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The magnitude of ans is correct but the sign is negative. Which is incorrect. But the procedure seem to be correct.

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  • $\begingroup$ I suspect that the problem lies in the resolution of $\cos^{-1}(\sin(u))$ vs. $\cos^{-1}(\cos(u-\pi/2))$, but I'm not certain. $\endgroup$
    – Arthur
    Jan 20, 2017 at 18:02
  • $\begingroup$ What is your question? $\endgroup$
    – R.W
    Jan 20, 2017 at 18:04
  • $\begingroup$ related: math.stackexchange.com/questions/1543142/… $\endgroup$
    – MrYouMath
    Jan 20, 2017 at 18:06
  • $\begingroup$ The answer looks correct to me. The area under the curve $y=u-\pi/2$ between $u=0$ and $u=1$ IS negative. Draw the graph and see. $\endgroup$
    – Ian Taylor
    Jan 20, 2017 at 18:10

3 Answers 3

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The problem lies in your choice of trig identity. Cosine is an even function, so $\cos(\frac{\pi}{2}-x)$=$\cos(x-\frac{\pi}{2})$ The problem with your "cancelation" is that $y=\frac{\pi}{2}-x$ and $y=x-\frac{\pi}{2}$ are not the same graph. On interval $[0,1]$ one is above the axis and one below. That explains your answer off by a negative. So the best thing is to graph the original function first as to determine what trig identity should be used to rewrite the sine.

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  • $\begingroup$ Yes cos(π/2-x)=cos(x-π/2) but if you plot y=arccos(cos(π/2-x)) and y=arccos(cos(x-π/2)) it will be same. $\endgroup$
    – spaul
    Jan 20, 2017 at 18:16
  • $\begingroup$ @user408669. You are absolutely correct about that. HOWEVER: Once you enforce the cancellation property of the trig functions, things change. Try graphing the lines and see what happens! By the way, I upvoted your question, I think this is a great post... $\endgroup$
    – imranfat
    Jan 20, 2017 at 18:17
  • $\begingroup$ Okk. I understand your point. But how do you suggest doing this in an exam hall without Mathematica. $\endgroup$
    – spaul
    Jan 20, 2017 at 18:42
  • $\begingroup$ If this question is on an exam, then that professor is mean...I probably would have done it wrong too just because the identity issue would not even occurred to me. Without any graph, this question would trick a lot of people and had this been a multiple choice question, certainly lot of people would have picked the wrong answer... $\endgroup$
    – imranfat
    Jan 20, 2017 at 18:54
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I think this is because of the property of cos to absorb negative sign of angle.

If you write $\sin x = \cos(\frac π2 - x)$

You got $\frac π2 - \frac12$

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The problem is the range of arccosine, which is between $0$ and $\pi$. Therefore, $\cos^{-1}(\cos x)\ne x$ if $x$ is negative.

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  • $\begingroup$ This is a very reasonable explanation. Thankyou $\endgroup$
    – spaul
    Jan 20, 2017 at 18:49

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