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Math Help - Trigonometry Identities

  1. #1
    Newbie
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    Apr 2009
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    Trigonometry Identities

    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)
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  2. #2
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    Lexington, MA (USA)
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    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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