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

Question

asked Jul 3, 2020 101k views

1 Answer

Thomas Schneiter · Answer 1 · 2020-07-08T21:00:11+0000

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

$\cos^2x=\frac{1+\cos2x}2$

We get

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

Expand the numerator to apply the identity again:

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

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

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

$\cos^6x=\frac5{16}+\frac7{16}\cos2x+\frac3{16}\cos4x+\frac1{16}\cos2x\cos4x$

Finally, make use of the product identity for cosine:

$\cos2x\cos4x=\frac{\cos6x+\cos2x}2$

so that ultimately,

$\cos^6x=\frac5{16}+\frac7{16}\cos2x+\frac3{16}\cos4x+\frac1{32}\cos2x+\frac1{32}\cos6x$

$\cos^6x=\frac5{16}+(15)/(32)\cos2x+\frac3{16}\cos4x+\frac1{32}\cos6x$