given sinθ=-5/13 and π<θ<3π/2 find
sin2θ
cos( θ-4π/3)
sin(θ/2)
can some1 please help me with this?
Let's go through this again.
θ is in the 3rd quadrant, so x and y are both negative, and the hypotenuse h = 13
sinθ = -5/13
cosθ = -12/13
Use these values in your half-angle and double-angle and sum/difference formulas.
sin 2θ = 2 sinθ cosθ = 2(-5/13)(-12/13) = 120/169
cos(θ-4π/3) = cosθ cos4π/3 + sinθ sin4π/3
= (-12/13)(-1/2) + (-5/13)(-√3/2)
= 12/26 + 5√3/26
sin θ/2 = √((1-cosθ)/2)
= √((1 + 12/13)/2)
= √(25/26)
Use this same method for all your other similar problems.
To find the value of sin(2θ), we can use the double angle formula:
sin(2θ) = 2sin(θ)cos(θ)
Given that sin(θ) = -5/13, we can find cos(θ) using the Pythagorean identity:
cos^2(θ) + sin^2(θ) = 1
cos^2(θ) + (-5/13)^2 = 1
cos^2(θ) + 25/169 = 1
cos^2(θ) = 1 - 25/169
cos^2(θ) = 144/169
cos(θ) = √(144/169)
cos(θ) = 12/13 (taking the positive square root since θ is in the third quadrant)
Now substituting these values into the formula, we get:
sin(2θ) = 2sin(θ)cos(θ)
= 2 * (-5/13) * (12/13)
= -120/169
So, sin(2θ) = -120/169.
To find cos(θ - 4π/3), we use the difference of angles formula:
cos(θ - 4π/3) = cos(θ)cos(4π/3) + sin(θ)sin(4π/3)
We already know the values of cos(θ) and sin(θ) from the previous calculation. Now, we can find cos(4π/3) and sin(4π/3).
We know that cos(4π/3) = cos(π + π/3), which is the same as cos(π/3), and sin(4π/3) = sin(π + π/3), which is the same as sin(π/3).
Since π < θ < 3π/2, we are in the third quadrant where cos(θ) is negative and sin(θ) is negative.
Using the unit circle or trigonometric identities, we can find that cos(π/3) = 1/2 and sin(π/3) = √3/2.
Replacing these values in the formula, we get:
cos(θ - 4π/3) = (12/13) * (1/2) + (-5/13) * (√3/2)
= 12/26 - 5√3/26
= (12 - 5√3)/26
So, cos(θ - 4π/3) = (12 - 5√3)/26.
To find sin(θ/2), we can use the half-angle formula for sine:
sin(θ/2) = ±√((1 - cos(θ))/2)
Given that cos(θ) = 12/13 (positive value since θ is in the third quadrant), we can calculate:
sin(θ/2) = ±√((1 - 12/13)/2)
= ±√(1/26)
= ±1/√26
So, sin(θ/2) = ±1/√26.
Please note that the ± sign indicates that there are two possible solutions, which depend on the exact value of θ within the given range.