Given-the-function-f-cos2-sin-for-0-pi-plot-the-graph-for-intervals-of-pi-6-hence-find-the-value-of-cos2-sin-

Question Number 47586 by Rio Michael last updated on 11/Nov/18

Given the function f(θ)= cos2θ − sinθ. (for 0°≤θ≤π) plot the graph for intervals of (π/6). hence find the value of cos2θ=sinθ.

$${Given}\:{the}\:{function}\:{f}\left(\theta\right)=\:{cos}\mathrm{2}\theta\:−\:{sin}\theta.\:\left({for}\:\mathrm{0}°\leqslant\theta\leqslant\pi\right) \\ $$$${plot}\:{the}\:{graph}\:{for}\:\:{intervals}\:{of}\:\frac{\pi}{\mathrm{6}}. \\ $$$${hence}\:{find}\:{the}\:{value}\:{of}\:{cos}\mathrm{2}\theta={sin}\theta. \\ $$

Answered by Joel578 last updated on 12/Nov/18

Commented by Joel578 last updated on 12/Nov/18

cos 2θ = sin θ 1 − 2sin^2 θ = sin θ 2sin^2 θ + sin θ − 1 = 0 (2sin θ − 1)(sin θ + 1) = 0 sin θ = (1/2) ∨ sin θ = −1

$$\mathrm{cos}\:\mathrm{2}\theta\:=\:\mathrm{sin}\:\theta \\ $$$$\mathrm{1}\:−\:\mathrm{2sin}^{\mathrm{2}} \:\theta\:=\:\mathrm{sin}\:\theta \\ $$$$\mathrm{2sin}^{\mathrm{2}} \:\theta\:+\:\mathrm{sin}\:\theta\:−\:\mathrm{1}\:=\:\mathrm{0} \\ $$$$\left(\mathrm{2sin}\:\theta\:−\:\mathrm{1}\right)\left(\mathrm{sin}\:\theta\:+\:\mathrm{1}\right)\:=\:\mathrm{0} \\ $$$$\mathrm{sin}\:\theta\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\vee\:\:\mathrm{sin}\:\theta\:=\:−\mathrm{1} \\ $$

Commented by Rio Michael last updated on 13/Nov/18

thanks sir.But i thought cos2θ = sinθ could be located from the graph. That is at point where cos2θ−sinθ(f(θ))=0

$${thanks}\:{sir}.{But}\:{i}\:{thought}\: \\ $$$${cos}\mathrm{2}\theta\:=\:{sin}\theta\: \\ $$$${could}\:{be}\:{located}\:{from}\:{the}\:{graph}.\:{That}\:{is}\:{at}\:{point}\:{where} \\ $$$${cos}\mathrm{2}\theta−{sin}\theta\left({f}\left(\theta\right)\right)=\mathrm{0} \\ $$

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