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.\[\dfrac{\alpha }{2}\]
B.$\alpha $
C.$2\alpha $
D. \[\dfrac{\alpha }{6}\]

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. $\sin C + \sin D = 2\sin \left( {\dfrac{{C + D}}{2}} \right)\cos \left( {\dfrac{{C - D}}{2}} \right)$ 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,
\[\cos \left( { - q} \right){\text{ }} = {\text{ cos}}\left( q \right)\]

Complete answer:
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 $
\[ \Rightarrow \] $\sin A + \sin B = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)$
\[ \Rightarrow \] $\sin A - \sin B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)$
\[ \Rightarrow \]$\cos A - \cos B = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)$