# Thread: Trigonometric Identities, need help fast!

1. ## Trigonometric Identities, need help fast!

I'm reading on a distance, and this is the first time I just got totally stuck. I need help really fast!

It's probably really easy.

I understand that sin^2v + cos^2v = 1

Now for an example:
_________________________________________
What is Sin(v) if cos(v) = -(24/25)?
I'm doing sin^2(v) = 1-(-(24/25)^2) = 1.9216 which ofcourse is wrong (answer is -(7/25)).
_________________________________________

Example 2:

_________________________________________
1/cos^2(v) = 1 + tan^2(v)
This is how the solution is written in the book:
1+tan^2(v) = 1 + (sin^2(v)/cos^2(v)) = (cos^2(v) + sin^2(v))/cos^2(v)

For some reason I don't get how 1 becomes cos^2(v).
_________________________________________

What am I doing wrong? Appreciate help!!

2. Originally Posted by vane505
What is Sin(v) if cos(v) = -(24/25)?
I'm doing sin^2(v) = 1-(-(24/25)^2) = 1.9216 which ofcourse is wrong
Be a little more careful with your "minus" signs. If cos(v) = -24/25, then [cos(v)]^2 = [-24/25]^2, not -[24/25]^2. The square is on all of the value, not just the fraction!

Originally Posted by vane505
1/cos^2(v) = 1 + tan^2(v)
This is how the solution is written in the book:
1+tan^2(v) = 1 + (sin^2(v)/cos^2(v)) = (cos^2(v) + sin^2(v))/cos^2(v)
For some reason I don't get how 1 becomes cos^2(v).
It wasn't that the 1 "became" a cosine squared. It's that they converted the "1" to the common denominator:

. . . . .$\displaystyle 1\, +\, \frac{\sin^2{(v)}}{\cos^2{(v)}}$

. . . . .$\displaystyle \frac{\cos^2{(v)}}{\cos^2{(v)}}\, +\, \frac{\sin^2{(v)}}{\cos^2{(v)}}$

. . . . .$\displaystyle \frac{\cos^2{(v)}\, +\, \sin^2{(v)}}{\cos^2{(v)}}$

Hope that helps!

3. Thank you! Got both of them now