where did this come from.

Dr.Peterson said:
I'm presuming that this is an April Fool's joke, because I know you're more intelligent than this.

No one ever said that "C was tan C"; the number C is not the same as the tangent of C. And the tangent function was not introduced in the solution, but stated in the problem: "where [MATH]\tan C = \frac{b}{a}[/MATH]. That is, part of what you are to prove is that if you define C as an angle whose tangent is b/a, then ...
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I went over this problem again today step by step very carefully. I was able this time to follow the whole business from start to solution. Yes, tanC = b/a and so b=a tanC and C is an angle greater than -90 and less than 90. Part of my problem here I think was/is that I am being haunted by the ghost of a memory of having found that in the function form a sin(bx-c) the c was always a tan after the argument had been put in brackets and the b taken out of the parenthesis thus [ b(x-c/b)] after this operation I had found that the c/b was always a tangent. So I was conflating two types of problem that while they have similarities, are not the same. In this problem the author introduces the tangent so that he can use the sin/cos form for his purposes. I think what I should have done here is draw the triangle and label the sides and angles so that the thing is grounded in some kind of non-algebraic reality.
 
Jomo said:
Please state where they introduced C as tanC?

I promise you that they did not do that. if you could please tell me where you think that made this change I will show you what happened. You do remember that they said that a and b are connected by tan(C) = b/a. So then b=a tan(C). Then theny can replace any and all b's with tan(C). This is how tan(C) came into the equation. Never was C replaced with tan(C).
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Right. I went over this again today and saw how and why the tanC was being used here.
 
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