Question #9702

given that sin Ɵ = 2/55 and cos Φ = -2/5 with Ɵ and Φ in Q-II.

1. Determine cos Ɵ?

2. Find sin Φ?

3. What is sin (Ɵ+Φ)?

4. Find cos (Ɵ+Φ)?

5. Determine tan (Ɵ-Φ)?

6. Equivalent to sin (Ɵ-Φ)?
1

Expert's answer

2012-05-29T07:32:13-0400

sinθ=255\sin \theta = \frac{2}{55} and cosφ=25\cos \varphi = -\frac{2}{5} and θ\theta and φ\varphi-lies in the second quadrant.

1. Determine cosθ\cos \theta?


cos2θ+sin2θ=1\cos^2 \theta + \sin^2 \theta = 1cosθ=±1sin2θ\cos \theta = \pm \sqrt{1 - \sin^2 \theta}


If θ\theta lie in the second quadrant than cosθ<0\cos \theta < 0

cosθ=1(255)2=143025=302155\cos \theta = - \sqrt{1 - \left(\frac{2}{55}\right)^2} = - \sqrt{1 - \frac{4}{3025}} = - \frac{\sqrt{3021}}{55}


2. Find sinφ\sin \varphi

cos2φ+sin2φ=1\cos^2 \varphi + \sin^2 \varphi = 1sinφ=±1cos2φ\sin \varphi = \pm \sqrt{1 - \cos^2 \varphi}


If φ\varphi lie in the second quadrant than sinφ>0\sin \varphi > 0

sinφ=1cos2φ=1(25)2=2125=215\sin \varphi = \sqrt{1 - \cos^2 \varphi} = \sqrt{1 - \left(-\frac{2}{5}\right)^2} = \sqrt{\frac{21}{25}} = \frac{\sqrt{21}}{5}


3. What sin(φ+θ)\sin (\varphi + \theta)?


sin(φ+θ)=sinφcosθ+cosφsinθ\sin (\varphi + \theta) = \sin \varphi \cos \theta + \cos \varphi \sin \thetasin(φ+θ)=215(302155)+(25)255=37049+4275\sin (\varphi + \theta) = \frac{\sqrt{21}}{5} \left(- \frac{\sqrt{3021}}{55}\right) + \left(- \frac{2}{5}\right) \frac{2}{55} = - \frac{3 \sqrt{7049} + 4}{275}


4. Find cos(φ+θ)\cos (\varphi + \theta)?


cos(φ+θ)=(25)(302155)215255=23021221275\cos (\varphi + \theta) = \left(- \frac{2}{5}\right) \left(- \frac{\sqrt{3021}}{55}\right) - \frac{\sqrt{21}}{5} \frac{2}{55} = \frac{2 \sqrt{3021} - 2 \sqrt{21}}{275}


5. Determine tg(θφ)\operatorname{tg}(\theta - \varphi)

tg(θφ)=tgθ+tgφ1tgθtgφ\operatorname{tg}(\theta - \varphi) = \frac{\operatorname{tg} \theta + \operatorname{tg} \varphi}{1 - \operatorname{tg} \theta \operatorname{tg} \varphi}tgθ=sinθcosθ=255302155=23021tg\theta = \frac{\sin\theta}{\cos\theta} = \frac{\frac{2}{55}}{-\frac{\sqrt{3021}}{55}} = -\frac{2}{\sqrt{3021}}tgφ=sinφcosφ=21525=212tg\varphi = \frac{\sin\varphi}{\cos\varphi} = \frac{\frac{\sqrt{21}}{5}}{-\frac{2}{5}} = -\frac{\sqrt{21}}{2}tg(θφ)=23021+212123021212=46344123021221\tg(\theta - \varphi) = \frac{-\frac{2}{\sqrt{3021}} + -\frac{\sqrt{21}}{2}}{1 - \frac{2}{\sqrt{3021}} \cdot \frac{\sqrt{21}}{2}} = \frac{-4 - \sqrt{63441}}{2\sqrt{3021} - 2\sqrt{21}}


6. Equivalent to sin(θφ)\sin(\theta - \varphi)?


sin(θφ)=sinθcosφcosθsinφ\sin(\theta - \varphi) = \sin\theta\cos\varphi - \cos\theta\sin\varphisin(θφ)=255(25)+302155215=4+37049275\sin(\theta - \varphi) = \frac{2}{55}\left(-\frac{2}{5}\right) + \frac{\sqrt{3021}}{55} \frac{\sqrt{21}}{5} = \frac{-4 + 3\sqrt{7049}}{275}

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