Trigonometry question

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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:

$A=\arcsin(x)$

then:

$<br /> x=\sin (A)<br />$

but:

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

so:

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

Rearranging gives:

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

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:

$<br /> \cos(A)= \sqrt{1-x^2}<br />$

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]

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

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

RonL

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