Trig Identities

**millerst** · November 12th 2008, 02:59 PM

Hi, I'm having no luck solving a couple of identities, could someone be of assistance?

1) (1-sin^2x)/(cosx) = sin2x/2sinx
2) cscx/cosx = tanx + cosx

**skeeter** · November 12th 2008, 03:23 PM

Originally Posted by millerst

Hi, I'm having no luck solving a couple of identities, could someone be of assistance?

1) (1-sin^2x)/(cosx) = sin2x/2sinx
2) cscx/cosx = tanx + cosx

hints for #1 ...

$1-\sin^2{x} = \cos^2{x}$

$\sin(2x) = 2\sin{x}\cos{x}$

#2 is not an identity

**Soroban** · November 12th 2008, 03:31 PM

Hello, millerst!

The second one isn't true . . .

$1)\;\;\frac{1-\sin^2\!x}{\cos x} \:=\:\frac{\sin2x}{2\sin x}$

The left side is: . $\frac{\overbrace{1-\sin^2\!x}^{\text{This is }\cos^2\!x}}{\cos x} \;=\;\frac{\cos^2\!x}{\cos x} \;=\;\cos x$

$\text{Multiply by }\frac{2\sin x}{2\sin x}\!:\;\;\cos x\cdot\frac{2\sin x}{2\sin x} \;=\;\frac{\overbrace{2\sin x\cos x}^{\text{This is }\sin2x}}{2\sin x} \;=\;\frac{\sin2x}{2\sin x}$

**millerst** · November 12th 2008, 03:38 PM

Just wondering, I find these things rather hard, and normally I pick of concepts quickly. Is there anything I can do to make these things easier?

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