for f(x)=2 sin 4x - 3 cos 4x find a, b, and h so that f(x)≈a sin (b(x - h)). a≈□, b=□, and h≈□ (simplify…

for f(x)=2 sin 4x - 3 cos 4x find a, b, and h so that f(x)≈a sin (b(x - h)). a≈□, b=□, and h≈□ (simplify your answers. round to three decimal places as needed.)

for f(x)=2 sin 4x - 3 cos 4x find a, b, and h so that f(x)≈a sin (b(x - h)). a≈□, b=□, and h≈□ (simplify your answers. round to three decimal places as needed.)

Answer

Explanation:

Step1: Recall the formula

We know that $A\sin\alpha - B\cos\alpha=\sqrt{A^{2}+B^{2}}\sin(\alpha-\varphi)$ where $\tan\varphi=\frac{B}{A}$. Here $A = 2$, $B = 3$ and $\alpha=4x$.

Step2: Calculate the value of $a$

Using the formula $a=\sqrt{A^{2}+B^{2}}$, substituting $A = 2$ and $B = 3$, we get $a=\sqrt{2^{2}+3^{2}}=\sqrt{4 + 9}=\sqrt{13}\approx3.606$.

Step3: Determine the value of $b$

Comparing $f(x)=a\sin(b(x - h))$ with $f(x)=2\sin4x-3\cos4x$, we can see that $b = 4$.

Step4: Calculate the value of $h$

We know that $\tan\varphi=\frac{B}{A}$, so $\tan h=\frac{3}{2}$. Then $h=\arctan(\frac{3}{2})\approx0.983$.

Answer:

$a\approx3.606$, $b = 4$, $h\approx0.983$