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
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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 ... $\displaystyle \csc{x} = \dfrac{1}{\sin{x}}$ $\displaystyle \tan{x} = \dfrac{\sin{x}}{\cos{x}}$ $\displaystyle \sec{x} = \dfrac{1}{\cos{x}}$
It is like this (1/sinx)(sinx/cosx)= sec (1/cosx)= secx secx=secx
I can help $\displaystyle \displaystyle \csc \theta \tan \theta$ $\displaystyle \displaystyle = \frac{1}{\sin \theta}\times \frac{\sin \theta}{\cos \theta}$ $\displaystyle \displaystyle =\frac{1}{\cos \theta}$ $\displaystyle \displaystyle =\sec\theta$
okay thanks! what about this problem? 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
Originally Posted by TacticalPro okay thanks! what about this problem? 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.
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
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 $\displaystyle sec\theta cos\theta\;\;?$
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
Originally Posted by TacticalPro Well, isn't it 1? because: cosθ/cosθ Correct.
ok so now i'm left with: 1/-sinθ = -cscθ cscθ = cscθ
Originally Posted by TacticalPro ok so now i'm left with: 1/-sinθ = -cscθ cscθ = cscθ Why not just say $\displaystyle \displaystyle\frac{1}{-\sin}=-\csc=\text{RHS}$
You're probably overanalysing the two sides being equal. You only need evaluate the left hand side to see what it eventually yields... $\displaystyle \displaystyle\frac{1}{-sin\theta}=-\frac{1}{sin\theta}=-csc\theta$
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