Asked by Lola
(sinx-cosx)^2=sin2x
Answers
Answered by
Reiny
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
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
Answered by
Dexi
thnx thnx so much
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.