Using Sum-to-Product Formulas Solve the equation by first using a Sum-to-Product Formula. sin⁡θ+sin⁡3θ=0

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Using Sum-to-Product Formulas Solve the equation by first using a Sum-to-Product Formula.  sin⁡θ+sin⁡3θ=0

hajavaF · Accepted Answer

Approach:  The range of the trigonometric functions of sin⁡θ is lie between [−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,  sin⁡u+sin⁡v=2sin⁡u+v2cos⁡u+v2  Cosine function has period 2π, thus find the solution in any interval of length 2π. Sine function is positive in first and second quadrant.  Calculation:  Consider the equation.  sin⁡θ+sin⁡3θ=0  Use Sum-to-Product formulas in the above equation,  sin⁡θ+sin⁡3θ=0  2sin⁡θ+3θ2cos⁡θ−3θ2=0  2sin⁡2θcos⁡θ=0  Use the zero product property,  sin⁡2θ=0…(1)  cos⁡θ=0…(2)  Consider equation (1).  sin⁡2θ=0  Taking sine inverse both sides,  sin−1sin⁡2θ=sin−1(0)  2θ=sin−1(0)  2θ=0,π  The solution of the equation is obtained by adding in the integer multiples of π,  2θ=kπ  θ=kπ2  Consider equation (2).  cos⁡θ=0  Taking cos−1 both sides,  cos−1cos⁡θ=cos−1(0)  θ=cos−1(0)  θ=π2  The solution of the equation is obtained by adding in the integer multiples of π,  θ=π2+kπ  The compact general solution is θ=πk2.  Therefore, the solution of the trigonometry equation sin⁡θ+sin⁡3θ=0 is θ=πk2

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

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