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How do you evaluate #[cos(arccos (3/5) - arcsin (4/5))]#?

Answer 1

Quinn Anderson

#arccos(3/5) = arcsin(4/5)# so this is just #cos(0) = 1#

#3^2+4^2=5^2#

So a #3,4,5# triangle is a right angled triangle.

#arccos(3/5) = B = arcsin(4/5)#

So: #cos(arccos(3/5)-arcsin(4/5)) = cos(0) = 1#

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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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