Please help with this Trig identity!!!!!

**amanda0603** · March 23rd 2009, 05:02 PM

For some reason I CANT get the answer for

(1-sin^2X)/(sinX-cscX)= -sinX

PLEASE PLEASE PLEASE!

**e^(i*pi)** · March 23rd 2009, 05:05 PM

Originally Posted by amanda0603

For some reason I CANT get the answer for

(1-sin^2X)/(sinX-cscX)= -sinX

PLEASE PLEASE PLEASE!

$sin(x) - csc(x) = sin(x) - \frac{1}{sin(x)}$

set the same denominator:

$\frac{sin^2(x) -1}{sin(x)}$

$1-sin^2(x) \times \frac{sin(x)}{sin^2(x) -1}$

Note that $1-sin^2(x) = -(sin^2(x) - 1)$

$-(sin^2(x)-1) \times \frac{sin(x)}{sin^2(x) -1}$

which cancels to $-sin(x)$

**BCHurricane89** · March 23rd 2009, 05:07 PM

well, $sin^2(x)+cos^2(x)=1$ , so $1-sin^2(x)=$ $cos^2(x)$

**Soroban** · March 23rd 2009, 05:25 PM

Hello, Amanda!

Another approach . . .

$\frac{1-\sin^2\!x}{\sin x-\csc x} \:=\: -\sin x$

$\text{We have: }\;\frac{1-\sin^2\!x}{\sin x-\csc x} \:=\:\frac{1-\sin^2\!x}{\sin x - \frac{1}{\sin x}}$

Multiply by $\frac{\sin x}{\sin x}\!:\;\;\frac{\sin x}{\sin x}\cdot\frac{1-\sin^2\!x}{\sin x - \frac{1}{\sin x}} \;=\;\frac{\sin x(1 - \sin^2\!x)}{\sin^2\!x - 1}$

Factor and reduce: . $\frac{-\sin x\,({\color{red}\rlap{////////}}\sin^2\!x-1)}{{\color{red}\rlap{////////}}\sin^2\!x-1} \;=\; -\sin x$

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