Sine and Cosine Problem

Question

Hello, I am having trouble with the following problem:

If A and B are both acute angles and sin(A)=6/13 and cos(B)=48/49 find the value of cos9-A)-sin(-B). 

Write the answer as a simplified fraction.

Answer 1

 

The answer you are trying to find can be written like this:

    cos(-A) - sin(-B) = cos(A) + sin(B)

 

because cosine is an odd function that "eats" the negative sign

and sine is an even function that "spits out" the negative sign.

Research the definition of even and odd functions for more info.

Here's a link: http://www2.clarku.edu/~djoyce/trig/functions.html

 

Because sin(A) = 6/13, opposite =6 and hypotnuses = 13

That means, by pythagorean theorem, adjacent = square-root(133)

So cos(A) = adjacent/hypotneuse = square-root(133)/13

 

LIkewise , since cos(B) = 48/49, adjacent = 48 and hypotneuese=49,

again by pythagorean theorem, opposite = square-root(97).

So sin(B) = opposite/hypotneuse = square-root(97)/49

 

Adding them together: square-root(133)/13 + square-root(97)/49

 

                              = 49*square-root(133) + 13*square-root(97)

                                 -------------------------------------------------- 

                                                          637

 

 

Answer 2

cos(-A) - sin(-B)

 

cos(0 - A) = cos0cosA + sin0sinB = cos(A)

sin(0 - B) = sin0cosB - cos0sinB = -sin(B)

 

 

Therefore,

 

cos(A) + sin(B)

 

 

Use the Pythagorean identity to find cos(A) and sin(B).

 

 

sin²A + cos²A = 1             and              sin²B + cos²B = 1

 

 

Finish up here.