Question 236887: Find (2x) given that sin(x)=2/5 and x is in quadrant II. Give exact value for sin(2x) Found 2 solutions by Alan3354, nyc_function:Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! Find (2x) given that sin(x)=2/5 and x is in quadrant II. Give exact value for sin(2x)
----------------
-----------
sin(2x) = 2sin(x)cos(x)
= 2*(2/5)*sqrt(21)/5
= 4sqrt(21)/25
cos(x) is negative in Q2, so use the - sqrt
sin(2x) = -4sqrt(21)/25
You can put this solution on YOUR website! Sine is positive in quadrants 1 and 2.
sin2x is one of our double trig identities.
sin2x = 2sinxcosx
Given sinx = 2/5, we know that opposite side of right triangle in quadrant 2 is 2 and the hypotenuse is 5. We also know that sine = opposite/hypotenuse. We need to find the adjacent side.
Let A = adjacent side
A^2 + 2^2 = 5^2
A^2 + 4 = 25
A^2 = 25 - 4
A^2 = 21
A = sqrt{21}.
Now that we know the value of the adjacent side, we can find cosx.
cosx = adjacent side over hypotenuse.
cosx = sqrt{21}/5
Of course, cosine is negative in quadrant 2.
So, the correct answer for cosx = -sqrt{21}/5
We can now plug the values for sinx and cosx into the double trig identity for sin2x and simplify.
sin2x = 2((2/5)(-sqrt{21}/5)
sin2x = -4(sqrt{21})/25
