Find a, b, and h so that f(x) a sin (b(x - h)). Round to two decimal places if necessary.
-$$f ( x ) = \sin 2 x - 3 \cos 2 x$$
A) $\mathrm { a } = 1.41 , \mathrm {~b} = 2 , \mathrm {~h} = 0.62$
B) $\mathrm { a } = 3.16 , \mathrm {~b} = 4 , \mathrm {~h} = 0.95$
C) $a = 3.16 , b = 2 , h = 0.62$
D) $a = 0.62 , b = 4 , h = 3.16$

Question

Quizplus · Accepted Answer

The given function $f(x) = \sin 2x - 3\cos 2x$ can be rewritten using the identity $\sin(A - B) = \sin A \cos B - \cos A \sin B$. To match the given form, we need to express the function in terms of a single sine function with a phase shift. This can be achieved by recognizing that the coefficients of $\sin 2x$ and $\cos 2x$ can be seen as the components of a vector in polar coordinates, where the magnitude $a$ and the angle $h$ (phase shift) can be determined as follows:1. The magnitude $a$ is found by the formula $a = \sqrt{1^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10} \approx 3.16$.2. The angle $h$ is found by $\tan h = \frac{-3}{1}$, which gives $h \approx 0.62$ radians when considering the correct quadrant for the angle based on the signs of the sine and cosine coefficients.Therefore, the correct choice is $a = 3.16$, $b = 2$, and $h = 0.62$, which matches option C.

Find A, B, and H So That F(x) a Sin f(x)=sin⁡2x−3cos⁡2xf ( x ) = \sin 2 x - 3 \cos 2 xf(x)=sin2x−3cos2x

Find A, B, and H So That F(x) a Sin $f ( x ) = \sin 2 x - 3 \cos 2 x$