How do you calculate $sin (a r c sec (- 3)) cos (a r c sec (- 3))$ $sin(arcsec(-3))cos(arcsec(-3))$ ?

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Answer 1 · 2018-03-16T16:31:21

The answer is $= - \frac{2}{9} \sqrt{2}$ $=-2/9sqrt2$

Let

$a r c sec (- 3) = x$ $arcsec(-3)=x$

Then,

$sec x = - 3$ $secx=-3$

$cos x = - \frac{1}{3}$ $cosx=-1/3$

$sin x = \sqrt{1 - {(\frac{1}{3})}^{2}} = \sqrt{\frac{8}{9}} = \frac{\sqrt{8}}{3}$ $sinx=sqrt(1-(1/3)^2)=sqrt(8/9)=sqrt8/3$

Therefore,

$sin (a r c sec (- 3)) \cdot cos (a r c sec (- 3)) = \frac{\sqrt{8}}{3} \cdot (- \frac{1}{3})$ $sin(arcsec(-3))*cos(arcsec(-3))=sqrt8/3*(-1/3)$

$= - \frac{2}{9} \sqrt{2}$ $=-2/9sqrt2$

How do you calculate sin(arcsec(−3))cos(arcsec(−3))sin(arcsec(-3))cos(arcsec(-3))?