If $\sin x+\csc x=2\tan x$, Find value of $\cos^9x+\cot^9x+\sin^7x$

Solution: \begin{align*}&\sin x+\csc x=2\tan x \\ &\sin x+\frac{1}{\sin x}=2\frac{\sin x}{\cos x} \\ &\sin^2x+1=2\frac{\sin^2x}{\cos x} \\ &\sin^2x\cos x+\cos x=2\sin^2x \\ &(1-\cos^2x)\cos x+\cos x=2(1-\cos^2x) \\ &\cos^3x-2\cos^2x-2\cos x+2=0\end{align*}

Am I doing right ?

How to do further ?

  • $\begingroup$ Where did this problem come from? You can obviously solve the cubic equation you have got for $\cos x$ numerically (it has only one real root in the range [-1,1]). Note that you will get two possible values for $\sin x$ and hence two different possible values for $\cos^9x+\cot^9x+\sin^7x$. But you may be expected to do some clever manipulation. It depends on the source of the problem. $\endgroup$ – almagest Jun 14 '16 at 6:25
  • $\begingroup$ Its an objective question. Options are a) $1$ b) $0$ c) $-1$ and d) $2$ $\endgroup$ – rst Jun 14 '16 at 6:47
  • $\begingroup$ I don't think any of those options are correct. Are you sure that the statement of the problem is correctly written? $\endgroup$ – mickep Jun 14 '16 at 6:55
  • $\begingroup$ yeah, statement of question is correct. $\endgroup$ – rst Jun 14 '16 at 7:12

This is not really an answer, but a comment with image indicating that something is wrong with the problem in case the options are $1$, $0$, $-1$ and $2$ as stated in a comment above.

In the picture below I have let Mathematica draw the graphs of $\sin x+\csc x$ (blue), $2\tan x$ (yellow), $\cos^9x+\cot^9x+\sin^7x$ (green), $1$ (red) and $-1$ (purple).

It should be clear that, when the blue and yellow graphs meet each other, the green is not at any of the values stated in the problem.


To make it more evident, here is a zoomed graph around the first of the two points where the graphs meet:

enter image description here

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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