Question

If $cos2\theta =\cos 3\theta$ and $\theta$ is an acute angle, then $\sin \theta$ is equal to

• A. $\frac{\sqrt{10+2\sqrt{5}}}{4}$
  
• B. $\frac{\sqrt{10-2\sqrt{5}}}{4}$
• C. $0$
• D. None of these

Answer 1

Answer: (A)

$\frac{\sqrt{10+2\sqrt{5}}}{4}$

Given

$\cos 2\theta =\cos 3\theta$

$\cos 2\theta =2{\cos }^2\theta -1\cdots \left(1\right)$

$\cos 3\theta =4{\cos }^3\theta -3\cos \theta \cdots \left(2\right)$

According to Question $\left(1\right)=\left(2\right)$

Equating $\left(1\right)$ and $\left(2\right)$

$⟹2{\cos }^2\theta -1=4{\cos }^3\theta -3\cos \theta$

$⟹4{\cos }^3\theta -2{\cos }^2\theta -3\cos \theta +1=0$

Considering the equation as $4x^3-2x^2-3x+1=0$

Using Synthetic Division

$\begin{array}{c}1\\ \\ \end{array}|\begin{array}{cccc}4& -2& -3& 1\\ 0& 4& 2& -1\\ ----& ----& ----& ----\\ 4& 2& -1& 0\end{array}$

$⟹$ Root of equation is $1$

$⟹\cos x=1$

The equation can be rewritten as

$(\cos x-1)(4{\cos }^2\theta +2\cos \theta -1)=0$

The next root is given by

$4{\cos }^2\theta +2\cos \theta -1=0$

Conidering the quadratic eqaution with $a=4,b=2,c=-1$

By Quadratic formula

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$⟹\cos \theta =\frac{-2\pm \sqrt{(-2{)}^2-4(4\left)\right(-1)}}{2\left(4\right)}$

$⟹\cos \theta =\frac{-2\pm \sqrt{4+16}}{8}$

$⟹\cos \theta =\frac{-2\pm \sqrt{20}}{8}$

$⟹\cos \theta =\frac{-2\pm 2\sqrt{5}}{8}$

$⟹\cos \theta =\frac{-1\pm \sqrt{5}}{4}$

Squaring on both sides.

$⟹{\cos }^2\theta ={\left(\frac{-1\pm \sqrt{5}}{4}\right)}^2$

$⟹{\cos }^2\theta =\frac{5+1-2\sqrt{5}}{16}$

$⟹{\cos }^2\theta =\frac{6-2\sqrt{5}}{16}$

$⟹{\sin }^2\theta =1-\frac{6-2\sqrt{5}}{16}$

$({\sin }^2\theta =1-{\cos }^2\theta )$

$⟹{\sin }^2\theta =\frac{16-6+2\sqrt{5}}{16}$

$⟹{\sin }^2\theta =\frac{10+2\sqrt{5}}{16}$

$⟹\sin \theta =\frac{\sqrt{10+2\sqrt{5}}}{4}$