Question

If $\sin \left(\theta \right)+\sin \left(2\theta \right)+\sin \left(3\theta \right)=\sin \left(\alpha \right)$ and $\cos \left(\theta \right)+\cos \left(2\theta \right)+\cos \left(3\theta \right)=\cos \left(\alpha \right)$, then $\theta$ is equal to

• A. $\frac{\alpha }{2}$
• B. $\alpha$
• C. $2\alpha$
• D. $\frac{\alpha }{6}$

Answer 1

The correct option is A

$\frac{\alpha }{2}$

Explanation for the correct option

Step 1: Simplify the equation $\mathbf{sin}\left(\theta \right)\mathbf{+}\mathbf{sin}\left(2\theta \right)\mathbf{+}\mathbf{sin}\left(3\theta \right)\mathbf{=}\mathbf{sin}\left(\alpha \right)$

In the given equation we can apply the formula,

$\mathbf{sin}\left(A\right)\mathbf{+}\mathbf{sin}\left(B\right)\mathbf{=}\mathbf{2}\mathbf{sin}\left(\frac{A+B}{2}\right)\mathbf{cos}\left(\frac{A-B}{2}\right)$

Let $A=\theta \text{and}B=3\theta$, then we get

$\begin{array}{rcl}\therefore \sin \left(\alpha \right)& =& 2\sin \left(\frac{3\theta +\theta }{2}\right)\cos \left(\frac{3\theta -\theta }{2}\right)+\sin \left(2\theta \right)\\ & =& 2\sin \left(\frac{4\theta }{2}\right)\cos \left(\frac{2\theta }{2}\right)+\sin \left(2\theta \right)\\ & =& 2\sin \left(2\theta \right)\cos \left(\theta \right)+\sin \left(2\theta \right)\\ & =& \sin \left(2\theta \right)\left(2\cos \left(\theta \right)+1\right)\end{array}$

Here, we created the first equation,

$\sin \left(\alpha \right)=\sin \begin{array}{l}\left(2\theta \right)\end{array}\begin{array}{l}\left(2\cos \left(\theta \right)+1\right)\end{array}\dots \left(1\right)$

Step 2: Step 1: Simplify the equation $\mathbf{cos}\left(\theta \right)\mathbf{+}\mathbf{cos}\left(2\theta \right)\mathbf{+}\mathbf{cos}\left(3\theta \right)\mathbf{=}\mathbf{cos}\left(\alpha \right)$,

To simplify this equation, we need to use the formula given as

$\mathbf{cos}\left(A\right)\mathbf{+}\mathbf{cos}\left(B\right)\mathbf{=}\mathbf{2}\mathbf{cos}\left(\frac{A-B}{2}\right)\mathbf{cos}\left(\frac{A+B}{2}\right)$

Let $A=\theta \text{and}B=3\theta$, then we get

$\begin{array}{rcl}\therefore \cos \left(\alpha \right)& =& 2\cos \left(\frac{3\theta -\theta }{2}\right)\cos \left(\frac{3\theta +\theta }{2}\right)+\cos \left(2\theta \right)\\ & =& 2\cos \left(\frac{2\theta }{2}\right)\cos \left(\frac{4\theta }{2}\right)+\cos \left(2\theta \right)\\ & =& 2\cos \left(\theta \right)\cos \left(2\theta \right)+\cos \left(2\theta \right)\\ & =& \cos \left(2\theta \right)\left(2\cos \left(\theta \right)+1\right)\end{array}$

The equation created here is,

$\cos \left(\alpha \right)=\cos \left(2\theta \right)(\cos \left(\theta \right)+1)\dots \left(2\right)$

Step 3: Determination of $\mathit{\theta }$

To find $\theta$, divide this equation $\left(1\right)$ by $\left(2\right)$, we get

$\begin{array}{rcl}\therefore \frac{\sin \left(\alpha \right)}{\cos \left(\alpha \right)}& =& \frac{\sin \left(2\theta \right)\left(2\cos \left(\theta \right)+1\right)}{\cos \left(2\theta \right)\left(2\cos \left(\theta \right)+1\right)}\\ & \Rightarrow & \tan \left(\alpha \right)=\frac{\sin \left(2\theta \right)}{\cos \left(2\theta \right)}\\ & =& \tan \left(2\theta \right)\end{array}$

From this, we have

$\begin{array}{rcl}\therefore 2\theta & =& \alpha \\ & \Rightarrow & \theta =\frac{\alpha }{2}\end{array}$

Hence, the correct option is (A).