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# 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 toA.$\dfrac{\alpha }{2}$ B.$\alpha$ C.$2\alpha$ D. $\dfrac{\alpha }{6}$

Last updated date: 07th Sep 2024
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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. $\sin C + \sin D = 2\sin \left( {\dfrac{{C + D}}{2}} \right)\cos \left( {\dfrac{{C - D}}{2}} \right)$ and $\cos A + \cos B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)$.
The cosine function is even; therefore,
$\cos \left( { - q} \right){\text{ }} = {\text{ cos}}\left( q \right)$

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

Correct Answer: A.$\dfrac{\alpha }{2}$

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.
$\Rightarrow$ ${\sin ^2}\theta + {\cos ^2}\theta = 1$
$\Rightarrow$ ${\tan ^2}\theta + 1 = {\sec ^2}\theta$
$\Rightarrow$ $1 + {\cot ^2}\theta = {\csc ^2}\theta$
$\Rightarrow$ $\sin A + \sin B = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)$
$\Rightarrow$ $\sin A - \sin B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)$
$\Rightarrow$$\cos A - \cos B = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)$