How do you evaluate #sin(2arccos(3/5))#?

Answer 1

It's #24/25#

We need to make use of the property that #cos(arccos(x))=x#. At the moment we have a sine instead of a cosine. We also know the formula: #cos^2x+sin^2x=1#. To make use of both of these, we need to square the expression and immediately take the square root (so we don't do anything illegal):

#sqrt(sin^2(2arccos(3/5)))=sqrt(1-cos^2(2arccos(3/5))#

The only problem that we have now is the #2# in front of the #arccos#. We can solve this by ussing the double angle formula:

#cos(2a)=cos^2a-sin^2a=2cos^2a-1=1-2sin^2a#

We need it in therms of the cosine, so let's take the second one:

#sqrt(1-(2cos^2(arccos(3/5))-1)^2# Simplifying this: #sqrt(1-4*cos^4(arccos(3/5))+4cos^2(arccos(3/5))-1)# Now we can replace #cos(arccos(x))# by #x#: #sqrt(4*(3/5)^2-4*(3/5)^4)=sqrt(4*(3/5)^2*(1-(3/5)^2)# #=2*3/5*sqrt(1-9/25)=6/5*sqrt(16/25)=6/5*4/5=24/25#

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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.