Find y such that cos y=cos(90+y) and −90degrees <= y <= 0degrees?

[iamsmooth]

Homework Statement

Find y such that cos y=cos(90+y) and −90degrees <= y <= 0degrees?

Homework Equations

cos(a+b)=sin(a)cos(b)+cos(a)sin(y)

The Attempt at a Solution

This is a question that came up in my linear algebra class. We just learned about dot product, but this doesn't seem like it's related. It almost seems like a straight trig question.

Anyways, is this a trick question? Wouldn't the answer be 0 since you can't take the cosine of a negative number?

 

[Mark44]

This is a straight trig problem, and isn't a trick question.

The answer is NOT zero; the cosine and sine functions are defined for all real numbers. Use your relevant equations to find the solution.

 

[iamsmooth]

Okay, so plugged in:

cos y = sin(90)cos(y)+cos(90)sin(y)

cos(90) = 0 so that leaves

cos y = sin(90)cos(y)

cos(y)/cos(y) = sin(90)
1 = sin(90), but that's already given...

I don't know what I'm doing... don't know how to do this if there's an unknown. What do I do from here ?
Is it just simple algebra?

 

[iamsmooth]

I'm stupid, sorry. I used the wrong identity ><

I got it

 

[Mark44]

So what did you get for the solution?

 

[iamsmooth]

cos(a+b) = cos(a)cos(b)-sin(a)sin(b)

So cos(90+y) = cos(90)cos(y)-sin(90)sin(y)

cos(y) = (0)cos(y) - (1)sin(y)

cos(y) = -sin(y)

1 = -sin(y)/cos(y)

sin(y)/cos(y) = -1

sin(y)/cos(y) = tan(y) = -1

tan(-45) = -1, so -45

I had to use a calculator to get the last part though. I guess I need to memorize my trig and unit circle at this point of linear algebra.

 

[Mark44]

Yes, you should memorize the sine and cosine of 0, 30, 45, 60, and 90 degrees (0, pi/6, pi/4, pi/3, pi/2).

The next-to-last line you have is tan(y) = -1, so y = tan^-1(-1) = -45 degrees. When you start with an equation with an unknown to find (y in this case), your last line should have the unknown in it; i.e., y = -45 degrees. Units are good, too.