What is $Cot(arcsin (-5/13))$ ?

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Answer 1 · 2015-07-21T21:08:28

$cot(arcsin(-5/13))= -12/5$

Let $" "theta=arcsin(-5/13)$

This means that we are now looking for $color(red)cottheta!$

$=>sin(theta)=-5/13$

Use the identity,

$cos^2theta+sin^2theta=1$

NB : $sintheta$ is negative so $theta$ is also negative.

We shall the importance of this info later.

$=>(cos^2theta+sin^2theta)/sin^2theta=1/sin^2theta$

$=>cos^2theta/sin^2theta+1=1/sin^2theta$

$=>cot^2theta+1=1/sin^2theta$

$=>cot^2theta=1/sin^2x-1$

$=> cottheta=+-sqrt(1/sin^2(theta)-1)$

$=>cottheta=+-sqrt(1/(-5/13)^2-1)=+-sqrt(169/25-1)=+-sqrt(144/25)=+-12/5$

WE saw the evidence previously that $theta$ should be negative only.

And since $cottheta$ is odd $=>cott(-A)=-cot(A)$ Where $A$ is a positive angle.

So, it becomes clear that $cottheta=color(blue)+12/5$

REMEMBER what we called $theta$ was actually $arcsin(-15/13)$

$=>cot(arcsin(-5/13)) = color(blue)(12/5)$

What is Cot(arcsin (-5/13)) Cot(arcsin (-5/13)) ?