Simplify the expression: $\sin x (1+\cot^2 x)$
Please help. I have tried all identities and still can't figure it out. Thanks.
$\begingroup$
$\endgroup$
3
-
$\begingroup$ cosec x. What's wrong with your attempt ? $\endgroup$– creativeCommented Apr 10, 2017 at 5:54
-
$\begingroup$ I recommend you for next time to include all the work you've done so far so that we don't repeat what you already know. $\endgroup$– projectilemotionCommented Apr 10, 2017 at 6:10
-
$\begingroup$ Do you mean $(\sin x)(1+\cot^2x)$ or $\sin(x(1+\cot^2x))$? $\endgroup$– JRNCommented Apr 10, 2017 at 6:22
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
3
Note the following identity: $$\sin^2{x}+\cos^2{x}\equiv 1$$ If you divide both sides by $\sin^2{x}$, you can derive another identity: $$1+\cot^2{x}\equiv \csc^2{x}$$ Can you continue?
answered Apr 10, 2017 at 5:57
-
$\begingroup$ yes i got there originally but got stuck at sinx x 1/sin^2 x $\endgroup$– DANCommented Apr 10, 2017 at 5:59
-
$\begingroup$ thats where i got that i showed you im just not sure how to multiply with the sin and the sin^2 $\endgroup$– DANCommented Apr 10, 2017 at 6:02
-
$\begingroup$ Maybe you don't know this? $$\sin^2{x}\equiv (\sin{x})^2$$ Dividing the numerator and denominator of the fraction by $\sin{x}$, you obtain: $$\frac{\sin{x}}{\sin^2{x}}=\frac{1}{\sin{x}}$$ $\endgroup$ Commented Apr 10, 2017 at 6:04
$\begingroup$
$\endgroup$
2
I assume you mean the expression
$\sin{x} \left(1 + \cot^2{x} \right) = \sin{x} \left(1 + \frac{1}{\tan^2{x}} \right) = \sin{x} \left(1 + \frac{\cos^2{x}}{\sin^2{x}} \right) = \sin{x} + \frac{\cos^2{x}}{\sin{x}} = \sin{x} + \cos{x}\cot{x}$.
However, it's not becoming much simpler. There are many forms that you can give for this expression. Ask Wolfram Alpha for more expressions. According to Wolfram, it can also be $\csc(x)$.
-
-
$\begingroup$ $\sin{x} \left(1 + \cot^2{x} \right)=\sin x \csc^2 x=\frac{\sin x}{\sin^2 x}=\frac{1}{\sin x}=\csc x$ $\endgroup$– creativeCommented Apr 10, 2017 at 6:02