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

MATH110 December 2012

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Question 07 (b)
Let $f(x)=\sin x-\cos x$. Find all values of $x$ satisfying the equation $\displaystyle f'(x)=0$ in the interval $[0,\pi )$. Justify your answer.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Solving the equation $f'(x)=0$ will be similar to the previous question - you'll want to know the values of $\sin x$ and $\cos x$ at certain angles. However the answer will be different from before - how is this equation different than the previous equation?

Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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 $x=3\pi /4$.