Question #199ee

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Answer 1 · 2017-08-07T22:40:24

Write the function as:

$F (x) = sin x - 2 cos x = \sqrt{5} (\frac{1}{\sqrt{5}} sin x - \frac{2}{\sqrt{5}} cos x)$ $F(x) = sinx-2cosx = sqrt5(1/sqrt5sinx-2/sqrt5cosx)$

As ${(\frac{1}{\sqrt{5}})}^{2} + {(\frac{2}{\sqrt{5}})}^{2} = 1$ $(1/sqrt5)^2+(2/sqrt5)^2 = 1$

we can put: $cos ϕ = \frac{1}{\sqrt{5}}$ $cos phi = 1/sqrt5$ and $sin ϕ = \frac{2}{\sqrt{5}}$ $sin phi = 2/sqrt5$

so that:

$F (x) = \sqrt{5} (cos ϕ sin x - sin ϕ cos x) = \sqrt{5} cos (x + ϕ)$ $F(x) = sqrt5(cosphisinx-sinphicosx) = sqrt(5)cos(x+phi)$

with $tan ϕ = \frac{\frac{2}{\sqrt{5}}}{\frac{1}{\sqrt{5}}} = 2$ $tan phi = (2/sqrt5)/(1/sqrt5) = 2$ , so $ϕ = arctan 2$ $phi = arctan 2$

Thus the function is a sinusoid of amplitude $\sqrt{5}$ $sqrt5$ and phase $ϕ$ $phi$ . Accordingly it is decreasing when:

$0 < x < π - ϕ$ $0 < x < pi-phi$

increasing when:

$π - ϕ < x < 2 π - ϕ$ $pi-phi < x < 2pi-phi$

and again decreasing for

$2 π - ϕ < x < 2 π$ $2pi-phi < x < 2pi$

It is concave up for:

$\frac{π}{2} - ϕ < x < \frac{3 π}{2} - ϕ$ $pi/2 -phi < x < (3pi)/2-phi$

and concave down in the rest of the interval.

graph{sqrt5cos(x+1.1071487178) [-0.1, 6.3, -3, 3]}