given sinθ=-5/13 and π<θ<3π/2 find
sin2θ
cos( θ-4π/3)
sin(θ/2)
can some1 explain to me how to do these?
sinA = -5/13 = Y/r,
X^2 + Y^2 = r^2,
X^2 + (-5)^2 = (13)^2,
X^2 + 25 = 169,
X^2 = 169 - 25 = 144,
X = +- 12.
X = -12 Because it places our resultant
vector in the required quadrant(Q3).
(X,Y) = (-12,-5).
tanA = Y/X = --5/-12 = 0.416666,
A = 22.62 Deg. ,
A = 22.62 + 180 = 202.62(Q3).
180 < 202.62 < 270. Q3.
sin(2A) = sin(405.24) = 0.710.
cos(A-4pi/3) = cos(202.62-240) = 0.7946
sin(A/2) = +- sqrt((1-cosA)/2) =
sqrt((1-cos202.62)/2) = 0.9806
Correction:
sinA = -5/13 = Y/r,
A = 157.38 Deg.
sin(2A) = sin(314.76) = -0.710.
cos(A-4pi/3)=cos(157.38-240)=0.1284
sin(A/2) = +- sqrt((1-cosA)/2) =
sqrt((1-cos157.38)/2 = 0.9806.
To find the values of sin(2θ), cos(θ-4π/3), and sin(θ/2) given sinθ = -5/13 and π < θ < 3π/2, follow these steps:
1. Start by finding the value of cosθ using the given information. Since sinθ = -5/13, you can use the Pythagorean identity: cosθ = ±√(1 - sin^2θ).
In this case, since θ is in the second quadrant (where sinθ is negative), cosθ is positive.
⇒ cosθ = √(1 - (-5/13)^2) = √(1 - 25/169) = √(144/169) = 12/13 (positive value).
2. Now, use the trigonometric identities to find sin(2θ):
sin(2θ) = 2sinθcosθ
Plug in the known values: sin(2θ) = 2(-5/13)(12/13) = -120/169.
3. Next, calculate cos(θ - 4π/3):
To find cos(θ - 4π/3), you need to determine the value of θ - 4π/3. Since θ is between π and 3π/2, we can rewrite 4π/3 as 12π/6, which is also equal to 2π.
⇒ θ - 4π/3 = θ - 2π
cos(θ - 4π/3) = cos(θ - 2π)
Recall that cosθ = 12/13 (positive value) from step 1.
cos(θ - 4π/3) = cosθ (since subtracting 2π does not change the value of the cosine) = 12/13.
4. Finally, find sin(θ/2):
To find sin(θ/2), you divide θ by 2.
sin(θ/2) = sin(θ) / (2√(1 + cosθ))
Plug in the known values: sin(θ/2) = (-5/13) / (2√(1 + 12/13))
Simplify the expression: sin(θ/2) = (-5/13) / (2√(25/13)) = (-5/13) / (2(5/13)) = -1/2.
Therefore, the values are:
sin(2θ) = -120/169
cos(θ - 4π/3) = 12/13
sin(θ/2) = -1/2.