Please Help me solve this equation. Im thinking double angles are used.
2007-12-19 10:33:52 · 4 answers · asked by Abel T 2 in Science & Mathematics ➔ Mathematics
sin(3x)=cos(90-3x)=cosx
90-3x=x
90=4x
x=90/4=22.5
x=22.5
2007-12-19 10:37:34 · answer #1 · answered by Omega 3 · 1⤊ 0⤋
You don't need to use double angles.
sin3X = cosX = sin(90 - X + 360n) = sin(90 + X + 360)
Therefore, we have
3X = 90 - X + 360n => x = 22.5 + 90n, degrees, where n can be any integer.
3X = 90 + X + 360n => x = 45 + 180n, degrees, where n can be any integer.
2007-12-19 10:39:06 · answer #2 · answered by sahsjing 7 · 0⤊ 0⤋
Use the identity:
Sin(A+B) = Sin(A)Cos(B) + Cos(A)Cos(B)
Cos(A+B) = Cos(A)Cos(B) - Sin(A)Sin(B) (note the sign reversal).
begin with A=X and B=2X (then A+B=3X)
You'll be left with a Sin(2X) and a Cos(2X)
Use the identities again, this time with A=X and B=X
Some simplifications will occur.
You may need to use [Sin(X)]^2 + [Cos(X)]^2 = 1
2007-12-19 10:40:04 · answer #3 · answered by Raymond 7 · 2⤊ 0⤋
tsk, tsk, tsk
2007-12-20 12:59:44 · answer #4 · answered by Robel S 2 · 0⤊ 0⤋