sin θ = 2 55 \sin \theta = \frac{2}{55} sin θ = 55 2 cos φ = − 2 5 \cos \varphi = -\frac{2}{5} cos φ = − 5 2 θ \theta θ φ \varphi φ
1. Determine cos θ \cos \theta cos θ
cos 2 θ + sin 2 θ = 1 \cos^2 \theta + \sin^2 \theta = 1 cos 2 θ + sin 2 θ = 1 cos θ = ± 1 − sin 2 θ \cos \theta = \pm \sqrt{1 - \sin^2 \theta} cos θ = ± 1 − sin 2 θ
If θ \theta θ cos θ < 0 \cos \theta < 0 cos θ < 0
cos θ = − 1 − ( 2 55 ) 2 = − 1 − 4 3025 = − 3021 55 \cos \theta = - \sqrt{1 - \left(\frac{2}{55}\right)^2} = - \sqrt{1 - \frac{4}{3025}} = - \frac{\sqrt{3021}}{55} cos θ = − 1 − ( 55 2 ) 2 = − 1 − 3025 4 = − 55 3021
2. Find sin φ \sin \varphi sin φ
cos 2 φ + sin 2 φ = 1 \cos^2 \varphi + \sin^2 \varphi = 1 cos 2 φ + sin 2 φ = 1 sin φ = ± 1 − cos 2 φ \sin \varphi = \pm \sqrt{1 - \cos^2 \varphi} sin φ = ± 1 − cos 2 φ
If φ \varphi φ sin φ > 0 \sin \varphi > 0 sin φ > 0
sin φ = 1 − cos 2 φ = 1 − ( − 2 5 ) 2 = 21 25 = 21 5 \sin \varphi = \sqrt{1 - \cos^2 \varphi} = \sqrt{1 - \left(-\frac{2}{5}\right)^2} = \sqrt{\frac{21}{25}} = \frac{\sqrt{21}}{5} sin φ = 1 − cos 2 φ = 1 − ( − 5 2 ) 2 = 25 21 = 5 21
3. What sin ( φ + θ ) \sin (\varphi + \theta) sin ( φ + θ )
sin ( φ + θ ) = sin φ cos θ + cos φ sin θ \sin (\varphi + \theta) = \sin \varphi \cos \theta + \cos \varphi \sin \theta sin ( φ + θ ) = sin φ cos θ + cos φ sin θ sin ( φ + θ ) = 21 5 ( − 3021 55 ) + ( − 2 5 ) 2 55 = − 3 7049 + 4 275 \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} sin ( φ + θ ) = 5 21 ( − 55 3021 ) + ( − 5 2 ) 55 2 = − 275 3 7049 + 4
4. Find cos ( φ + θ ) \cos (\varphi + \theta) cos ( φ + θ )
cos ( φ + θ ) = ( − 2 5 ) ( − 3021 55 ) − 21 5 2 55 = 2 3021 − 2 21 275 \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} cos ( φ + θ ) = ( − 5 2 ) ( − 55 3021 ) − 5 21 55 2 = 275 2 3021 − 2 21
5. Determine tg ( θ − φ ) \operatorname{tg}(\theta - \varphi) tg ( θ − φ )
tg ( θ − φ ) = tg θ + tg φ 1 − tg θ tg φ \operatorname{tg}(\theta - \varphi) = \frac{\operatorname{tg} \theta + \operatorname{tg} \varphi}{1 - \operatorname{tg} \theta \operatorname{tg} \varphi} tg ( θ − φ ) = 1 − tg θ tg φ tg θ + tg φ t g θ = sin θ cos θ = 2 55 − 3021 55 = − 2 3021 tg\theta = \frac{\sin\theta}{\cos\theta} = \frac{\frac{2}{55}}{-\frac{\sqrt{3021}}{55}} = -\frac{2}{\sqrt{3021}} t g θ = cos θ sin θ = − 55 3021 55 2 = − 3021 2 t g φ = sin φ cos φ = 21 5 − 2 5 = − 21 2 tg\varphi = \frac{\sin\varphi}{\cos\varphi} = \frac{\frac{\sqrt{21}}{5}}{-\frac{2}{5}} = -\frac{\sqrt{21}}{2} t g φ = cos φ sin φ = − 5 2 5 21 = − 2 21 tg ( θ − φ ) = − 2 3021 + − 21 2 1 − 2 3021 ⋅ 21 2 = − 4 − 63441 2 3021 − 2 21 \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}} tg ( θ − φ ) = 1 − 3021 2 ⋅ 2 21 − 3021 2 + − 2 21 = 2 3021 − 2 21 − 4 − 63441
6. Equivalent to sin ( θ − φ ) \sin(\theta - \varphi) sin ( θ − φ )
sin ( θ − φ ) = sin θ cos φ − cos θ sin φ \sin(\theta - \varphi) = \sin\theta\cos\varphi - \cos\theta\sin\varphi sin ( θ − φ ) = sin θ cos φ − cos θ sin φ sin ( θ − φ ) = 2 55 ( − 2 5 ) + 3021 55 21 5 = − 4 + 3 7049 275 \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} sin ( θ − φ ) = 55 2 ( − 5 2 ) + 55 3021 5 21 = 275 − 4 + 3 7049 
#339914 on Dec 2023