# trigoo

(sinx-cosx)^2=sin2x

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1. What are we doing? Solving?

of so, then square both sides

sin^2 - 2sinxcosx + cos^2 = sin2x
1 - sin2x = sin2x
2sin2x = 1
sin2x = 1/2
2x = 30° or 150°
x = 15° or 75°

since we squared, all answers must be checked ..
if x=15°
LS = (sin15 - cos15)^2 = .5
RS = sin 30 = .5 = Ls, so x=15° works

if x = 75°
Ls = (sin75-cos75)^2 = .5
RS = sin 150 = .5 , so x = 75° works

In radians your solution would be
x = π/12 and 5π/12

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2. thnx thnx so much

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