MHB Express 3sin(3x)-4cos(3x) in the form Rcos(3x+\alpha)

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To express 3sin(3x) - 4cos(3x) in the form Rcos(3x + α), the coefficients a and b are identified as 3 and -4, respectively. The transformation involves calculating R as the square root of the sum of the squares of a and b, which results in R = 5. The angle α is determined using the arctangent function, leading to α = tan^(-1)(-4/3). Finally, to find the smallest x for which 3sin(3x) - 4cos(3x) equals 4, the equation can be solved using the derived expression.
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Tried simplifying it of course, but didn't get far. Here's tbe problem:

''Express 3sin(3x)-4cos(3x) in the form Rcos(3x+\alpha),\alpha\ge0;R>0. Hence, find the smallest possible value of x for which 3sin(3x)-4cos(3x)=4.''

Bit confusing for me, especially the last part. How do you solve this, lads?
 
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ThomsonKevin said:
Tried simplifying it of course, but didn't get far. Here's tbe problem:

''Express 3sin(3x)-4cos(3x) in the form Rcos(3x+\alpha),\alpha\ge0;R>0. Hence, find the smallest possible value of x for which 3sin(3x)-4cos(3x)=4.''

Bit confusing for me, especially the last part. How do you solve this, lads?
[math]a~cos(3x) + b~sin(3x) = \sqrt{a^2 + b^2}~cos \left ( 3x - tan^{-1} \left ( \frac{b}{a} \right ) \right )[/math]
where a = 3, b = -4. (Warning: We have to take x > 0 for this.)

You can find that identity (and many others) here.

-Dan
 
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