Trigonometric Identities

  • Thread starter Peter G.
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  • #1
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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
 

Answers and Replies

  • #2
35,393
7,271
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
 
  • #3
442
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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!
 
  • #4
35,393
7,271
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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