135°$ $\endgroup$ – ZAF Aug 31 '19 at 14:46
  • $\begingroup$ @ZAF $4x \ne 2x$, so your result does not apply. $\endgroup$ – Xander Henderson♦ Aug 31 '19 at 14:50
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    $\begingroup$ $(\cos 2x)^2 = 1 - (\sin 2x)^2$ And that's true for all quadrants. So you have (if you've done your trig correctly; I did not double check) that $1-2(\sin 2x)^2 = \sin 2x$ which is simply a quadratic equation. $\endgroup$ – fleablood Aug 31 '19 at 15:02
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Rearrange

$$\dfrac{\cos 5 \theta}{\sin \theta} + \dfrac{\sin 5 \theta}{\cos \theta} = 2$$

to get

$$\cos(4\theta) = \sin(2\theta)$$

Then, use $\cos 2x=1-2\sin^2 x$

$$1-2\sin^2(2\theta)= \sin(2\theta)$$

$$\sin(2\theta) = -1, \space \sin(2\theta)=1/2$$

$$\theta = 15^\circ,\space \theta=75^\circ$$

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  • $\begingroup$ First case: $2 \theta = \arcsin (-1) = - \dfrac{\pi}{2} $... Shouldn't $\theta$ in that case be $- \dfrac{\pi}{4}.$ ? $\endgroup$ – Aleksandra Asanin Aug 31 '19 at 15:06
  • $\begingroup$ Your $\theta$ is restricted to $\theta \in (0^{\circ}, 135^{\circ})$. So, $-45^\circ$ is excluded $\endgroup$ – Quanto Aug 31 '19 at 15:08
  • $\begingroup$ Now I understand, both solutions are for the second case. Thanks. $\endgroup$ – Aleksandra Asanin Aug 31 '19 at 15:13
  • $\begingroup$ What about $2\theta = 1.5 \pi $? $\endgroup$ – Jaroslaw Matlak Aug 31 '19 at 15:55
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    $\begingroup$ You get $135^\circ$, which is outside the domain specified. $\endgroup$ – Quanto Aug 31 '19 at 15:57
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Hint $$\cos 4x = \cos^2 2x - \sin^2 2x = 1-\sin^2 2x - \sin^2 2x = 1-2\sin^2 2x$$

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Since Cos4X = Sin 2X ,$$ Cos 4X = Cos (90 - 2X) $$ 4X = 2nπ $\pm$(90-2X) $$ Hence \,X = 15 , 75 $$

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Multiplying by $$\sin(x)\cos(x)$$ we get $$\cos(5x)\cos(x)+\sin(5x)\sin(x)=2\sin(x)\cos(x)$$ and this is $$\cos(5x-x)=2\sin(x)\cos(x)$$ $$\cos(4x)=\sin(2x)$$ $$\cos^2(2x)-\sin^2(2x)=\sin(2x)$$

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