Trigonometric Identities

[Peter G.]

Hi,

I am given that, for π/2 < x < π, sin x = 2/√13

a) Find Cos x
b) Find tan 2 x

So, what I did was: I drew a triangle and found that the missing side was equal to 3. From then, I deduced that cos x was equal to 3/√13

The problem was however that the angle must lie between the values given above. What I did was I simply added a negative sign. Is that right?

For part b I did sin 2 x / cos 2 x = tan 2 x and solved. Is that right? I got a negative answer too, which makes sense in terms of the unit circle.

Thanks,
Peter

[Mark44]

Hi,

I am given that, for π/2 < x < π, sin x = 2/√13

a) Find Cos x
b) Find tan 2 x

So, what I did was: I drew a triangle and found that the missing side was equal to 3. From then, I deduced that cos x was equal to 3/√13

The problem was however that the angle must lie between the values given above. What I did was I simply added a negative sign. Is that right?
Yes.


For part b I did sin 2 x / cos 2 x = tan 2 x and solved. Is that right?
I don't think so. You know sin(x) and you have found cos(x), but you don't know sin(2x) or cos(2x).

Use the double angle identity for tangent: tan(2x) = 2tanx/(1 - tan2x).
I got a negative answer too, which makes sense in terms of the unit circle.

Thanks,
Peter

[Peter G.]

Ah ok. I did the sin 2(x)/cos2(x) because I hadn't learned the tan identity and therefore didn't have it in my formula booklet. Maybe I had to know it and I didn't :redface:

Thanks!

[Mark44]

Actually, what you started to do would have worked. Since you know both sin(x) and cos(x) you could have used them to get sin(2x) and cos(2x), and then evaluated sin(2x)/cos(2x). What I suggested is just more direct.