How do you evaluate $arccos (sin (13 \frac{π}{7}))$ $arccos(sin(13pi/7))$ ?

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Answer 1 · 2016-07-06T16:33:03

$(- \frac{19}{14}) π$ $(-19/14)pi$

Use $sin (\frac{π}{2} - a) = cos a and a r c cos (cos b) = b$ $sin (pi/2-a)=cos a and arc cos (cos b) =b$ .

Here,

$sin ((\frac{13}{7}) π) = cos (\frac{π}{2} - (\frac{13}{7}) π) = cos (- (\frac{19}{14}) π)$ $sin ((13/7)pi)=cos (pi/2-(13/7)pi)=cos(-(19/14)pi)$ .

The given expression

$= a r c cos (cos (- (\frac{19}{14}) π)$ $= arc cos (cos(-(19/14)pi)$

$= - (\frac{19}{14}) π = - {244.3}^{o}$ $=-(19/14)pi=-244.3^o$

Note that, for any arbitrary a, $sin a = cos (\frac{π}{2} - a)$ $sin a = cos (pi/2-a)$ .

Do not use cos (-b)=cos b, in any intermediate step..

Here, for the start angle $(\frac{13}{7}) π$ $(13/7)pi$ (not the principal value), the end

angle is $- (\frac{19}{14}) π$ $-(19/14)pi$

Had we used calculator, we would have got the answer as

${114.7}^{o} = \frac{9 π}{14} = (2 π - (\frac{19}{14}) π)$ $114.7^o = (9pi)/14 =(2pi-(19/14)pi)$ ..

How do you evaluate arccos(sin(13pi/7))arccos(sin(13π7))arccos(sin(13pi/7))?