Posted by **Anonymous** on Saturday, November 17, 2007 at 5:37pm.

Prove:

sin2x / 1 - cos2x = cotx

My Attempt:

LS:

= 2sinxcosx / - 1 - (1 - 2sin^2x)

= 2sinxcosx / - 1 + 2sin^2x

= cosx / sinx - 1

= cosx / sinx - 1/1

= cosx / sinx - sinx / sinx

--

Prove:

2sin(x+y)sin(x-y) = cos2y - cos2x

My Attempt:

RS:

= 1 - 2sin^2y - 1 - 2sin^2x

= 1 - 1 - 2sin^2y - 2sin^2x

= -2sin^2y - 2sin^2x

- Follow-up -
**Anonymous**, Saturday, November 17, 2007 at 5:51pm
Solved the first problem, I know what I did wrong...

LS:

= 2sinxcosx /1 - (1 - 2sin^2x)

= 2sinxcosx / 1 - 1 + 2sin^2x

= 2sinxcosx / 2sin^2x

= cosx / sinx

= cotx

- Math - Trig - Double Angles -
**Npgm**, Saturday, November 17, 2007 at 5:52pm
sin2x / 1 - cos2x = cotx

2sinxcosx / 1 - (1-2sin^2x)

2sinxcosx / 2sin^2x

cosx/sinx = cotx

- Math - Trig - Double Angles -
**Anonymous**, Saturday, November 17, 2007 at 5:55pm
I know how to solve the first question.

- Math - Trig - Double Angles -
**Npgm**, Saturday, November 17, 2007 at 6:10pm
2sin(x+y)sin(x-y) = cos2y - cos2x

lhs

2(sinx cosy + cosx siny) (sin x cos y cosx siny)

2( sin^2xcos^2y sinxcosycosxsiny + sinxcosycosxsiny cos^2xsin^2y)

2(sin^2xcos^2y cos^2xsin^2y)

2[(1-cos^2x)cos^2y (1-cos^2y)cos^2x]

2[cos^2y-cos^2xcos^2y cos^2x + cos^2xcos^2y]

2[cos^2y-cos^2x]

rhs

2 cos^2y - 1 - 2cos^2x+1

2[cos^2y cos^2x]

- Math - Trig - Double Angles -
**Anonymous**, Saturday, November 17, 2007 at 6:17pm
thanks

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