# Help me with trig functions!!

• Feb 6th 2011, 11:59 AM
TacticalPro
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
• Feb 6th 2011, 12:04 PM
skeeter
Quote:

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 ...

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

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

$\sec{x} = \dfrac{1}{\cos{x}}$
• Feb 6th 2011, 12:06 PM
homeylova223
It is like this
(1/sinx)(sinx/cosx)= sec
(1/cosx)= secx
secx=secx
• Feb 6th 2011, 12:07 PM
pickslides
I can help

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

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

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

$\displaystyle =\sec\theta$
• Feb 6th 2011, 12:26 PM
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
• Feb 6th 2011, 12:32 PM
Quote:

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

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

Now write tan in terms of sine and cosine and use sin(-A)=-sinA, cos(-A)=cosA.
• Feb 6th 2011, 12:45 PM
TacticalPro
Quote:

No,

you should have

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

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
• Feb 6th 2011, 01:02 PM
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 $sec\theta cos\theta\;\;?$
• Feb 6th 2011, 02:19 PM
TacticalPro
Quote:

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 $sec\theta cos\theta\;\;?$

Well, isn't it 1? because:

cosθ/cosθ
• Feb 6th 2011, 02:21 PM
dwsmith
Quote:

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

cosθ/cosθ

Correct.
• Feb 6th 2011, 02:24 PM
TacticalPro
ok so now i'm left with:

1/-sinθ = -cscθ

cscθ = cscθ
• Feb 6th 2011, 02:26 PM
dwsmith
Quote:

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

1/-sinθ = -cscθ

cscθ = cscθ

Why not just say

$\displaystyle\frac{1}{-\sin}=-\csc=\text{RHS}$
• Feb 6th 2011, 03:07 PM
$\displaystyle\frac{1}{-sin\theta}=-\frac{1}{sin\theta}=-csc\theta$