How do you simplify #sqrt((1-costheta)(1+costheta))#?

2 Answers
Konstantinos Michailidis
Jul 23, 2016

It is

#sqrt((1-costheta)(1+costheta))=sqrt[1-cos^2 theta]= sqrt[sin^2 theta]=abs(sin theta)#

where #abs# is the absolute value.

We used the fact that #sin^2 theta+cos^2 theta=1#

Alexander
Jul 23, 2016

#sqrt((1-costheta)(1+costheta)) = |sintheta|#

Explanation:

For this problem, we can use the difference of squares methods, which tells us that

#(a-b)(a+b) = a^2-b^2#

Applying this method, we then get

#sqrt((1-costheta)(1+costheta)) = sqrt(1-cos^2theta)#

Also note that #sin^2theta + cos^2theta = 1 -> sin^2theta = 1-cos^2theta#

Thus we can simplify this even further, giving us

#sqrt(1-cos^2theta) = sqrt(sin^2theta)#

We have to be careful here, because the root implies that we should have to answers. In fact, if we would have made the mistake of saying that #sqrt(sin^2theta) = sin x#, we would have the following graphs:

Graph of #sqrt(sin^2theta)#:
graph{sqrt(sin x * sinx) [-5.504, 5.596, -2.153, 3.396]}
Graph of #sin theta#:
graph{sin x [-5.504, 5.596, -2.153, 3.396]}

These functions are noticeably different, but they're only off by a little bit. In this problem, we could say that

#sqrt((1-costheta)(1+costheta)) #

#= sqrt(sin^2theta) = -sintheta# and #sin theta -> |sintheta|#

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