Pythagorean Identities - finding sin and cos ?

[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.

[Dick]

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

[nukeman]

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

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)

?

[Dick]

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)

?

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

[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.

[Dick]

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.

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).

[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.

[SteamKing]

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.

[HallsofIvy]

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.

[tiny-tim]

hi nukeman! :smile:

(have a theta: θ and try using the X2 icon just above the Reply box :wink:)
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)

?

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 :smile: 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 :wink: