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
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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