If Ѳ is in Quadrant II, and cos Ѳ=-3/4 , find an exact value for sin 2Ѳ.
sin2A= 2sinAcosA
now we know that cosA=-3/4, and theta is in the second quadrant so you sketch a triangle in the second quadrant, remember the bowtie method
-3 is a leg and 4 is the hypotenuse based on sohcahtoa
plug these two into the pythagorean theorem to find the missing leg.
then you know that the missing leg is sqrt of 7
so sinA is sqrt7/4
2sinAcosA= sin2A
2(sqrt7/4)(-3/4)= sin2A
= -6(sqrt7)/16
= -3(sqrt7)/8
To find the exact value of sin 2Ѳ, we can use the double angle identity for sine:
sin 2Ѳ = 2sin Ѳ * cos Ѳ
Since we are given that Ѳ is in Quadrant II, we know that sine is positive in Quadrant II. We are also given that cos Ѳ = -3/4.
Using the Pythagorean identity, we can find the value of sin Ѳ:
sin^2 Ѳ = 1 - cos^2 Ѳ
sin^2 Ѳ = 1 - (-3/4)^2
sin^2 Ѳ = 1 - 9/16
sin^2 Ѳ = 16/16 - 9/16
sin^2 Ѳ = 7/16
Taking the square root of both sides, we get:
sin Ѳ = √7/4
Now, we can substitute the values of sin Ѳ and cos Ѳ into the double angle identity:
sin 2Ѳ = 2(√7/4)(-3/4)
sin 2Ѳ = -6√7/16
Therefore, the exact value of sin 2Ѳ is -6√7/16.
To find an exact value for sin 2Ѳ, we need to use trigonometric identities. One such identity is the double-angle identity for sin:
sin 2Ѳ = 2sinѲcosѲ
Since we know cos Ѳ = -3/4, let's find sin Ѳ first. To do that, we can use the Pythagorean identity:
sin²Ѳ + cos²Ѳ = 1
Replacing cosѲ with -3/4, we get:
sin²Ѳ + (-3/4)² = 1
sin²Ѳ + 9/16 = 1
sin²Ѳ = 1 - 9/16
sin²Ѳ = 16/16 - 9/16
sin²Ѳ = 7/16
Taking the square root of both sides (since we want an exact value), we get:
sinѲ = ±√(7/16)
Since we are in Quadrant II, where sin is positive, we take the positive square root:
sinѲ = √(7/16)
= √7/√16
= √7/4
Now that we have sinѲ, we can compute sin 2Ѳ:
sin 2Ѳ = 2sinѲcosѲ
= 2(√7/4)(-3/4)
= -6√7/16
So, the exact value of sin 2Ѳ when cosѲ = -3/4 is -6√7/16.
Make a sketch of your triangle in quadrant II to see that
x = -3, r = 4 and from Pythagoras that y = √7 in II
so sinØ = √7/4
sin 2Ø = 2sinØcosØ
= 2((√7/4)(-3/4)
= -3√7/8