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(xy) = cos2y  cos2x
My Attempt:
RS:
= 1  2sin^2y  1  2sin^2x
= 1  1  2sin^2y  2sin^2x
= 2sin^2y  2sin^2x

Followup  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  (12sin^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(xy) = 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[(1cos^2x)cos^2y – (1cos^2y)cos^2x]
2[cos^2ycos^2xcos^2y – cos^2x + cos^2xcos^2y]
2[cos^2ycos^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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