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

Question

5027 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-16T16:31:21

The answer is $=-2/9sqrt2$

Let

$arcsec(-3)=x$

Then,

$secx=-3$

$cosx=-1/3$

$sinx=sqrt(1-(1/3)^2)=sqrt(8/9)=sqrt8/3$

Therefore,

$sin(arcsec(-3))*cos(arcsec(-3))=sqrt8/3*(-1/3)$

$=-2/9sqrt2$

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