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How do you express #sin^4theta-cos^3(pi/2-theta ) # in terms of non-exponential trigonometric functions?

Answer 1

Carter Anderson

#=1/4(1-2cos2theta+1/2(1+cos4theta) -3sintheta+sin3theta)#

Given #=sin^4theta-cos^3(pi/2-theta)#

#=sin^4theta-sin^3theta#

Now we know #sin^2theta =1/2(1-cos2theta)# and #sin^3theta=1/4(3sintheta-sin3theta)#

Inserting these values the given expression becomes

#=(1/2(1-cos2theta))^2-1/4(3sintheta-sin3theta)#

#=1/4(1-2cos2theta+cos^2 2theta -3sintheta+sin3theta)#

#=1/4(1-2cos2theta+1/2(1+cos4theta) -3sintheta+sin3theta)#

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Answer 2

Josiah Anderson

[ \sin^4(\theta) - \cos^3\left(\frac{\pi}{2} - \theta\right) = \sin^4(\theta) - \sin^3(\theta) ]

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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.