sin4x = (sin2x)(sin2x) = (1/4)(1-cos2x)(1-cos2x) = (1/4)(1 -2cos2x + cos2(2x))
= (1/4)(1 - 2cos2x + 1/2 + (1/2)cos4x) = (1/4)(3/2 - 2cos2x + cos4x/2)
So, 17sin4x = (17/4)(3/2 -2cos2x + cos4x/2)
Bern S.
asked 11/21/19Use the power-reducing formulas to rewrite the expression as an equivalent expression that does not contain powers of trigonometric functions greater than 1.
17sin4x
sin4x = (sin2x)(sin2x) = (1/4)(1-cos2x)(1-cos2x) = (1/4)(1 -2cos2x + cos2(2x))
= (1/4)(1 - 2cos2x + 1/2 + (1/2)cos4x) = (1/4)(3/2 - 2cos2x + cos4x/2)
So, 17sin4x = (17/4)(3/2 -2cos2x + cos4x/2)
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