How do you express #cos(4theta)# in terms of #cos(2theta)#?

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Answer 1 · 2015-06-10T23:05:38

#cos(4theta) = 2(cos(2theta))^2-1#

Start by replacing #4theta# with #2theta+2theta#

#cos(4theta) = cos(2theta+2theta)#

Knowing that #cos(a+b) = cos(a)cos(b)-sin(a)sin(b)# then

#cos(2theta+2theta) = (cos(2theta))^2-(sin(2theta))^2#

Knowing that #(cos(x))^2+(sin(x))^2 = 1# then

#(sin(x))^2 = 1-(cos(x))^2#

#rarr cos(4theta) = (cos(2theta))^2-(1-(cos(2theta))^2)#

# = 2(cos(2theta))^2-1#