Equation:
cosθ= -12/13 for π <θ<3π /2
prompt:
find sin 2θ, cos 2θ, and tan 2θ
1 answer:
Cosθ = -12/13.
For π <θ<3π /2 means 180° <θ< 270°. That is the third quadrant.
Let us just have the positive value of Cosθ = 12/13
Cosθ = Adjacent / Hypotenuse = 12 / 13
So we imagine a right angled triangle with adjacent side = 12, and Hypotenuse = 13.
To get the opposite side we apply Pythagoras' Theorem. Let the opposite side be x.
x² + 12² = 13²
x² + 144 = 169
x² = 169 - 144
x² = 25
x = √25
x = 5.
Sinθ = Opposite / Hypotenuse = 5 / 13
Tanθ = Opposite / Adjacent = 5 / 12
Recall the angle is in the third quadrant, and in the third quadrant, only Tangent is positive, Cosine and Sine are both negative.
Therefore
Cosθ = -12/13 Sinθ = -5/13 Tanθ = 5/12
Solving:
i) Sin2θ = 2SinθCosθ. By Trigonometric Identity.
= 2*(-5/13)*(-12/13)
= 120/169
ii) Cos2θ = 2Cos²θ - 1
= 2*(-12/13)(-12/13) - 1
= 288/169 - 1
= (288 - 169) / 288
= 119/288
Tan2θ = 2Tanθ /(1 - Tan²θ)
= 2*(5/12) / ( 1- (5/12)²)
= (5/6) / ( 1 - 25/144)
= (5/6) / ( (144 -25)/144)
= (5/6) / (169/25)
= (5/6) * (25/169)
= 125/1014
I hope this helps.
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