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What is #cos^5theta-6sin^3theta# in terms of non-exponential trigonometric functions?

Answer 1

Use these trig identities to transform #cos^5 x# and #sin^3 x# #cos 2x = 2cos^2 x - 1 = 1 - 2sin^2 x# #sin^2 x + cos ^2 x = 1#

(1) --> #cos^5 x = cos x(cos^2 x)(cos^2 x) =# #= cos x(1 - sin x)(1 + sin x)((1 + cos 2x)/2)# (2) --> #6sin^3 x = 6sin x(sin^2 x) = 6sin x(1 - cos x)(1 + cos x).# f(x) = (1) + (2) #f(x) = (1/2)cos x(1 - sin x)(1 + sin x)(1 + cos 2x) - 6sin x(1 - cos x)(1 + cos x)#

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Answer 2

James Adkins

cos^5(theta) - 6sin^3(theta) can be expressed as:

cos^2(theta) * (cos^3(theta) - 6sin^2(theta))

Using the Pythagorean identity cos^2(theta) = 1 - sin^2(theta), we substitute:

(1 - sin^2(theta)) * (cos^3(theta) - 6sin^2(theta))

Expanding:

cos^3(theta) - 6sin^2(theta) - sin^2(theta) * (cos^3(theta) - 6sin^2(theta))

Further simplifying:

cos^3(theta) - 6sin^2(theta) - sin^2(theta) * cos^3(theta) + 6sin^4(theta)

Finally, combining like terms:

cos^3(theta) - sin^2(theta) * cos^3(theta) - 6sin^2(theta) + 6sin^4(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.