find sin2x, cos2x, and tan2x if sinx= -1/sqrt 5 and x terminates in quadrant II
To find sin(2x), cos(2x), and tan(2x), we first need to find sin(x), cos(x), and tan(x).
Given that sin(x) = -1/√5 and x terminates in the second quadrant, we can use the Pythagorean Identity to find cos(x):
cos(x) = √(1 - sin^2(x))
= √(1 - (-1/√5)^2)
= √(1 - 1/5)
= √(4/5)
= 2√5 / 5
Now we can calculate sin(2x), cos(2x), and tan(2x) using double-angle formulas:
sin(2x) = 2sin(x)cos(x)
= 2 * (-1/√5) * (2√5 / 5)
= -4/5
cos(2x) = cos^2(x) - sin^2(x)
= (2√5/5)^2 - (-1/√5)^2
= 4/5 - 1/5
= 3/5
tan(2x) = sin(2x) / cos(2x)
= (-4/5) / (3/5)
= -4/3
Therefore, sin(2x) = -4/5, cos(2x) = 3/5, and tan(2x) = -4/3.
To find the values of sin(2x), cos(2x), and tan(2x) given that sin(x) = -1/√5 and x terminates in quadrant II, we can use various trigonometric identities.
First, let's determine the value of cos(x) using the Pythagorean identity:
sin(x) = -1/√5
cos^2(x) = 1 - sin^2(x)
cos^2(x) = 1 - (-1/√5)^2
cos^2(x) = 1 - 1/5
cos^2(x) = 4/5
Since x terminates in quadrant II, the cosine value is negative. Therefore, cos(x) = -√(4/5) = -2/√5.
Now, for the double-angle formulas:
1. sin(2x) = 2sin(x)cos(x)
2. cos(2x) = cos^2(x) - sin^2(x)
3. tan(2x) = sin(2x) / cos(2x)
Let's plug in the values:
1. sin(2x) = 2(-1/√5)(-2/√5) = 2/5
2. cos(2x) = (-2/√5)^2 - (-1/√5)^2 = 4/5 - 1/5 = 3/5
3. tan(2x) = (2/5) / (3/5) = 2/3
Therefore, sin(2x) = 2/5, cos(2x) = 3/5, and tan(2x) = 2/3.
sorry. sinx is not negative in QII.
But draw your triangle. Recall that
sinx = y/r
cosx = x/r
Then just apply your double angle formulas.
Assuming that you meant QIII, then you have
cosx = -2/√5
sin2x = 2sinx cosx = 2(-1/√5)(-2/√5) = 4/5
and so on for the other functions.