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Math  /  Trigonometry

QuestionUse the power reducing formulas to rewrite sin⁡4x\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. sin⁡4x=\sin ^{4} x= □\square □\square □\square x+x+ □\square □\square x sin⁡4x=\sin ^{4} x= □\square ++ □\square □cos⁡\square \cos □\square ～x+x+ □\square cos⁡\cos □\square 7x7 x sin⁡4x=\sin ^{4} x= □\square - □\square cos⁡\cos □\square I xx sin⁡4x=\sin ^{4} x= □\square ++ □\square cos⁡\cos □\square x

Studdy Solution

STEP 1

1. We need to express sin⁡4x\sin^4 x using the power-reducing formulas.
2. The goal is to rewrite the expression in terms of the first power of cosine.
3. The power-reducing formulas for sine and cosine will be used to simplify the expression.

STEP 2

1. Apply the power-reducing formula for sin⁡2x\sin^2 x.
2. Use the result from step 1 to express sin⁡4x\sin^4 x.
3. Simplify the expression to match one of the given forms.

STEP 3

Apply the power-reducing formula for sin⁡2x\sin^2 x:
sin⁡2x=1−cos⁡(2x)2\sin^2 x = \frac{1 - \cos(2x)}{2}

STEP 4

Express sin⁡4x\sin^4 x using sin⁡2x\sin^2 x:
sin⁡4x=(sin⁡2x)2=(1−cos⁡(2x)2)2\sin^4 x = (\sin^2 x)^2 = \left(\frac{1 - \cos(2x)}{2}\right)^2

STEP 5

Simplify the expression:
sin⁡4x=(1−cos⁡(2x)2)2=(1−cos⁡(2x))24\sin^4 x = \left(\frac{1 - \cos(2x)}{2}\right)^2 = \frac{(1 - \cos(2x))^2}{4}
Expand (1−cos⁡(2x))2(1 - \cos(2x))^2:
(1−cos⁡(2x))2=1−2cos⁡(2x)+cos⁡2(2x)(1 - \cos(2x))^2 = 1 - 2\cos(2x) + \cos^2(2x)
Use the power-reducing formula for cos⁡2(2x)\cos^2(2x):
cos⁡2(2x)=1+cos⁡(4x)2\cos^2(2x) = \frac{1 + \cos(4x)}{2}
Substitute back:
sin⁡4x=1−2cos⁡(2x)+1+cos⁡(4x)24\sin^4 x = \frac{1 - 2\cos(2x) + \frac{1 + \cos(4x)}{2}}{4}
Simplify:
sin⁡4x=2−4cos⁡(2x)+1+cos⁡(4x)8\sin^4 x = \frac{2 - 4\cos(2x) + 1 + \cos(4x)}{8}
sin⁡4x=38−12cos⁡(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:
sin⁡4x=38−12cos⁡(2x)+18cos⁡(4x)\sin^4 x = \frac{3}{8} - \frac{1}{2}\cos(2x) + \frac{1}{8}\cos(4x)

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