# Thread: Help me with trig functions!!

1. ## Help me with trig functions!!

I'm having a hard time understanding how to verify trig functions using trig identities.

can someone walk me through how to verify this equation? thanks

(cscθ)(tanθ) = secθ

2. Originally Posted by TacticalPro
I'm having a hard time understanding how to verify trig functions using trig identities.

can someone walk me through how to verify this equation? thanks

(cscθ)(tanθ) = secθ
basic identities which you should know ...

$\displaystyle \csc{x} = \dfrac{1}{\sin{x}}$

$\displaystyle \tan{x} = \dfrac{\sin{x}}{\cos{x}}$

$\displaystyle \sec{x} = \dfrac{1}{\cos{x}}$

3. It is like this
(1/sinx)(sinx/cosx)= sec
(1/cosx)= secx
secx=secx

4. I can help

$\displaystyle \displaystyle \csc \theta \tan \theta$

$\displaystyle \displaystyle = \frac{1}{\sin \theta}\times \frac{\sin \theta}{\cos \theta}$

$\displaystyle \displaystyle =\frac{1}{\cos \theta}$

$\displaystyle \displaystyle =\sec\theta$

csc[(pi/2)-θ] / tan(-θ) = -cscθ

i reduced that down to

secθ / tanθ = cscθ

and i'm stuck. can anyone help me with what I do next? thanks

6. Originally Posted by TacticalPro

csc[(pi/2)-θ] / tan(-θ) = -cscθ

i reduced that down to

secθ / tanθ = cscθ

and i'm stuck. can anyone help me with what I do next? thanks
No,

you should have

$\displaystyle \displaystyle\frac{sec\theta}{tan(-\theta)}=csc\theta}$

Now write tan in terms of sine and cosine and use sin(-A)=-sinA, cos(-A)=cosA.

7. Originally Posted by Archie Meade
No,

you should have

$\displaystyle \displaystyle\frac{sec\theta}{tan(-\theta)}=csc\theta}$

Now write tan in terms of sine and cosine and use sin(-A)=-sinA, cos(-A)=cosA.
I'm assuming you forgot to put a negative sign (-) in front of cscθ?

anyway, i did what u suggested and I got:

secθ / (-sinθ/cosθ) = - cscθ

and I rewrote it as

secθcosθ / -sinθ = - cscθ

and now I'm stuck again

8. No, I didn't forget.
The minus sign only comes at the last step.
I'm working from the left and evaluating that.

Now, what is $\displaystyle sec\theta cos\theta\;\;?$

9. Originally Posted by Archie Meade
No, I didn't forget.
The minus sign only comes at the last step.
I'm working from the left and evaluating that.

Now, what is $\displaystyle sec\theta cos\theta\;\;?$
Well, isn't it 1? because:

cosθ/cosθ

10. Originally Posted by TacticalPro
Well, isn't it 1? because:

cosθ/cosθ
Correct.

11. ok so now i'm left with:

1/-sinθ = -cscθ

cscθ = cscθ

12. Originally Posted by TacticalPro
ok so now i'm left with:

1/-sinθ = -cscθ

cscθ = cscθ
Why not just say

$\displaystyle \displaystyle\frac{1}{-\sin}=-\csc=\text{RHS}$

13. You're probably overanalysing the two sides being equal.

You only need evaluate the left hand side to see what it eventually yields...

$\displaystyle \displaystyle\frac{1}{-sin\theta}=-\frac{1}{sin\theta}=-csc\theta$