Using Sum-to-Product Formulas Solve the equation by first using a Sum-to-Product Formula. cos⁡5θ−cos⁡7θ=0cos⁡5θ−cos⁡7θ=0

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Using Sum-to-Product Formulas Solve the equation by first using a Sum-to-Product Formula.  cos⁡5θ−cos⁡7θ=0cos⁡5θ−cos⁡7θ=0

cyhuddwyr9 · Accepted Answer

Approach: The range of the trigonometric functions of sin⁡θ 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, cos⁡u+cos⁡v=−2sin⁡u+v2sin⁡u+v2cos⁡u−cos⁡v=−2sin⁡u+v2sin⁡u−v2 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. cos⁡5θ−cos⁡7θ=0cos⁡5θ−cos⁡7θ=0 Use Sum-to-Product formulas in the above equation, cos⁡5θ−cos⁡7θ=0 −2sin⁡5θ+7θ2sin⁡5θ+7θ2=0cos⁡5θ−cos⁡7θ=0−2sin⁡5θ+7θ2sin⁡5θ−7θ2=0−2sin⁡6θsin⁡(−θ)=0 −2sin⁡6θsin⁡(−θ)=0 Use the zero product property, sin⁡6θ=0sin⁡6θ=0…(1)…(1) sin⁡θ=0sin⁡θ=0…(2)…(2) Consider equation (1). sin⁡6θ=0sin⁡6θ=0 Taking sine inverse both sides, sin−1⁡sin⁡6θ=sin−1⁡(0) 6θ=sin−1(0)sin−1sin⁡6θ=sin−1(0)6θ=sin−1(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). sin⁡θ=0sin⁡θ=0 Taking sin⁡ inverse both sides, sin−1sin⁡θ=sin−1(0) θ=sin−1(0)sin−1sin⁡θ=sin−1(0)θ=sin−1(0)θ=π

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

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