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December 21, 2014

December 21, 2014

Posted by **Sarah** on Wednesday, May 26, 2010 at 9:39pm.

- math -
**Reiny**, Wednesday, May 26, 2010 at 9:55pmOK, B+C = 180-A

so

LS = 2sin(B+C)sinA

= 2sin(180 - A)sinA

= 2(-sinA)sinA

= -2 sin^2 A

RS = 1 - cos 2A

= 1 -(2sin^2 A - 1)

= -2sin^2 A

= LS

- math -
**Mike**, Wednesday, May 26, 2010 at 10:00pmin part 1, how does sin(180-A) equal -sinA from line 4-5

- Mike you are right - math -
**Reiny**, Wednesday, May 26, 2010 at 10:12pmActually I have 2 errors in my solution

1.

sin(180-A) = sinA , not -sinA

e.g. sin (150) = sin(180-150) = sin 30

2. cos 2A = 1 - 2sin^2 A, I had it backwards

so here is the correct solution

LS = 2sin(B+C)sinA

= 2sin(180 - A)sinA

= 2(sinA)sinA

= 2 sin^2 A

RS = 1 - cos 2A

= 1 -(1 - 2sin^2 A)

= 2sin^2 A

= LS

sorry about that

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