Trigonometry Identities

**Giggly2** · April 29th 2009, 07:32 PM

Hey I am also having a lot of trouble with these identities and I'm wondering if anyone knows how to work them??

secX/sinX - 2sinX/cosX = cotX - tanX

tanX - sinX / sin(cubed)X = secX / 1 + cosX

1 + 3sin(squared)X secX(to 4th power)X = sec(to the 6th power)X -tan(to the 6th power)X

tanX + cotX / secXcscX times tanX/sinX = secX

cosX + 1 / secX - tanX - 1 - cosx / secX + tanX = 2(1 + tanX)

**Soroban** · April 30th 2009, 03:39 AM

Hello, Giggly2!

We can't understand what you're typing.
You need more parentheses.

I think I've deciphered the 4th one . . .

$\frac{\tan x + \cot x}{\sec x\csc x}\cdot\frac{\tan x}{\sin x} \:=\: \sec x$

$\text{The left side is: }\;\frac{(\tan x + \cot x)\cdot\tan x}{(\sec x\csc x)\cdot\sin x} \;=\;\frac{\tan^2\!x + \overbrace{\cot x\tan x}^{\text{This is 1}}}{\sec x\underbrace{(\csc x\sin x)}_{\text{This is 1}}}$

. . . . $= \;\frac{\overbrace{\tan^2\!x + 1}^{\text{This is }\sec^2\!x}}{\sec x} \;=\;\frac{\sec^2\!x}{\sec x} \;=\;\sec x$

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