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

Aneeka Hunt 2021-08-04 Answered
Using Sum-to-Product Formulas Solve the equation by first using a Sum-to-Product Formula.
cos5θcos7θ=0cos5θcos7θ=0
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Expert Answer

cyhuddwyr9
Answered 2021-08-05 Author has 90 answers

Approach:
The range of the trigonometric functions of sinθ is lie between [1,1][1,1]. No solution exists beyond this range.
Simplify the equation.
Obtain the factors of the equation.
The sum-to-product formulas for cosine is,
cosu+cosv=2sinu+v2sinu+v2cosucosv=2sinu+v2sinuv2
Cosine function has period 2π2π, thus find the solution in any interval of length 2π2π. Sine function is positive in first and second quadrant.
Calculation:
Consider the equation.
cos5θcos7θ=0cos5θcos7θ=0
Use Sum-to-Product formulas in the above equation,
cos5θcos7θ=0
2sin5θ+7θ2sin5θ+7θ2=0cos5θcos7θ=02sin5θ+7θ2sin5θ7θ2=02sin6θsin(θ)=0
2sin6θsin(θ)=0
Use the zero product property,
sin6θ=0sin6θ=0(1)(1)
sinθ=0sinθ=0(2)(2)
Consider equation (1).
sin6θ=0sin6θ=0
Taking sine inverse both sides,
sin1sin6θ=sin1(0)
6θ=sin1(0)sin1sin6θ=sin1(0)6θ=sin1(0)=0,π
=0,π
The solution of the equation is obtained by adding in the integer multiples of π,
6θ=kπ
θ=kπ66θ=kπθ=kπ6
Consider equation (2).
sinθ=0sinθ=0
Taking sin inverse both sides,
sin1sinθ=sin1(0)
θ=sin1(0)sin1sinθ=sin1(0)θ=sin1(0)θ=π

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