Expert: Sherman D. - 3/14/2008

Question
How would you put:
f(x) = 3sin(2x) - cos(2x)
into the form of:
f(x) = a sin(bx - c)
?

Answer
f(x) = asin(bx - c)
f(x) = a(sin(bx)cos(c) - sin(c)cos(bx))

f(x) = 3sin(2x) - cos(2x)
f(x) = 6sin(x)cos(x) - (1 - 2sin(x)^2)
f(x) = 6sin(x)cos(x) - 1 + 2sin(x)^2
f(x) = 6sin(x)cos(x) - sin(x)^2 - cos(x)^2 + 2sin(x)^2
f(x) = 6sin(x)cos(x) + sin(x)^2 - cos(x)^2

f(x) = 6sin(x)cos(x) - (2cos(x)^2 - 1)
f(x) = 6sin(x)cos(x) - 2cos(x)^2 + 1
f(x) = 6sin(x)cos(x) - 2cos(x)^2 + sin(x)^2 + cos(x)^2
f(x) = 6sin(x)cos(x) - cos(x)^2 + sin(x)^2

f(x) = 6sin(x)cos(x) - (cos(x)^2 - sin(x)^2)
f(x) = 6sin(x)cos(x) - cos(x)^2 + sin(x)^2

i can't seem to get it in that form, check with answers.yahoo.com

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as the title, i would put 3sin(2x)-cos(2x) into a*sin(bx-c)