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Simplify the following trigonometric expression

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Pleast help me with the problem.

Simplify the following trigonometric expression

Multiply cosθ/cscθ+1 by cscθ-1/cscθ-1

Thank you! :)

Jul 6, 2017
edited by KarmaKizuna  Jul 6, 2017
edited by KarmaKizuna  Jul 6, 2017

#1
+3

Simplify the following trigonometric expression

Multiply cosθ/cscθ+1 by cscθ-1/cscθ-1

$$\begin{array}{|rcll|} \hline && \left( \frac{ \cos(\phi) } {\csc(\phi)+1 } \right) \cdot \left( \frac{ \csc(\phi)-1 } { \csc(\phi)-1 } \right ) \\\\ &=& \frac{ \cos(\phi)\cdot \Big(\csc(\phi)-1 \Big) } { \csc^2(\phi)-1 } \quad & | \quad \csc(\phi) = \frac{1}{\sin(\phi)} \\\\ &=& \frac{ \cos(\phi)\cdot \Big(\frac{1}{\sin(\phi)}-1 \Big) } { \frac{1}{\sin^2(\phi)}-1 } \\\\ &=& \frac{ \cos(\phi)\cdot \Big(\frac{1}{\sin(\phi)}-1 \Big) } { \frac{1-\sin^2(\phi)}{\sin^2(\phi)} } \\\\ &=& \frac{ \cos(\phi)\cdot \Big(\frac{1}{\sin(\phi)}-1 \Big)\cdot \sin^2(\phi) } { 1-\sin^2(\phi) } \quad & | \quad 1-\sin^2(\phi) = \cos^2(\phi) \\\\ &=& \frac{ \cos(\phi)\cdot \Big(\frac{1}{\sin(\phi)}-1 \Big)\cdot \sin^2(\phi) } { \cos^2(\phi) } \\\\ &=& \frac{ \Big(\frac{1}{\sin(\phi)}-1 \Big)\cdot \sin^2(\phi) } { \cos(\phi) } \\\\ &=& \frac{ \Big(\frac{1-\sin(\phi)}{\sin(\phi)}\Big)\cdot \sin^2(\phi) } { \cos(\phi) } \\\\ &=& \frac{ \Big(1-\sin(\phi)\Big)\cdot \sin(\phi) } { \cos(\phi) } \\\\ &=& \Big(1-\sin(\phi)\Big) \cdot \tan(\phi) \\\\ &=& \tan(\phi)-\tan(\phi)\cdot \sin(\phi) \\ \hline \end{array}$$ Jul 6, 2017
edited by heureka  Jul 6, 2017
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Hello Heureka,

Pleast check your message center. Thank you for helping me solve the problem. :)

KarmaKizuna  Jul 6, 2017
#3
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$$\dfrac{\cos \theta}{\csc\theta +1}\cdot \dfrac{\csc\theta - 1}{\csc\theta - 1}\\ =\dfrac{\cos \theta}{\csc\theta + 1}\\ =\dfrac{\sin 2\theta}{2(1+\sin \theta)}$$

The second fraction just have the same numerator and denominator...

Jul 6, 2017
#4
+1

Hello Max,

I got the solution to the problem. It is tanθ - tanθ sinθ

I did not know the method used to solve this problem.

KarmaKizuna  Jul 6, 2017
#5
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Hi KarmaKizuna,

Welcome to Web2.0calc forum, I hope you learn lots here :)

You can write people private messages if you want to but please use it as a way to direct them to your query on the forum.

Make sure you include the link to your question or the post you want them to answer or discuss.

The question itself should be public, I mean it should be on the forum.

It seems you queried Heureka's answer privately. I would have like to have seen what that query was.

When you ask a question, it is there for us all to answer, or learn from, so please do not hide it away in a private message :)

Jul 6, 2017
#6
+2

Going from Max's answer, we have

sin 2θ / [2 (1 + sinθ)]

2sinθcosθ / [2 ( 1 + sin θ) ]

sinθcosθ / (1 + sinθ)

sinθcosθ ( 1 - sinθ) / [1 - sin^2θ]

sinθcosθ [ 1 - sinθ] / cos^2θ

sinθ[1 - sinθ]/cosθ

sinθ/cosθ  - sin^2θ/cosθ

tanθ - tanθsinθ

The moral here is that with these trig expressions, there may not be just one "correct" answer !!!!   Jul 6, 2017