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What is #4sin^5theta# in terms of non-exponential trigonometric functions?

Answer 1

Nolan Alexander

Use this identity ...

#sin^2x = (1-cos2x)/2#

#4sin^5x=4xx(sin^2x)xx(sin^2x)xxsinx#

#=4xx(1-cos2x)/2 xx (1-cos2x)/2 xx sinx#

#=(1-cos2x)(1-cos2x)(sinx)#

Hope that helped

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Answer 2

Asher Friedman

[ 4 \sin^5(\theta) = 4(\sin(\theta))^5 ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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