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# Prove any linear combination of sine and cosine can be expressed solely in terms of sine

With reference to the previous exercise, prove that if then there exist with such that Proof. Let Then, and . Further, we have, since Hence, we know there exists such that But, if , then Thus, ### 5 comments

1. Kim says:

How do we know that there exists alpha? Shouldn’t we prove the continuity of cosine or something like that?

• Mihajlo says:

I agree.

• Anonymous says:

True, but it’s not so hard to make a geometric argument like Apostol does in the text. (Proof using continuity will have to come later.) Since values of cosine and sine are coordinates of points on the units circle, just define alpha to be twice the area of the circular sector described by the points (1, 0) and (A/C, B/C). It’s clear that this area exists, hence we can choose an alpah such that cos alpha = A/C and sin alpha = B/C.

2. I think your solution assumed for granted that the codomain of sin(x) is [0,1]. Which has no analytic proof in prior. So I think this is not a valid proof. Do you have any way around this?

• Awamoki says:

The codomain of sin(x) is [-1,1] cause the Pythagorean identity sin^2(theta)+cos^2(theta)=1 and so his solution is legit.