13pi/12 lies between pi and 2pi, which means sin(13pi/12) < 0
Recall the double angle identity,
sin^2(x) = (1 - cos(2x))/2
If we let x = 13pi/12, then
sin(13pi/12) = - sqrt[(1 - cos(13pi/6))/2]
where we took the negative square root because we expect a negative value.
Now, because cosine has a period of 2pi, we have
cos(13pi/6) = cos(2pi + pi/6) = cos(pi/6) = sqrt[3]/2
Then
sin(13pi/12) = - sqrt[(1 - sqrt[3]/2)/2]
sin(13pi/12) = - sqrt[2 - sqrt[3]]/2
Answer:
Step-by-step explanation:
Answer:
The answer is in the photo
Step-by-step explanation:
Add y to both sides in 1st equation
-2x=y
sub -2x for y in2nd equation
3x-(-2x)=10
3x+2x=10
5x=10
divide 5
x=2
sub back
-2x=y
-2(2)=y
-4=y
(x,y)
(2,-4)
C is ans
