Question says simplify cosec x + cot x

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godfreyjh

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Dear Friends
Could you be so kind to give us some assistance with the following


The following question is asked under a section on Trig Identities
Simply states

Simplify cosec x + cot x

We managed to get this far: 1sin⁡ x + cos⁡sin⁡ x\displaystyle \dfrac{1}{\sin}\, x\, +\, \dfrac{\cos}{\sin}\, x

Which we simplified down to

. . .1 + cos⁡(x)sin⁡(x)\displaystyle \small{\dfrac{1\, +\, \cos(x)}{\sin(x)}}

Given that the question is 3 marks
Surely there is more to this than the above

We cannot see further simplification or we wondering are we even on the right track. Maybe missing the point completely. Some assistance and guidance would be greatly appreciated.


Thank you for your time and assistance
regards
Godfrey
 

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Well, if I'm being honest, I really really dislike these type of vague problem statements. With trig functions, expressions can typically be rewritten in many different forms, and then it becomes an exercise in semantics to determine if one is "more simplified" than the others. In this case, I'll go out on a limb and assume that the "simplified" version is one containing the fewest terms. Under this definition, you can go one step further to simplify the expression down to one containing just a single term.

To accomplish this, I'd brush up on the double-angle and half-angle identities. If you're having difficulties finding these in your notes and/or your textbook, this websitehttp://www.dummies.com/education/math/trigonometry/the-origin-of-the-half-angle-identities-for-sine/ has the information. Specifically, recall that csc(x)=1sin(x)\displaystyle csc(x)=\dfrac{1}{sin(x)} and cot(x)=1tan(x)\displaystyle cot(x)=\dfrac{1}{tan(x)}

As an aside, judging from your notation, you also appear to be suffering from a very common confusion about what trig functions mean and what they are. As part of your working, you include:

1sin x+cossin x\displaystyle \dfrac{1}{sin} \: x + \dfrac{cos}{sin} \: x

But this is utter nonsense. Sine (sin) is a function and must have an argument. sin(x) does not indicate sin times x. Rather, is it read as "the sine of x." And similarly with cosine.
 
ksdhart2 said:
As an aside, judging from your notation, you also appear to be suffering from a very common confusion about what trig functions mean and what they are. As part of your working, you include:

1sin x+cossin x\displaystyle \dfrac{1}{sin} \: x + \dfrac{cos}{sin} \: x

But this is utter nonsense. Sine (sin) is a function and must have an argument. sin(x) does not indicate sin times x. Rather, is it read as "the sine of x." And similarly with cosine.
Click to expand...

Yes you are right utter nonsense and I did not see it. This is the correct notation 1sin (x)+cos (x)sin (x)\displaystyle \dfrac{1}{sin\,(x)}+ \dfrac{cos\,(x)}{sin\,(x)} I apologise for the error in layout which I did not see.

I will ask tomorrow for the solution in class and see what the lecturer says.

Thanks again

Regards.
 
Last edited:
Subhotosh Khan said:
cosec(x) + cot(x) = 1/sin(x) + cos(x)/sin(x)

= [1 + cos(x)]/sin(x) ................. one possible simplification↓

= 2 * cos2(x/2)/[2 * sin(x/2) * cos(x/2)]

= cos(x/2)/sin(x/2)

= cot(x/2)
Click to expand...

Thank you for your hints and later your simplification Subhotosh your assistance is greatly appreciated.

Regards
 
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