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:
Divide each side of the equation by 2.
Step-by-step explanation:
(I took the test so i know its right) Basically, when doing the this by completing the square we need y^2 by itself, so we divide 2y^2-16y=6 by two first.
Answer:
ok
Step-by-step explanation:
what
Answer:
10 to the power of 250
Step-by-step explanation:
i have no clue it is a guess
