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I have been trying to solve this problem for 15 minutes and I’m not any closer to a solution (in an exam, we are expected to do questions like this in 6 minutes or less which is why I’m asking now). I can’t reach the expression shown in the question. Any help would be greatly appreciated. Just to clarify, you also need to find the value of kenter image description here

So far I have tried to integrate (1-x^2)^-0.5 dx and to perform the same action for the other term, but I get an answer containing arc sins and sins, and I’m not sure if this is the right course of action or not, it’s just that it doesn’t help me get to the expression required in the question. I keep getting y = sin(C+arcsin(x)).

AOD
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  • You give up after only 15 minutes?! ... What have you tried? Please show your attempt. – sammy gerbil Nov 11 '20 at 22:15
  • In an exam, I wouldn’t have 15 minutes to spend on a question like this, and I’m trying to get to the point where I can answer a question like this in less than 6 minutes. – AOD Nov 12 '20 at 00:09
  • mathjax/latex? typing it up instead of image? – BCLC Jan 12 '21 at 08:20

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You are almost there. Let $x=\sin\theta$ so that $\arcsin x=\theta$. Then $$y=\sin(C+\theta)=\sin C\cos\theta+\cos C\sin\theta=\sin C \sqrt{1-x^2}+x\cos C$$

Notice that if we make $\sin C=\frac12$ then $\cos C=\frac12\sqrt3$ so $$2y=\sqrt{1-x^2}+\sqrt3 x$$

Comparing with the solution given we see that $k=3$.

Check: When $y=\frac12\sqrt3, x=\frac12$ then LHS$=\sqrt3$ and RHS$=\frac12\sqrt3+\frac12\sqrt3=\sqrt3$ so LHS=RHS.

sammy gerbil
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