Answer on Question #48027 – Math – Trigonometry
given: sin Θ = 4 / 5 cos Φ = 12 / 13 \sin \Theta = 4/5 \cos \Phi = 12/13 sin Θ = 4/5 cos Φ = 12/13 sin ( Θ + Φ ) \sin (\Theta + \Phi) sin ( Θ + Φ )
Solution:
sin θ = 4 5 \sin \theta = \frac{4}{5} sin θ = 5 4 cos Φ = 12 13 \cos \Phi = \frac{12}{13} cos Φ = 13 12
Hence, Θ ∈ [ 0 ∘ ; 180 ∘ ] \Theta \in [0{}^{\circ}; 180{}^{\circ}] Θ ∈ [ 0 ∘ ; 180 ∘ ] Θ \Theta Θ Φ ∈ [ 0 ∘ ; 90 ∘ ] \Phi \in [0{}^{\circ}; 90{}^{\circ}] Φ ∈ [ 0 ∘ ; 90 ∘ ] Φ ∈ [ 270 ∘ ; 360 ∘ ] \Phi \in [270{}^{\circ}; 360{}^{\circ}] Φ ∈ [ 270 ∘ ; 360 ∘ ] Φ \Phi Φ
Relationship between the sine and the cosine (Pythagorean identity):
sin 2 θ + cos 2 θ = 1 \sin^2 \theta + \cos^2 \theta = 1 sin 2 θ + cos 2 θ = 1
A. Suppose that Θ ∈ [ 0 ∘ ; 90 ∘ ] \Theta \in [0{}^{\circ}; 90{}^{\circ}] Θ ∈ [ 0 ∘ ; 90 ∘ ] Φ ∈ [ 0 ∘ ; 90 ∘ ] \Phi \in [0{}^{\circ}; 90{}^{\circ}] Φ ∈ [ 0 ∘ ; 90 ∘ ]
cos θ = 1 − sin 2 θ = 1 − ( 4 5 ) 2 = 3 5 \cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \left(\frac{4}{5}\right)^2} = \frac{3}{5} cos θ = 1 − sin 2 θ = 1 − ( 5 4 ) 2 = 5 3 sin 2 Φ + cos 2 Φ = 1 \sin^2 \Phi + \cos^2 \Phi = 1 sin 2 Φ + cos 2 Φ = 1 sin Φ = 1 − cos 2 Φ = 1 − ( 12 13 ) 2 = 5 13 \sin \Phi = \sqrt{1 - \cos^2 \Phi} = \sqrt{1 - \left(\frac{12}{13}\right)^2} = \frac{5}{13} sin Φ = 1 − cos 2 Φ = 1 − ( 13 12 ) 2 = 13 5
Sine of sum:
sin ( θ + Φ ) = sin θ cos Φ + cos θ sin Φ = 4 5 ⋅ 12 13 + 3 5 ⋅ 5 13 = 63 65 . \sin (\theta + \Phi) = \sin \theta \cos \Phi + \cos \theta \sin \Phi = \frac{4}{5} \cdot \frac{12}{13} + \frac{3}{5} \cdot \frac{5}{13} = \frac{63}{65}. sin ( θ + Φ ) = sin θ cos Φ + cos θ sin Φ = 5 4 ⋅ 13 12 + 5 3 ⋅ 13 5 = 65 63 .
B. Suppose that angles Θ ∈ [ 0 ∘ ; 90 ∘ ] \Theta \in [0{}^{\circ}; 90{}^{\circ}] Θ ∈ [ 0 ∘ ; 90 ∘ ] Φ ∈ [ 270 ∘ ; 360 ∘ ] \Phi \in [270{}^{\circ}; 360{}^{\circ}] Φ ∈ [ 270 ∘ ; 360 ∘ ]
cos θ = 1 − sin 2 θ = 1 − ( 4 5 ) 2 = 3 5 \cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \left(\frac{4}{5}\right)^2} = \frac{3}{5} cos θ = 1 − sin 2 θ = 1 − ( 5 4 ) 2 = 5 3 sin 2 Φ + cos 2 Φ = 1 \sin^2 \Phi + \cos^2 \Phi = 1 sin 2 Φ + cos 2 Φ = 1 sin Φ = − 1 − cos 2 Φ = − 1 − ( 12 13 ) 2 = − 5 13 \sin \Phi = -\sqrt{1 - \cos^2 \Phi} = -\sqrt{1 - \left(\frac{12}{13}\right)^2} = -\frac{5}{13} sin Φ = − 1 − cos 2 Φ = − 1 − ( 13 12 ) 2 = − 13 5
Sine of sum:
sin ( θ + Φ ) = sin θ cos Φ + cos θ sin Φ = 4 5 ⋅ 12 13 − 3 5 ⋅ 5 13 = 33 65 . \sin (\theta + \Phi) = \sin \theta \cos \Phi + \cos \theta \sin \Phi = \frac{4}{5} \cdot \frac{12}{13} - \frac{3}{5} \cdot \frac{5}{13} = \frac{33}{65}. sin ( θ + Φ ) = sin θ cos Φ + cos θ sin Φ = 5 4 ⋅ 13 12 − 5 3 ⋅ 13 5 = 65 33 .
