Trigonometric Identities: How to Find Cos x and Tan 2 x given Sin x = 2/√13

AI Thread Summary
Given sin x = 2/√13 for π/2 < x < π, cos x is determined to be -3/√13 after considering the angle's quadrant. The user initially attempted to find tan 2x using sin 2x/cos 2x but was advised to use the double angle identity for tangent instead. The correct approach involves calculating sin(2x) and cos(2x) from known values of sin(x) and cos(x). The discussion highlights the importance of understanding trigonometric identities and their applications in different quadrants. Overall, the conversation emphasizes the need for proper formulas when solving trigonometric problems.
Peter G.
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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
 
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Peter G. said:
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.
Peter G. said:
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).
Peter G. said:
I got a negative answer too, which makes sense in terms of the unit circle.

Thanks,
Peter
 
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!
 
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.
 
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