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goldenfox [79]
3 years ago
11

Lowering powers write in terms of first power of cosine. Cos^6

Mathematics
1 answer:
yaroslaw [1]3 years ago
4 0

The main identity you need is the double angle one for cosine:

\cos^2x=\dfrac{1+\cos2x}2

We get

\cos^6x=(\cos^2x)^3=\left(\dfrac{1+\cos2x}2\right)^3=\dfrac{(1+\cos2x)^3}8

Expand the numerator to apply the identity again:

\cos^6x=\dfrac{1+3\cos2x+3\cos^22x+\cos^32x}8

\cos^6x=\dfrac{1+3\cos2x+3\left(\frac{1+\cos2(2x)}2\right)+\cos2x\left(\frac{1+\cos2(2x)}2\right)}8

\cos^6x=\dfrac{1+3\cos2x+\frac32+\frac32\cos4x+\frac12\cos2x(1+\cos4x)}8

\cos^6x=\dfrac5{16}+\dfrac7{16}\cos2x+\dfrac3{16}\cos4x+\dfrac1{16}\cos2x\cos4x

Finally, make use of the product identity for cosine:

\cos2x\cos4x=\dfrac{\cos6x+\cos2x}2

so that ultimately,

\cos^6x=\dfrac5{16}+\dfrac7{16}\cos2x+\dfrac3{16}\cos4x+\dfrac1{32}\cos2x+\dfrac1{32}\cos6x

\cos^6x=\dfrac5{16}+\dfrac{15}{32}\cos2x+\dfrac3{16}\cos4x+\dfrac1{32}\cos6x

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