C. Suppose that Θ ∈ [ 90 ∘ ; 180 ∘ ] \Theta \in [90{}^{\circ}; 180{}^{\circ}] Θ ∈ [ 90 ∘ ; 180 ∘ ] Φ ∈ [ 0 ∘ ; 90 ∘ ] \Phi \in [0{}^{\circ}; 90{}^{\circ}] Φ ∈ [ 0 ∘ ; 90 ∘ ]
cos θ = − 1 − sin 2 θ = − 1 − ( 4 5 ) 2 = − 3 5 \cos \theta = -\sqrt{1 - \sin^2 \theta} = -\sqrt{1 - \left(\frac{4}{5}\right)^2} = -\frac{3}{5} cos θ = − 1 − sin 2 θ = − 1 − ( 5 4 ) 2 = − 5 3 sin 2 Φ + cos 2 Φ = 1 \sin^2 \Phi + \cos^2 \Phi = 1 sin 2 Φ + cos 2 Φ = 1 sin Φ = 1 − cos 2 Φ = 1 − ( 12 13 ) 2 = 5 13 \sin \Phi = \sqrt{1 - \cos^2 \Phi} = \sqrt{1 - \left(\frac{12}{13}\right)^2} = \frac{5}{13} sin Φ = 1 − cos 2 Φ = 1 − ( 13 12 ) 2 = 13 5
Sine of sum:
sin ( θ + Φ ) = sin θ cos Φ + cos θ sin Φ = 4 5 ⋅ 12 13 − 3 5 ⋅ 5 13 = 33 65 . \sin (\theta + \Phi) = \sin \theta \cos \Phi + \cos \theta \sin \Phi = \frac{4}{5} \cdot \frac{12}{13} - \frac{3}{5} \cdot \frac{5}{13} = \frac{33}{65}. sin ( θ + Φ ) = sin θ cos Φ + cos θ sin Φ = 5 4 ⋅ 13 12 − 5 3 ⋅ 13 5 = 65 33 .
D. Suppose that Θ ∈ [ 90 ∘ ; 180 ∘ ] \Theta \in [90{}^{\circ}; 180{}^{\circ}] Θ ∈ [ 90 ∘ ; 180 ∘ ] Φ ∈ [ 270 ∘ ; 360 ∘ ] \Phi \in [270{}^{\circ}; 360{}^{\circ}] Φ ∈ [ 270 ∘ ; 360 ∘ ]
cos θ = − 1 − sin 2 θ = − 1 − ( 4 5 ) 2 = − 3 5 \cos \theta = - \sqrt {1 - \sin^ {2} \theta} = - \sqrt {1 - \left(\frac {4}{5}\right) ^ {2}} = - \frac {3}{5} cos θ = − 1 − sin 2 θ = − 1 − ( 5 4 ) 2 = − 5 3 sin 2 Φ + cos 2 Φ = 1 \sin^ {2} \Phi + \cos^ {2} \Phi = 1 sin 2 Φ + cos 2 Φ = 1 sin Φ = − 1 − cos 2 Φ = − 1 − ( 12 13 ) 2 = − 5 13 \sin \Phi = - \sqrt {1 - \cos^ {2} \Phi} = - \sqrt {1 - \left(\frac {12}{13}\right) ^ {2}} = - \frac {5}{13} sin Φ = − 1 − cos 2 Φ = − 1 − ( 13 12 ) 2 = − 13 5
Sine of sum:
sin ( θ + Φ ) = sin θ cos Φ + cos θ sin Φ = 4 5 ⋅ 12 13 + 3 5 ⋅ 5 13 = 63 65 . \sin (\theta + \Phi) = \sin \theta \cos \Phi + \cos \theta \sin \Phi = \frac {4}{5} \cdot \frac {12}{13} + \frac {3}{5} \cdot \frac {5}{13} = \frac {63}{65}. sin ( θ + Φ ) = sin θ cos Φ + cos θ sin Φ = 5 4 ⋅ 13 12 + 5 3 ⋅ 13 5 = 65 63 .
Answer:
sin ( θ + Φ ) = 63 65 , when Θ ∈ [ 0 ∘ ; 90 ∘ ] , Φ ∈ [ 0 ∘ ; 90 ∘ ] or Θ ∈ [ 90 ∘ ; 180 ∘ ] , \sin (\theta + \Phi) = \frac {63}{65}, \text{ when } \Theta \in [0{}^{\circ}; 90{}^{\circ}], \Phi \in [0{}^{\circ}; 90{}^{\circ}] \text{ or } \Theta \in [90{}^{\circ}; 180{}^{\circ}], sin ( θ + Φ ) = 65 63 , when Θ ∈ [ 0 ∘ ; 90 ∘ ] , Φ ∈ [ 0 ∘ ; 90 ∘ ] or Θ ∈ [ 90 ∘ ; 180 ∘ ] , Φ ∈ [ 270 ∘ ; 360 ∘ ] ; \Phi \in [270{}^{\circ}; 360{}^{\circ}]; Φ ∈ [ 270 ∘ ; 360 ∘ ] ; sin ( θ + Φ ) = 33 65 , when Θ ∈ [ 0 ∘ ; 90 ∘ ] , Φ ∈ [ 270 ∘ ; 360 ∘ ] or Θ ∈ [ 90 ∘ ; 180 ∘ ] , \sin (\theta + \Phi) = \frac {33}{65}, \text{ when } \Theta \in [0{}^{\circ}; 90{}^{\circ}], \Phi \in [270{}^{\circ}; 360{}^{\circ}] \text{ or } \Theta \in [90{}^{\circ}; 180{}^{\circ}], sin ( θ + Φ ) = 65 33 , when Θ ∈ [ 0 ∘ ; 90 ∘ ] , Φ ∈ [ 270 ∘ ; 360 ∘ ] or Θ ∈ [ 90 ∘ ; 180 ∘ ] , Φ ∈ [ 90 ∘ ; 180 ∘ ] , Φ ∈ [ 0 ∘ ; 90 ∘ ] . \Phi \in [90{}^{\circ}; 180{}^{\circ}], \Phi \in [0{}^{\circ}; 90{}^{\circ}]. Φ ∈ [ 90 ∘ ; 180 ∘ ] , Φ ∈ [ 0 ∘ ; 90 ∘ ] .
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