Trigonometry question

**kingkaisai2** · May 23rd 2006, 01:21 AM

[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]

**CaptainBlack** · May 23rd 2006, 02:41 AM

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:

$A=\arcsin(x)$

then:

$ x=\sin (A) $

but:

$(\sin(A))^2+(\cos(A))^2=1$ ,

so:

$x^2+(\cos(A))^2=1$ .

Rearranging gives:

$ \cos(A)=\pm \sqrt{1-x^2} $

Now as $x>0$ , $A\in (0,\pi)$ , and so $\cos(A)>0$ , so we
want the positive square root in the last equation so:

$ \cos(A)= \sqrt{1-x^2} $

RonL

**CaptainBlack** · May 23rd 2006, 02:46 AM

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]

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

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

RonL

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