Math  /  Trigonometry

QuestionUse the power reducing formulas to rewrite sin4x\sin ^{4} x in terms of the first power of cosine. Simplify your answer as much as possible: To indicate your answer, first choose one of the four forms below. Then fill in the blanks with the appropriace numbers. sin4x=\sin ^{4} x= \square \square \square x+x+ \square \square x sin4x=\sin ^{4} x= \square ++ \square cos\square \cos \squarex+x+ \square cos\cos \square 7x7 x sin4x=\sin ^{4} x= \square - \square cos\cos \square I xx sin4x=\sin ^{4} x= \square ++ \square cos\cos \square x

Studdy Solution
Simplify the expression:
sin4x=(1cos(2x)2)2=(1cos(2x))24\sin^4 x = \left(\frac{1 - \cos(2x)}{2}\right)^2 = \frac{(1 - \cos(2x))^2}{4}
Expand (1cos(2x))2(1 - \cos(2x))^2:
(1cos(2x))2=12cos(2x)+cos2(2x)(1 - \cos(2x))^2 = 1 - 2\cos(2x) + \cos^2(2x)
Use the power-reducing formula for cos2(2x)\cos^2(2x):
cos2(2x)=1+cos(4x)2\cos^2(2x) = \frac{1 + \cos(4x)}{2}
Substitute back:
sin4x=12cos(2x)+1+cos(4x)24\sin^4 x = \frac{1 - 2\cos(2x) + \frac{1 + \cos(4x)}{2}}{4}
Simplify:
sin4x=24cos(2x)+1+cos(4x)8\sin^4 x = \frac{2 - 4\cos(2x) + 1 + \cos(4x)}{8}
sin4x=3812cos(2x)+18cos(4x)\sin^4 x = \frac{3}{8} - \frac{1}{2}\cos(2x) + \frac{1}{8}\cos(4x)
The simplified expression matches the form:
sin4x=3812cos(2x)+18cos(4x)\sin^4 x = \frac{3}{8} - \frac{1}{2}\cos(2x) + \frac{1}{8}\cos(4x)

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