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Evaluate #sin^4 15^@ + cos^4 15^@#?

Answer 1

Daniel Acosta

The expression gives a result of #7/8#

It is evident that

#(sin^2x + cos^2x)^2 = (sin^2x + cos^2x)(sin^2x+ cos^2x) = sin^4x + cos^4x + 2cos^2xsin^2x#

Consequently,

#sin^4x + cos^4x = (sin^2x + cos^2x)^2 - 2cos^2xsin^2x#

Our goal is to locate:

#(sin^2(15˚) + cos^2(15˚))^2 - 2cos^2(15˚)sin^2(15˚) = sin^4(15˚) + cos^4(15˚)#

We know that #sin^2x+ cos^2x = 1#, therefore,

#1^2 - 2cos^2(15˚)sin^2(15˚) = sin^4(15˚) + cos^4(15˚)#

Now we notice that #sin^2(2x) = 4sin^2xcos^2x#, therefore,

#1^2 - 1/2sin^2(2(15˚)) = sin^4(15˚) + cos^4(15˚)#

#1^2 - 1/2sin^2(30˚) = sin^4(15˚) + cos^4(15˚)#

#1 - 1/2(1/2)^2 = sin^4(15˚) + cos^4(15˚)#

#7/8 = sin^4(15˚) + cos^4(15˚)#

And if we check by calculator, we see that we do indeed get #7/8#.

I hope this is useful!

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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.