This question already has an answer here:

$\cot x=\sin x \sin(\pi/2 -x) + \cos^2x \cot x$

I'm having difficulty with figuring out how to prove trigonometric identities. I know that in order to do these you need to use the trig ratios reciprocal, quotient identities and compound angle formulas. If someone could guide me through these kind of questions. That would be really appreciated!


marked as duplicate by N. F. Taussig, user26486, Joel Reyes Noche, Johanna, user99914 Apr 5 '15 at 5:17

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ @Ash did you not just ask this exact same question? $\endgroup$ – Quality Apr 5 '15 at 0:35
  • $\begingroup$ Yes, similarily but this time I would like someone to go through each step with me rather than simply posting the answers $\endgroup$ – Ash Apr 5 '15 at 0:54
  • $\begingroup$ @Ash The question structures don't differ. If you were not completely satisfied with the answer, don't accept it. Ask additional questions there in the comments instead of trying to figure it out by posting a separate question that is almost identical to the other one. At the very least you should point out what precisely you want that the answer did not provide. Either way this question is inevitably going to be closed as duplicate. $\endgroup$ – user26486 Apr 5 '15 at 1:00
  • $\begingroup$ Oh.. well would it be possible if the first version is closed since I find this one is helping me better? $\endgroup$ – Ash Apr 5 '15 at 1:01

$\sin(x)\sin(\frac{\pi}{2}-x) + \cos^2(x)\cot(x) = \sin(x)\cos(x) + \cos^2(x)\frac{\cos(x)}{\sin(x)}$ and now we want to find a common denominator to add the fractions so we get $\frac{\sin^2(x)\cos(x)}{\sin(x)} + \frac{\cos^2(x)\cos(x)}{\sin(x)} = \frac{(\sin^2(x)+\cos^2(x))\cos(x)}{\sin(x)} = \frac{\cos(x)}{\sin(x)} = \cot(x)$

  • $\begingroup$ So this is all influencing the LHS (left-hand side of the equation? $\endgroup$ – Ash Apr 5 '15 at 1:01
  • $\begingroup$ all of this is manipulating the RHS and then the final step is showing that it is equal to the LHS $\endgroup$ – nosyarg Apr 5 '15 at 1:40

First, note that using the angle addition formula of the sine function reveals $$\begin{align} \sin (\pi /2 -x)&=\sin(\pi /2)\cos(x)-\sin(x)\cos(\pi /2)\\ &=\cos(x) \end{align}$$

Then, $\sin x \sin (\pi /2 -x) =\sin x \cos x$.

Next, $\cos^2x \cot x =\frac{\cos^3x}{\sin x}$.

Putting it all together $$\begin{align} \sin x \sin (\pi /2 -x) +\sin x \cos^2x \cot x&=\cos(x)+\frac{\cos^3x}{\sin x}\\ &=\cot(x) \left(\sin^2 x +\cos^2 x \right)\\ &=\cot x \end{align}$$

as was to be shown!

  • $\begingroup$ wait, so where did the sinxcosx come from. Because aren't sin and cos treated as different variables? $\endgroup$ – Ash Apr 5 '15 at 1:19
  • $\begingroup$ They are multiplied in the first term $\sin x \sin (\pi /2 -x) = \sin x \cos x$. $\endgroup$ – Mark Viola Apr 5 '15 at 1:32
  • $\begingroup$ I'm still not quite sure if I follow. Where does the cosx portion come from? $\endgroup$ – Ash Apr 5 '15 at 1:48
  • $\begingroup$ @Ash He shows in the first line that $\sin(\pi/2-x)=\cos x$. So then $(\sin x)(\sin (\pi/2-x))=(\sin x)(\cos x)$. $\endgroup$ – user26486 Apr 5 '15 at 13:33
  • $\begingroup$ @user31415 Thank you for helping. Much appreciative! $\endgroup$ – Mark Viola Apr 5 '15 at 19:03

Not the answer you're looking for? Browse other questions tagged or ask your own question.