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Solve $\arccos(11/14)+\arcsin(-1/7).$

I'm using $\cos(a+b)=\cos(a)\cos(b)+\sin(a)\sin(b).$

Attempt: sheet1 and sheet2

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1 Answer 1

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Hint:

$$ \begin{align} &\sin\left(\cos^{-1}\left(\frac{11}{14}\right)-\sin^{-1}\left(\frac{1}{7}\right)\right)\\ &=\sin\left(\cos^{-1}\left(\frac{11}{14}\right)\right)\cos\left(\sin^{-1}\left(\frac{1}{7}\right)\right) -\cos\left(\cos^{-1}\left(\frac{11}{14}\right)\right)\sin\left(\sin^{-1}\left(\frac{1}{7}\right)\right)\\ &=\frac{5\sqrt3}{14}\frac{4\sqrt3}{7}-\frac{11}{14}\frac17\\ &=\frac{49}{98} \end{align} $$


Another Hint: $$ \begin{align} &\cos\left(\cos^{-1}\left(\frac{11}{14}\right)-\sin^{-1}\left(\frac{1}{7}\right)\right)\\ &=\cos\left(\cos^{-1}\left(\frac{11}{14}\right)\right)\cos\left(\sin^{-1}\left(\frac{1}{7}\right)\right) +\sin\left(\cos^{-1}\left(\frac{11}{14}\right)\right)\sin\left(\sin^{-1}\left(\frac{1}{7}\right)\right)\\ &=\frac{11}{14}\frac{4\sqrt3}{7}+\frac{5\sqrt3}{14}\frac17\\ &=\frac{49\sqrt3}{98} \end{align} $$

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    $\begingroup$ You are doing the whole computation, except noting that $49/98=1/2$? :) $\endgroup$ Sep 24, 2013 at 17:12
  • $\begingroup$ @ThomasAndrews: I was thinking of leaving out the last line. Perhaps I should have. Of course, I didn't solve for $x$ :-) $\endgroup$
    – robjohn
    Sep 24, 2013 at 17:13
  • $\begingroup$ You solved it for me, thanks. Where do I go wrong in my calculation? $\endgroup$
    – jacob
    Sep 24, 2013 at 18:11
  • $\begingroup$ The problem is that I get sin(acos(11/14)) * cos(acos(11/14)) = 5(√3)/14 * 11/14 and cos(asin(-1/7)) * sin(asin(-1/7) = 4(√3)/7 * (-1/7). See also photo $\endgroup$
    – jacob
    Sep 24, 2013 at 18:24
  • $\begingroup$ @jacob: That should work, too. See the addition to my answer. $\endgroup$
    – robjohn
    Sep 24, 2013 at 18:38

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