Answer to Question #243483 in Calculus for Anand

Question

find the maximum height of the curve
y = 4sin x - 3cos x above the x axis.

Answer 1

Find the maximum height of the curve $y = 4\sin x - 3\cos x$ above the $x$ axis.

$$\sqrt{(4)^2+(3)^2}=5$$

$$y =5( \dfrac{4}{5}\sin x - \dfrac{3}{5}\cos x)$$

Let $\cos \theta=\dfrac{4}{5}, \sin\theta=\dfrac{3}{5}, \theta=\sin^{-1}(\dfrac{3}{5}).$

Then

$$\dfrac{4}{5}\sin x - \dfrac{3}{5}\cos x=\sin(x-\theta), \theta=\sin^{-1}(\dfrac{3}{5})$$

$$y=5\sin^{-1}(\dfrac{3}{5})$$

$$-1\leq\sin^{-1}(\dfrac{3}{5})\leq 1$$

$$-5\leq y\leq 5$$

The maximum height of the curve of the curve $y = 4\sin x - 3\cos x$ above the $x$ axis is $5.$