How do you verify the identity #(sin^4beta-2sin^2beta+1)cosbeta=cos^5beta#?

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Answer 1 · 2017-01-26T19:43:25

see below

Left Hand Side:
#=(sin^4beta-2sin^2beta+1)cosbeta#

Note that #sin^4beta-2sin^2beta+1# looks like #x^4-2x^2+1# so we can factor it.

The factors are:
#(sin^2 beta-1)(sin^2beta-1) = (sin^2beta-1)^2 #

Hence,
#=(sin^2 beta -1)^2 cos beta#

#=[-1(1-sin^2 beta)]^2 cos beta#

#=(1-sin^2beta)^2cos beta#; Use identity #sin^2 beta + cos^2 beta =1#

#=(cos^2 beta)^2 cos beta#

#=cos^4 beta cos beta#

#=cos^5 beta#

#:. =# Right Hand Side