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December 19, 2014

December 19, 2014

Posted by **sh** on Saturday, February 27, 2010 at 11:22pm.

(1-sin2x)/cos2x = cos2x/(1+sin2x)

I tried starting from the right side,

RS:

=(cos²x-sin²x)/(1+2sinxcosx)

=(cos²x-(1-cos²x))/(1+2sinxcosx)

and the right side just goes in circle. May I get a hint to start off?

- Calculus -
**Reiny**, Saturday, February 27, 2010 at 11:51pmFirst of all since the angle is 2x throughout, let's just use y for 2x

Secondly, you probably want to prove it as an identity, rather than solve it

RS

= cosy/(1+siny)**[(1-siny)/(1-siny)]**

= cosy(1-siny)/(1- sin^2 y)

= cosy(1-siny)/cos^2y

= (1-siny)/cosy)

= (1- sin 2x)/cos 2x

= LS

- Calculus -
**drwls**, Saturday, February 27, 2010 at 11:51pmCross multiply.

cos^2(2x) = 1- sin^2(2x) = cos^2 2x

q.e.d.

x can be anything.

That is why it is called an identity

- Calculus -
**sh**, Sunday, February 28, 2010 at 12:00amGot it, thanks!

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