Explore sums involving sine and cosine functions of the form:
\[ f(x) = a\sin(kx) + b\cos(kx) \]It can be shown analytically that this sum can be rewritten as a single cosine function with a phase shift:
\[ a\sin(kx) + b\cos(kx) = A\cos(kx - C) \]This transformation is useful for analyzing amplitude, period, and phase shift in trigonometric expressions.
To express \(a\sin(kx) + b\cos(kx)\) as \(A\cos(kx - C)\), we use the cosine subtraction formula:
\[ \cos(\alpha - \beta) = \cos\alpha\cos\beta + \sin\alpha\sin\beta \]Setting \(\alpha = kx\) and \(\beta = C\), we expand:
\[ A\cos(kx - C) = A\cos(kx)\cos(C) + A\sin(kx)\sin(C)\]For this to equal \(a\sin(kx) + b\cos(kx)\), the coefficients of \(\sin(kx)\) and \(\cos(kx)\) must match:
Square and add the two equations:
\[ \begin{aligned} [A\sin(C)]^2 + [A\cos(C)]^2 &= a^2 + b^2 \\ A^2(\sin^2(C) + \cos^2(C)) &= a^2 + b^2 \\ A^2 &= a^2 + b^2 \quad \text{(using } \sin^2(C) + \cos^2(C) = 1\text{)} \end{aligned} \]Taking the positive root (amplitude is positive):
\[ \boxed{A = \sqrt{a^2 + b^2}} \]Divide the first equation by the second (assuming \(b \neq 0\)):
\[ \frac{A\sin(C)}{A\cos(C)} = \frac{a}{b} \quad \Rightarrow \quad \tan(C) = \frac{a}{b} \]Thus: \[ \boxed{\tan(C) = \frac{a}{b}} \]
Since \(\tan(C)\) has period \(\pi\), the calculator's \(\arctan(a/b)\) gives a value between \(-\pi/2\) and \(\pi/2\). The signs of \(a\) and \(b\) determine the actual quadrant:
Here are three examples demonstrating the transformation step-by-step.
Rewrite \(3\sin(2x) + 4\cos(2x)\) as \(A\cos(2x - C)\).
Step 1: Compute amplitude \(A\):
\[ A = \sqrt{3^2 + 4^2} = \sqrt{25} = 5 \]Step 2: Compute phase shift \(C\):
\[ \tan(C) = \frac{3}{4} = 0.75 \quad \Rightarrow \quad C = \arctan(0.75) \approx 0.6435 \text{ radians} \]Since \(a>0, b>0\), \(C\) is in Quadrant I, so no adjustment is needed.
Step 3: Write the result:
\[ 3\sin(2x) + 4\cos(2x) = 5\cos(2x - 0.6435) \]Verification: Expanding \(5\cos(2x - 0.6435)\) using the cosine subtraction formula yields \(4\cos(2x) + 3\sin(2x)\).
Rewrite \(\sin(x) - \sqrt{3}\cos(x)\) as \(A\cos(x - C)\).
Step 1: Compute amplitude \(A\):
\[ A = \sqrt{1^2 + (-\sqrt{3})^2} = \sqrt{4} = 2 \]Step 2: Compute phase shift \(C\):
\[ \tan(C) = \frac{1}{-\sqrt{3}} = -\frac{1}{\sqrt{3}} \approx -0.5774 \]Calculator gives \(C = \arctan(-1/\sqrt{3}) = -\pi/6\). Since \(a>0, b<0\), \(C\) is in Quadrant II, so add \(\pi\):
\[ C = -\frac{\pi}{6} + \pi = \frac{5\pi}{6} \text{ radians} \]Step 3: Write the result:
\[ \sin(x) - \sqrt{3}\cos(x) = 2\cos\left(x - \frac{5\pi}{6}\right) \]Rewrite \(-2\sin(3x) + 2\cos(3x)\) as \(A\cos(3x - C)\).
Step 1: Compute amplitude \(A\):
\[ A = \sqrt{(-2)^2 + 2^2} = \sqrt{8} = 2\sqrt{2} \]Step 2: Compute phase shift \(C\):
\[ \tan(C) = \frac{-2}{2} = -1 \]Calculator gives \(C = \arctan(-1) = -\pi/4\). Since \(a<0, b>0\), \(C\) is in Quadrant IV, so no adjustment is needed.
Step 3: Write the result:
\[ -2\sin(3x) + 2\cos(3x) = 2\sqrt{2}\cos\left(3x + \frac{\pi}{4}\right) \]Note: \(\cos(3x + \pi/4) = \cos(3x - (-\pi/4))\), so \(C = -\pi/4\).
The sum \(a\sin(kx) + b\cos(kx)\) can always be rewritten as a single cosine function with amplitude and phase shift:
\[ a\sin(kx) + b\cos(kx) = A\cos(kx - C) \]where:
\[ A = \sqrt{a^2 + b^2}, \quad \tan(C) = \frac{a}{b} \]The phase shift \(C\) must be chosen in the correct quadrant based on the signs of \(a\) and \(b\). This transformation simplifies analysis of the function's amplitude, period, and phase shift.