If $sin x = \frac{1}{6}$ $sinx=1/6$ , find $cos (x - \frac{π}{2})$ $cos(x-pi/2)$ ?

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Answer 1 · 2016-12-11T08:38:31

Given $sin x = \frac{1}{6}$ $sinx=1/6$

Now $cos (x - \frac{π}{2})$ $cos(x-pi/2)$

$\Rightarrow cos (- (\frac{π}{2} - x))$ $=>cos(-(pi/2-x))$

as $cos (- θ) = cos θ$ $cos(-theta)=costheta$

$\Rightarrow cos (\frac{π}{2} - x) = sin x = \frac{1}{6}$ $=>cos(pi/2-x)=sinx=1/6$

Answer 2 · 2016-12-11T08:41:15

$cos (x - \frac{π}{2}) = \frac{1}{6}$ $cos(x-pi/2)=1/6$

As $cos (- θ) = cos θ$ $cos(-theta)=costheta$ ,

we have $cos (x - \frac{π}{2}) = cos (- (\frac{π}{2} - x)) = cos (\frac{π}{2} - x)$ $cos(x-pi/2)=cos(-(pi/2-x))=cos(pi/2-x)$

But, $cos (\frac{π}{2} - x) = sin x = \frac{1}{6}$ $cos(pi/2-x)=sinx=1/6$

Hence, $cos (x - \frac{π}{2}) = \frac{1}{6}$ $cos(x-pi/2)=1/6$

If sinx=16sinx=1/6, find cos(x−π2)cos(x-pi/2)?