Question

If $$\sin \theta +\sin 2\theta +\sin 3\theta =\sin \alpha$$ and $$\cos \theta +\cos 2\theta +\cos 3\theta =\cos \alpha$$, then $\theta$ is equal to
A.$${\displaystyle \frac{\alpha }{2}}$$
B.$\alpha$
C.$2\alpha$
D. $${\displaystyle \frac{\alpha }{6}}$$

Answer 1

Hint: These types of questions are easy to solve if the correct trigonometric formula is used. Start solving the question by dividing the two equations. Use Trigonometric formula i.e. $\sin C+\sin D=2\sin \left({\displaystyle \frac{C+D}{2}}\right)\cos \left({\displaystyle \frac{C-D}{2}}\right)$ and $\cos A+\cos B=2\cos \left({\displaystyle \frac{A+B}{2}}\right)\cos \left({\displaystyle \frac{A-B}{2}}\right)$.
The cosine function is even; therefore,
$$\cos \left(-q\right)=\text{ cos}\left(q\right)$$

Complete answer:
Consider the equations,
$$\Rightarrow$$ $$\sin \theta +\sin 2\theta +\sin 3\theta =\sin \alpha \dots \dots (1)$$
$$\Rightarrow$$ $$\cos \theta +\cos 2\theta +\cos 3\theta =\cos \alpha \dots \dots (2)$$
Divide each side of the equation $$(1)$$ by the equation $(2)$.
$$\Rightarrow$$ $${\displaystyle \frac{\sin \theta +\sin 2\theta +\sin 3\theta }{\cos \theta +\cos 2\theta +\cos 3\theta }}={\displaystyle \frac{\sin \alpha }{\cos \alpha }}$$
Apply the trigonometric formula, $${\displaystyle \frac{\sin \theta }{\cos \theta }}=\tan \theta$$. Here, $\theta =\alpha$ .
$$\Rightarrow$$ $${\displaystyle \frac{\sin \theta +\sin 2\theta +\sin 3\theta }{\cos \theta +\cos 2\theta +\cos 3\theta }}=\tan \alpha$$
Combine the terms whose sum is the even number so we can use the formulas,
$$\Rightarrow$$ $${\displaystyle \frac{(\sin \theta +\sin 3\theta )+\sin 2\theta }{(\cos \theta +\cos 3\theta )+\cos 2\theta }}=\tan \alpha$$
Apply the trigonometric formula ; $\sin A+\sin B=2\sin \left({\displaystyle \frac{A+B}{2}}\right)\cos \left({\displaystyle \frac{A-B}{2}}\right)$ and $\cos A+\cos B=2\cos \left({\displaystyle \frac{A+B}{2}}\right)\cos \left({\displaystyle \frac{A-B}{2}}\right)$
Here, $A=\theta$ and $B=3\theta$. Substitute values of $A$and $B$ into the formula.
$$\Rightarrow$$ $${\displaystyle \frac{2\sin \left({\displaystyle \frac{\theta +3\theta }{2}}\right)\cos \left({\displaystyle \frac{\theta -3\theta }{2}}\right)+\sin 2\theta }{2\cos \left({\displaystyle \frac{\theta +3\theta }{2}}\right)\cos \left({\displaystyle \frac{\theta -3\theta }{2}}\right)+\cos 2\theta }}=\tan \alpha$$
$$\Rightarrow$$ $${\displaystyle \frac{2\sin \left(2\theta \right)\cos \left(-\theta \right)+\sin 2\theta }{2\cos \left(2\theta \right)\cos \left(-\theta \right)+\cos 2\theta }}=\tan \alpha$$
Since cosine is an even function $\therefore$$$\cos \left(-\theta \right)=\cos \theta$$.
$$\Rightarrow {\displaystyle \frac{2\sin \left(2\theta \right)\cos \left(\theta \right)+\sin 2\theta }{2\cos \left(2\theta \right)\cos \left(\theta \right)+\cos 2\theta }}=\tan \alpha$$
$$\Rightarrow {\displaystyle \frac{\sin 2\theta (2\cos \left(\theta \right)+1)}{\cos 2\theta (2\cos \left(\theta \right)+1)}}=\tan \alpha$$
Cancel the common terms,
$$\Rightarrow {\displaystyle \frac{\sin 2\theta }{\cos 2\theta }}=\tan \alpha$$
Apply the trigonometric formula, $${\displaystyle \frac{\sin \theta }{\cos \theta }}=\tan \theta$$. Here, angle is $2\theta$ .
$$\Rightarrow \tan 2\theta =\tan \alpha$$
Comparing the angles we get,
$$\Rightarrow$$ $2\theta =\alpha$
$$\Rightarrow$$ $\theta ={\displaystyle \frac{\alpha }{2}}$

Correct Answer: A.$${\displaystyle \frac{\alpha }{2}}$$

Note:
The most important thing is to solve the question by remembering the trigonometric formula. The cosine function is the even function$$\cos \left(-\theta \right)=\cos \theta$$.
Always apply the correct trigonometric formula and think about the conversion from sine function to cosine function and vice versa i.e. $\cos ({90}^{\circ }-\theta )=\sin \theta$. Use the trigonometric formulas according to the question. Here are some useful formulas and identities are given below.
$$\Rightarrow$$ ${\sin }^2\theta +{\cos }^2\theta =1$
$$\Rightarrow$$ ${\tan }^2\theta +1={\sec }^2\theta$
$$\Rightarrow$$ $1+{\cot }^2\theta ={\csc }^2\theta$
$$\Rightarrow$$ $\sin A+\sin B=2\sin \left({\displaystyle \frac{A+B}{2}}\right)\cos \left({\displaystyle \frac{A-B}{2}}\right)$
$$\Rightarrow$$ $\sin A-\sin B=2\cos \left({\displaystyle \frac{A+B}{2}}\right)\sin \left({\displaystyle \frac{A-B}{2}}\right)$
$$\Rightarrow$$$\cos A-\cos B=2\sin \left({\displaystyle \frac{A+B}{2}}\right)\sin \left({\displaystyle \frac{A-B}{2}}\right)$