# Solve the equation by first using a Sum-to-Product Formula. \sin \theta + \sin 3 \theta = 0

Using Sum-to-Product Formulas Solve the equation by first using a Sum-to-Product Formula.
$\mathrm{sin}\theta +\mathrm{sin}3\theta =0$
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hajavaF
Approach:
The range of the trigonometric functions of $\mathrm{sin}\theta$ is lie between $\left[-1,1\right]$. No solution exists beyond this range.
Simplify the equation.
Obtain the factors of the equation.
The sum-to-product formulas for cosine is,
$\mathrm{sin}u+\mathrm{sin}v=2\mathrm{sin}\frac{u+v}{2}\mathrm{cos}\frac{u+v}{2}$
Cosine function has period $2\pi$, thus find the solution in any interval of length $2\pi$. Sine function is positive in first and second quadrant.
Calculation:
Consider the equation.
$\mathrm{sin}\theta +\mathrm{sin}3\theta =0$
Use Sum-to-Product formulas in the above equation,
$\mathrm{sin}\theta +\mathrm{sin}3\theta =0$
$2\mathrm{sin}\frac{\theta +3\theta }{2}\mathrm{cos}\frac{\theta -3\theta }{2}=0$
$2\mathrm{sin}2\theta \mathrm{cos}\theta =0$
Use the zero product property,
$\mathrm{sin}2\theta =0\dots \left(1\right)$
$\mathrm{cos}\theta =0\dots \left(2\right)$
Consider equation (1).
$\mathrm{sin}2\theta =0$
Taking sine inverse both sides,
${\mathrm{sin}}^{-1}\mathrm{sin}2\theta ={\mathrm{sin}}^{-1}\left(0\right)$
$2\theta ={\mathrm{sin}}^{-1}\left(0\right)$
$2\theta =0,\pi$
The solution of the equation is obtained by adding in the integer multiples of $\pi$,
$2\theta =k\pi$
$\theta =\frac{k\pi }{2}$
Consider equation (2).
$\mathrm{cos}\theta =0$
Taking ${\mathrm{cos}}^{-1}$ both sides,
${\mathrm{cos}}^{-1}\mathrm{cos}\theta ={\mathrm{cos}}^{-1}\left(0\right)$
$\theta ={\mathrm{cos}}^{-1}\left(0\right)$
$\theta =\frac{\pi }{2}$
The solution of the equation is obtained by adding in the integer multiples of $\pi$,
$\theta =\frac{\pi }{2}+k\pi$
The compact general solution is $\theta =\frac{\pi k}{2}$.
Therefore, the solution of the trigonometry equation $\mathrm{sin}\theta +\mathrm{sin}3\theta =0$ is $\theta =\frac{\pi k}{2}$