I tried by solving $f'(x)=0$ and then plugging those values in the initial function. I also noticed that $f(x)=\sin x+\sin\left(x+\frac{\pi}{2}\right)-2\sin x\sin \left(x+\frac{\pi }{2} \right)$. Perhaps there’s a more efficient way to find local extrema than looking for critical points?

  • 1
    $\begingroup$ Welcome to MSE. Typesetting hint: you only need \left(...\right) when you need the parentheses to grow as necessary (for example. when they contain fractions or other other tall expressions). Otherwise, it's enough to use (...). For example, $f'(x)=0$, not ${f}'\left( x \right)=0$. $\endgroup$ – Théophile Apr 7 at 22:14
  • $\begingroup$ Thanks, I'll keep that in mind. $\endgroup$ – Arianna Apr 7 at 22:16


To begin with, I would start with rearranging the proposed expression as follows:

\begin{align*} \sin(x) + (1 - 2\sin(x))\cos(x) & = \sin(x) + \cos(x) - 2\sin(x)\cos(x)\\\\ & = \sin(x) + \cos(x) - (1 + 2\sin(x)\cos(x)) + 1\\\\ & = \sin(x) + \cos(x) - (\sin(x) + \cos(x))^{2} + 1 \end{align*}

Now you can make the substitution $u = \sin(x) + \cos(x)$.

Can you take it from here?


I fully approve the expression of @APCorreia till

$$(\sin(x) + \cos(x)) - (\sin(x) + \cos(x))^{2} + 1$$

but I wouldn't advise to make the suggested substitution.

I think preferable to write the above expression under the form

$$f(x)=\tfrac54-((\cos x + \sin x)-\tfrac12)^2$$


which is easy to differentiate


Let $a=\operatorname{acos}\dfrac{\sqrt{2}}{4}.$

The derivative is zero for values:


(red points on the graphic below). It remains to discuss the sign of $f'$ on an interval with length $2 \pi$, due to the periodicity of $f$.

enter image description here

  • $\begingroup$ Strange that you haven't commented this solution. It's usual on this site that all people contributing in a significant way to answer a question get at least a little comment. $\endgroup$ – Jean Marie Apr 12 at 10:02

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.