Asked by james
If Ѳ is in Quadrant II, and cos Ѳ=-3/4 , find an exact value for sin 2Ѳ.
Answers
Answered by
Reiny
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
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
Answered by
Maryam
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
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
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