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1. Sin2A -1 = Cos2A
Well im stuck
Sin2A - 1 = 1 -2 Sin^2 A
Turning it into a quadratic won't work here since Sin2A = f(2x) .
2. Use sin 2A = 2sinAcosA

You'll have a quadratic in sin A.
3. (Original post by Bilbao)
Sin2A -1 = Cos2A
Well im stuck
Sin2A - 1 = 1 -2 Sin^2 A
Turning it into a quadratic won't work here since Sin2A = f(2x) .
which is

This then reduces nicely.
4. (Original post by BobbJo)
Use sin 2A = 2sinAcosA

You'll have a quadratic in sin A.
What do i do with the cosA?
5. (Original post by RDKGames)
which is

This then reduces nicely.
Ok ill try that
6. (Original post by Bilbao)
What do i do with the cosA?
You could write cos 2A = cos^2 A - sin^2 A instead, then use 1=sin^2 A + cos^2 A, the next step will become apparent
7. (Original post by Bilbao)
Sin2A -1 = Cos2A
Well im stuck
Sin2A - 1 = 1 -2 Sin^2 A
Turning it into a quadratic won't work here since Sin2A = f(2x) .
Dear Student you can proceed as follows.
Sin2A-1 = 2Cos^2 (A) -1
2Cos^2(A) -Sin2A = 0
2 Cos^2(A) - 2SinACosA = 0
2CosA(CosA-SinA) = 0
Either CosA = 0 or CosA - SinA = 0
A = InvCos(o) or TanA = 0
A = InvTan(0)
8. (Original post by Bilbao)
Sin2A -1 = Cos2A
Well im stuck
Sin2A - 1 = 1 -2 Sin^2 A
Turning it into a quadratic won't work here since Sin2A = f(2x) .

(Original post by Bilbao)
Sin2A -1 = Cos2A
Well im stuck
Sin2A - 1 = 1 -2 Sin^2 A
Turning it into a quadratic won't work here since Sin2A = f(2x) .
Dear Student you can proceed as follows.
Sin2A-1 = 2Cos^2 (A) -1
2Cos^2(A) -Sin2A = 0
2 Cos^2(A) - 2SinACosA = 0
2CosA(CosA-SinA) = 0
Either CosA = 0 or CosA - SinA = 0
A = InvCos(o) or TanA = 0
A = InvTan(0)

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