(sinx-cosx)^2=sin2x
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
thnx thnx so much
To prove that (sinx - cosx)^2 = sin2x, we can start by expanding the left-hand side of the equation and simplifying it.
Step 1: Expand (sinx - cosx)^2
Using the formula (a - b)^2 = a^2 - 2ab + b^2, we can expand (sinx - cosx)^2 as follows:
(sin x - cos x)^2 = sin^2x - 2(sin x)(cos x) + cos^2x
Step 2: Simplify the expression
Using the trigonometric identity sin^2x + cos^2x = 1, we can substitute that into the equation:
sin^2x - 2(sin x)(cos x) + cos^2x = 1 - 2(sin x)(cos x)
Step 3: Apply the double-angle formula for sine
The double-angle formula for sine states that sin2x = 2(sin x)(cos x). Substituting this into the equation gives:
1 - 2(sin x)(cos x) = sin2x
Therefore, (sin x - cos x)^2 = sin2x.