[SIZE=4]If A=sin(-1)x, where x>0 sin(-1)x means the shift sin in the calulator of angle x
Two part question
a)show that cosA=squareroot of 1-x^2
b)find expressions in terms of x for cosecA and cos2A[/SIZE]
Originally Posted by kingkaisai2
If:
$\displaystyle A=\arcsin(x)$
then:
$\displaystyle
x=\sin (A)
$
but:
$\displaystyle (\sin(A))^2+(\cos(A))^2=1$,
so:
$\displaystyle x^2+(\cos(A))^2=1$.
Rearranging gives:
$\displaystyle
\cos(A)=\pm \sqrt{1-x^2}
$
Now as $\displaystyle x>0$, $\displaystyle A\in (0,\pi)$, and so $\displaystyle \cos(A)>0$, so we
want the positive square root in the last equation so:
$\displaystyle
\cos(A)= \sqrt{1-x^2}
$
RonL