# Trigonometry question

• May 23rd 2006, 01:21 AM
kingkaisai2
Trigonometry question
[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
]
• May 23rd 2006, 02:41 AM
CaptainBlack
Quote:

Originally Posted by kingkaisai2
[SIZE=4]If A=sin(-1)x, where x>0 sin(-1)x means the shift sin in the calulator of angle x
[SIZE=5]Two part question
a)show that cosA=squareroot of 1-x^2

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
• May 23rd 2006, 02:46 AM
CaptainBlack
Quote:

Originally Posted by kingkaisai2
[SIZE=4]If A=sin(-1)x, where x>0 sin(-1)x means the shift sin in the calulator of angle x

b)find expressions in terms of x for cosecA and cos2A[/SIZE
]

$\displaystyle \csc(A)=1/ \sin(A)=1/x$.

$\displaystyle \cos(2A)=2 (cos(A))^2-1=2(1-x^2)-1=1-x^2$

RonL