Q:

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

Accepted Solution

A:
The main identity you need is the double angle one for cosine:[tex]\cos^2x=\dfrac{1+\cos2x}2[/tex]We get[tex]\cos^6x=(\cos^2x)^3=\left(\dfrac{1+\cos2x}2\right)^3=\dfrac{(1+\cos2x)^3}8[/tex]Expand the numerator to apply the identity again:[tex]\cos^6x=\dfrac{1+3\cos2x+3\cos^22x+\cos^32x}8[/tex][tex]\cos^6x=\dfrac{1+3\cos2x+3\left(\frac{1+\cos2(2x)}2\right)+\cos2x\left(\frac{1+\cos2(2x)}2\right)}8[/tex][tex]\cos^6x=\dfrac{1+3\cos2x+\frac32+\frac32\cos4x+\frac12\cos2x(1+\cos4x)}8[/tex][tex]\cos^6x=\dfrac5{16}+\dfrac7{16}\cos2x+\dfrac3{16}\cos4x+\dfrac1{16}\cos2x\cos4x[/tex]Finally, make use of the product identity for cosine:[tex]\cos2x\cos4x=\dfrac{\cos6x+\cos2x}2[/tex]so that ultimately,[tex]\cos^6x=\dfrac5{16}+\dfrac7{16}\cos2x+\dfrac3{16}\cos4x+\dfrac1{32}\cos2x+\dfrac1{32}\cos6x[/tex][tex]\cos^6x=\dfrac5{16}+\dfrac{15}{32}\cos2x+\dfrac3{16}\cos4x+\dfrac1{32}\cos6x[/tex]