# Solve the equation by first using a Sum-to-Product Formula. \cos 5 \theta - \cos 7 \theta = 0 \cos 5 \theta - \cos 7 \theta = 0

Using Sum-to-Product Formulas Solve the equation by first using a Sum-to-Product Formula.
$$\displaystyle{\cos{{5}}}\theta-{\cos{{7}}}\theta={0}{\cos{{5}}}\theta-{\cos{{7}}}\theta={0}$$

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cyhuddwyr9

Approach:
The range of the trigonometric functions of $$\displaystyle{\sin{\theta}}$$ is lie between $$\displaystyle{\left[-{1},{1}\right]}{\left[-{1},{1}\right]}$$. No solution exists beyond this range.
Simplify the equation.
Obtain the factors of the equation.
The sum-to-product formulas for cosine is,
$$\displaystyle{\cos{{u}}}+{\cos{{v}}}=-{2}{\sin{{\frac{{{u}+{v}}}{{{2}}}}}}{\sin{{\frac{{{u}+{v}}}{{{2}}}}}}{\cos{{u}}}-{\cos{{v}}}=-{2}{\sin{{u}}}+{v}{2}{\sin{{u}}}-{v}{2}$$
Cosine function has period $$\displaystyle{2}\pi{2}\pi$$, thus find the solution in any interval of length $$\displaystyle{2}\pi{2}\pi$$. Sine function is positive in first and second quadrant.
Calculation:
Consider the equation.
$$\displaystyle{\cos{{5}}}\theta-{\cos{{7}}}\theta={0}{\cos{{5}}}\theta-{\cos{{7}}}\theta={0}$$
Use Sum-to-Product formulas in the above equation,
$$\displaystyle{\cos{{5}}}\theta-{\cos{{7}}}\theta={0}$$
$$\displaystyle-{2}{\sin{{\frac{{{5}\theta+{7}\theta}}{{{2}}}}}}{\sin{{\frac{{{5}\theta+{7}\theta}}{{{2}}}}}}={0}{\cos{{5}}}\theta-{\cos{{7}}}\theta={0}-{2}{\sin{{5}}}\theta+{7}\theta{2}{\sin{{5}}}\theta-{7}\theta{2}={0}-{2}{\sin{{6}}}\theta{\sin{{\left(-\theta\right)}}}={0}$$
$$\displaystyle-{2}{\sin{{6}}}\theta{\sin{{\left(-\theta\right)}}}={0}$$
Use the zero product property,
$$\displaystyle{\sin{{6}}}\theta={0}{\sin{{6}}}\theta={0}\ldots{\left({1}\right)}\ldots{\left({1}\right)}$$
$$\displaystyle{\sin{\theta}}={0}{\sin{\theta}}={0}\ldots{\left({2}\right)}\ldots{\left({2}\right)}$$
Consider equation (1).
$$\displaystyle{\sin{{6}}}\theta={0}{\sin{{6}}}\theta={0}$$
Taking sine inverse both sides,
$$\sin^{-1} \sin 6 \theta = \sin^{-1} (0)$$
$$\displaystyle{6}\theta={{\sin}^{{-{1}}}{\left({0}\right)}}{\sin{-}}{1}{\sin{{6}}}\theta={\sin{-}}{1}{\left({0}\right)}{6}\theta={\sin{-}}{1}{\left({0}\right)}={0},\pi$$
$$\displaystyle={0},\pi$$
The solution of the equation is obtained by adding in the integer multiples of $$\displaystyle\pi$$,
$$\displaystyle{6}\theta={k}\pi$$
$$\displaystyle\theta={\frac{{{k}\pi}}{{{6}}}}{6}\theta={k}\pi\theta={k}\pi{6}$$
Consider equation (2).
$$\displaystyle{\sin{\theta}}={0}{\sin{\theta}}={0}$$
Taking $$\displaystyle{\sin{}}$$ inverse both sides,
$$\displaystyle{{\sin}^{{-{1}}}{\sin{\theta}}}={{\sin}^{{-{1}}}{\left({0}\right)}}$$
$$\displaystyle\theta={{\sin}^{{-{1}}}{\left({0}\right)}}{\sin{-}}{1}{\sin{\theta}}={\sin{-}}{1}{\left({0}\right)}\theta={\sin{-}}{1}{\left({0}\right)}\theta=\pi$$
$$\displaystyle\theta=\pi$$
The solution of the equation is obtained by adding in the integer multiples of $$\displaystyle\pi\pi$$,
$$\displaystyle\theta={k}\pi\theta={k}\pi$$
Therefore, the solutions of the trigonometry equation $$\displaystyle{\cos{{5}}}\theta-{\cos{{7}}}\theta={0}{\cos{{5}}}\theta-{\cos{{7}}}\theta={0}$$ is $$\displaystyle\theta=\pi{k},{\frac{{\pi{k}}}{{{6}}}}\theta=\pi{k},\pi{k}{6}$$.