Having Difficulties Simplifying an Expression Using Trigonometric Identities

**Mazurka** · May 20th 2012, 02:22 PM

Hello,

This question may seem easy to many. However, simplifying equations using trigonometric identities has always been a weakness of mine.

The expression to simplify is:

sin²(x) / cot(x)

I have tried a number of methods and used a number of different identities, to no avail. Often it seems like the expressions I am creating during simplification are more complex than the original expression itself.

Any help that can be provided is greatly appreciated. If someone can post the individual steps they took to solve this question, that would be even better.

Thanks!

Eric.

**skeeter** · May 20th 2012, 02:34 PM

not much one can do to "simplify" this expression ... maybe

$\sin^2{x} \cdot \tan{x}$ or $\frac{\sin^3{x}}{\cos{x}}$ ,

but that doesn't get you much in the way of "simplification".

was $\frac{\sin^2{x}}{\cot{x}}$ the original expression?

**Mazurka** · May 20th 2012, 03:10 PM

The expression posted above is the original expression.

I have tried plugging this expression into some "automatic solver" applications online, and the proposed answer is always tan(x). However, having an answer isn't very handy if I cannot figure out how to arrive at it.

**Reckoner** · May 20th 2012, 06:48 PM

Originally Posted by Mazurka

The expression posted above is the original expression.

I have tried plugging this expression into some "automatic solver" applications online, and the proposed answer is always tan(x). However, having an answer isn't very handy if I cannot figure out how to arrive at it.

$\frac{\sin^2x}{\cot x}$ is most certainly not equal to $\tan x$ for all $x$ .

For example, when $x=\frac\pi3$ ,

$\frac{\sin^2x}{\cot x}=\frac{\left(\sqrt3/2\right)^2}{\sqrt3/3}=\frac{3\sqrt3}4\ne\sqrt3=\tan x$ .

For the sake of clarity, could you post, word for word, the exact question you are being asked? I cannot come up with anything beyond what skeeter gave.

**Mazurka** · May 20th 2012, 07:05 PM

The question that is asked is:

"Given the functions f(x) = sin²(x) and g(x) = cot(x), determine the equation (f/g)(x) and simplify."

To my knowledge, (f/g)(x) = sin²(x) / cot(x). If I am incorrect, definitely let me know.

**Reckoner** · May 20th 2012, 07:35 PM

Originally Posted by Mazurka

The question that is asked is:

"Given the functions f(x) = sin²(x) and g(x) = cot(x), determine the equation (f/g)(x) and simplify."

To my knowledge, (f/g)(x) = sin²(x) / cot(x). If I am incorrect, definitely let me know.

You are correct, $(f/g)(x)=\frac{\sin^2x}{\cot x}$ (where $x\ne\frac{k\pi}2, k\in\mathbb{Z}$ ).

There isn't much simplification that can be done. I would probably convert the $\cot x$ to a $\tan x$ to get rid of the division,

$(f/g)(x)=\sin^2x\tan x$ ,

but there's not much else you can do. Don't over-think it, your answer is fine.

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