Question

The value of $\text{θ}$ satisfying $\text{sin7θ=sin4θ-sinθ}$ and $0<\text{θ<}\frac{\mathrm{\pi }}{2}$ are

Answer 1

Step 1: Simplify the given equation and use $\text{sinC+sinD}$ formula

The given equation is

$\text{sin7θ=sin4θ-sinθ}$

$\text{⇒sin7θ+sinθ=sin4θ ......(1)}$

Now we use the formula $\mathbf{\text{sinC+sinD=2sin}}\left(\frac{\text{C+D}}{2}\right)\mathbf{\text{cos}}\left(\frac{C-D}{2}\right)$

So

$$\begin{gathered} \text{sin7θ+sinθ=2sin}\left(\frac{7\theta +\theta }{2}\right)\text{cos}\left(\frac{7\theta -\theta }{2}\right) \\ =2\text{sin}\left(\frac{8\theta }{2}\right)\text{cos}\left(\frac{6\theta }{2}\right) \\ =2\text{sin4θ cos3θ ......(2)} \end{gathered}$$

From (1) and (2), we get

$\text{2 sin4θ cos3θ= sin4θ}$

Step 2: Solve the above equation for $\theta$

$$\begin{gathered} \text{⇒sin4θ(2cos3θ-1)=0} \\ \Rightarrow \text{sin4θ=0 or 2cos3θ-1=0} \\ \Rightarrow 4\theta =\text{nπ where n∈Z and 2cos3θ=1} \end{gathered}$$

$$\begin{gathered} \text{θ=}\frac{n\mathrm{\pi }}{4}\text{or cos3θ=}\frac{1}{2} \\ \theta =0,\frac{\mathrm{\pi }}{4},\frac{2\mathrm{\pi }}{4}\cdots \cdots \cdots (\text{when we put n=0,±1,±2 ⋯⋯⋯) or 3θ=}\frac{\mathrm{\pi }}{3} \end{gathered}$$

But $0<\theta <\frac{\mathrm{\pi }}{2}$ so we take some values of $\theta$ which lies in the given interval so

$\text{θ=}\frac{\mathrm{\pi }}{4}\text{and θ=}\frac{\mathrm{\pi }}{9}$

Hence values of $\text{θ=}\frac{\mathrm{\pi }}{4}\text{,}\frac{\mathrm{\pi }}{9}$