# Pythagorean Identities - finding sin and cos ?

1. Jun 26, 2011

### nukeman

1. The problem statement, all variables and given/known data

For tan(theta) = /2/3 (theta in quadrant 2) find sin(pheta) and cos(pheta) using the pythagorean identities.

2. Relevant equations

3. The attempt at a solution

Which 2 pythagorean identities would I use?

My trouble is with finding sin and cos from tan.

2. Jun 26, 2011

### Dick

You probably want to use tan(theta)=sin(theta)/cos(theta) and sin(theta)^2+cos(theta)^2=1. Now try it.

3. Jun 26, 2011

### nukeman

So, I get this one here: sin(theta)^2+cos(theta)^2=1 - worked out :)

But my other 2 options, according to my text's formula sheet, I can choose from the following formulas?

1+tan^2(theta) = sec^2(theta)

1 + cot^2(theta) = csc^2(theta)

?

4. Jun 26, 2011

### Dick

Since your problem involves tan(theta), I'd try working with the first one. sec(theta)=1/cos(theta), yes?

5. Jun 26, 2011

### nukeman

Dick, there is something I am not getting.

My formula sheet has the following 2 formulas to use, and the formula you are telling me is not one of them?

1+tan^2(theta) = sec^2(theta)

1 + cot^2(theta) = csc^2(theta)

Thanks so much for your help.

6. Jun 26, 2011

### Dick

sec(theta)=1/cos(theta) may not be in your list of pythogorean identities along with tan(theta)=sin(theta)/cos(theta), but I think you can use them anyway. They are sort of the definition of sec(theta) and tan(theta).

7. Jun 26, 2011

### eumyang

I guess you could start by using the
1 + tan2 θ = sec2 θ
identity to find sec θ, and by extension, cos θ. Then use another Pythagorean identity to find sin θ.

8. Jun 27, 2011

### SteamKing

Staff Emeritus
Take the Pythagorean Identity. Divide thru using the sin^2 term. Simplify.
Take the Pythagorean Identity. Divide thru using the cos^2 term. Simplify.
You should derive the two formulas on your formula sheet. They are both restatements of this most important trig identity.

9. Jun 27, 2011

### HallsofIvy

Staff Emeritus
The 'basic' Pythagorean identity is just the Pythagorean theorem- if a right triangle has legs of length a and b and hypotenuse of length c, then $c^2= a^2+ b^2$.

If tan(theta)= 2/3, then the triangle is similar to a triangle with "opposite side" of length 2 and "near side" of length 3. By the Pythagorean theorem, the hypotenuse, c, is given by $c^2= 2^2+ 3^2= 4+ 9= 13$ so the length of the hypotenuse is $\sqrt{13}$. Now, knowing that you have a right triangle with "opposite side" of length 2, "near side" of length 3, and "hypotenuse" of length $\sqrt{13}$ you can immediately calculate any of the trigonometric functions of this angle.

10. Jun 27, 2011

### tiny-tim

hi nukeman!

(have a theta: θ and try using the X2 icon just above the Reply box )
each of the three Pythagorean identities only relates two functions at a time …

one relates tan and sec, one relates cot and cosec, and one relates cos and sin …

to start from tan and get cos, you need to go from tan to sec (as eumyang says), then use sec = 1/cos

to start from tan and get sin, you either need to do al that, and then go from cos to sin, or you can start again, use cot = 1/tan, then go from cot to cosec, then use sin = 1/cosec