Find the maximum height of the curve y = 4 sin x − 3 cos x y = 4\sin x - 3\cos x y = 4 sin x − 3 cos x above the x x x axis.
( 4 ) 2 + ( 3 ) 2 = 5 \sqrt{(4)^2+(3)^2}=5 ( 4 ) 2 + ( 3 ) 2 = 5
y = 5 ( 4 5 sin x − 3 5 cos x ) y =5( \dfrac{4}{5}\sin x - \dfrac{3}{5}\cos x) y = 5 ( 5 4 sin x − 5 3 cos x ) Let cos θ = 4 5 , sin θ = 3 5 , θ = sin − 1 ( 3 5 ) . \cos \theta=\dfrac{4}{5}, \sin\theta=\dfrac{3}{5}, \theta=\sin^{-1}(\dfrac{3}{5}). cos θ = 5 4 , sin θ = 5 3 , θ = sin − 1 ( 5 3 ) .
Then
4 5 sin x − 3 5 cos x = sin ( x − θ ) , θ = sin − 1 ( 3 5 ) \dfrac{4}{5}\sin x - \dfrac{3}{5}\cos x=\sin(x-\theta), \theta=\sin^{-1}(\dfrac{3}{5}) 5 4 sin x − 5 3 cos x = sin ( x − θ ) , θ = sin − 1 ( 5 3 )
y = 5 sin − 1 ( 3 5 ) y=5\sin^{-1}(\dfrac{3}{5}) y = 5 sin − 1 ( 5 3 )
− 1 ≤ sin − 1 ( 3 5 ) ≤ 1 -1\leq\sin^{-1}(\dfrac{3}{5})\leq 1 − 1 ≤ sin − 1 ( 5 3 ) ≤ 1
− 5 ≤ y ≤ 5 -5\leq y\leq 5 − 5 ≤ y ≤ 5 The maximum height of the curve of the curve y = 4 sin x − 3 cos x y = 4\sin x - 3\cos x y = 4 sin x − 3 cos x above the x x x axis is 5. 5. 5.
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