Find sin2theta, cos2theta if csctheta=13/12 and lies in quadrant II.
Thank you!
If csc Ø = 13/12, the sinØ = 12/13
so we are dealing with the 12-5-13 right-angled triangle in II
then cosØ = -5/13
sin 2Ø = 2sinØcosØ
= 2(12/13)(-5/13)
= -120/169
cos 2Ø = cos^2 Ø - sin^2 Ø
= 25/169 - 144/169
= -119/169
To find sin(2θ) and cos(2θ), we first need to determine the value of sin(θ) and cos(θ) using the given information.
Given that csc(θ) = 13/12 and θ lies in quadrant II, we can determine the values of sin(θ) and cos(θ) by using the reciprocal trigonometric identities:
csc(θ) = 1/sin(θ)
Therefore, 1/sin(θ) = 13/12
To isolate sin(θ), we can take the reciprocal of both sides:
sin(θ) = 12/13
Since θ lies in quadrant II, sin(θ) is positive, but cos(θ) is negative based on the quadrant-specific signs.
Now that we have determined the values of sin(θ) and cos(θ), we can find sin(2θ) and cos(2θ) using the double-angle identities:
sin(2θ) = 2sin(θ)cos(θ)
cos(2θ) = cos^2(θ) - sin^2(θ)
Plugging in the values we obtained earlier:
sin(2θ) = 2(12/13)(-√(1 - (12/13)^2))
cos(2θ) = cos^2(θ) - sin^2(θ) = (-√(1 - (12/13)^2))^2 - (12/13)^2
Simplifying these expressions will give you the final values for sin(2θ) and cos(2θ).