How do you evaluate sin(2arccos(35))?

1 Answer
Nico Ekkart
Oct 17, 2015

It's 2425

Explanation:

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: cos2x+sin2x=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):

√sin2(2arccos(35))=√1−cos2(2arccos(35))

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)=cos2a−sin2a=2cos2a−1=1−2sin2a

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

√1−(2cos2(arccos(35))−1)2
Simplifying this:
√1−4⋅cos4(arccos(35))+4cos2(arccos(35))−1
Now we can replace cos(arccos(x)) by x:
√4⋅(35)2−4⋅(35)4= ⎷4⋅(35)2⋅(1−(35)2)
=2⋅35⋅√1−925=65⋅√1625=65⋅45=2425

Related questions
See all questions in Basic Inverse Trigonometric Functions
Impact of this question
20200 views around the world
You can reuse this answer
Creative Commons License