Help! I'm Struggling with Trig f(x) = -2sin3x - 4cos3x

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SUMMARY

The discussion focuses on transforming the trigonometric function f(x) = -2sin(3x) - 4cos(3x) into a single sine function format. The recommended approach involves first multiplying the entire expression by -1 to convert the coefficients to positive values. Subsequently, the expression can be rewritten as Rsin(3x - α) by determining R and α through the relationship R = √((-2)² + (-4)²) and using trigonometric identities to equate coefficients.

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ok i have no idea how to do this, and I'm pretty sure I should.
i have to express this in terms of the general sine funtion

f(x) = -2sin3x - 4cos3x

I don't even know where to start
 
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Expressions of that format [tex]a\sin x + b\cos x[/tex] can be rewritten in terms of a sine only for positive values of a and b.

However, you have negative values for a and b, so i'd start by multiplying throughout by -1.

Then you could say:

[tex]a\sin x + b\cos x \equiv R\sin(x - \alpha)[/tex]

Expand the right hand side of the equation and equate coefficients to find R and alpha.
 
Last edited:
Work out a right triangle with -2 and -4, where the hypothenuse is [itex]A = \sqrt{(-2)^2 + (-4)^2}[/itex] directed an angle [itex]\phi[/itex].
 

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