Science:Math Exam Resources/Courses/MATH110/December 2012/Question 07 (b)/Solution 1

This question is similar to the previous one, except we are using the derivative:

$\displaystyle f'(x)=\cos x-(-\sin x)=\cos x+\sin x$

Setting this equal to zero and moving a term over gives

$\sin x=-\cos x$ (or $\cos x=-\sin x$)

This means that $\sin x$ and $\cos x$ must have equal values but opposite signs. On the unit circle, $\cos \theta$ corresponds to x-values and $\sin \theta$ corresponds to y-values, so the opposite signs place us in quadrant II of the coordinate plane (the restriction of $[0,\pi )$ rules out quadrant IV). The angle where the values of $\sin \theta$ and $\cos \theta$ are equal in this quadrant is $3\pi /4$, so $x=3\pi /4$.