How do you find the exact value of arccos(sin(3*pi/2))arccos(sin(3π2))?

2 Answers
May 10, 2016

piπ plus other solutions.

Explanation:

You need to covert the expression involving the \sinsin inside the brackets into one involving a \coscos because \arccos(\cos x) =x arccos(cosx)=x.

There are always several ways to manipulate trig functions, however one of the most straight forward ways to covert an expression involving sine into one for cosine is to use the fact that they are the SAME FUNCTION just shifted over by 90^o90o or pi/2π2 radians, recall

\sin(x) = \cos(pi/2 - x)sin(x)=cos(π2x) .

So we replace \sin({3 \pi }/2) sin(3π2) with \cos(pi/2-{3 \pi }/2)cos(π23π2)
or = \cos (-{2pi}/2)=\cos (-pi)=cos(2π2)=cos(π)

\arccos( \sin({3 \pi }/2))=\arccos( \cos(-\pi ))=-piarccos(sin(3π2))=arccos(cos(π))=π.

There is the odd issue with multiple solutions to many expressions involving inverse trig functions. The most obvious relates to cos(x)= cos(-x)cos(x)=cos(x), so you can replace \cos(-pi)cos(π) with \cos(pi)cos(π) and repeat the above end up with \arccos( \sin({3 \pi }/2))=piarccos(sin(3π2))=π . Why?

Because of the periodicity of the cosine function with have cos(pi)=cos(2pi*k+pi)cos(π)=cos(2πk+π) , so there are even more answers! Infinity of them, \pm (2*k+1)pi±(2k+1)π, positive or negative odd multiples of piπ.

The real issue here is the inverse cosine, cosine is a function with has multiple y values so when you reverse it you actually get an infinite number of possible answers, when we use it we RESTRICT the values to a window of piπ size, 0 <= x <= pi 0xπ is a typical one (calculator often use this one). Others use - pi <= x <= 0 πx0 and pi <= x <= 2 pi πx2π is also valid. In each of these "windows" we only have one solution. I'm going to go with the calculator's answer for above.

Jun 14, 2016

pi.π.

Explanation:

We have, sin3pi/2=-1.sin3π2=1.

Hence, reqd. value = arccos(sin3pi/2)=arccos(-1) = theta,=arccos(sin3π2)=arccos(1)=θ, say.

Then, by defn. of arccos, costheta=-1=cos pi,arccos,cosθ=1=cosπ, where of course, theta in [0,pi].θ[0,π].
:. theta=pi, as cos fun. is one-one in [0,pi].