tanx = -3/4. x€ 2nd quadrant find sin (x/2), cos(x/2), tan(x/2)
Make your sketch of the corresponding right-angled triangle in quadrant II
you must have: x = -4, y = +3, r = 5
then sinx = 3/5, cosx = -4/5
recall: cos 2A = 1 - 2sin^2 A = 2 cos^2 A - 1
or
cos x = 1 - 2sin^2 (x/2) = 2cos^2 (x/2) - 1
1 - 2sin^2 (x/2) = -4/5
-2sin^2 (x/2) = -1 - 4/5
sin^2 (x/2) = 9/10
sin (x/2) = 3/√10 , (since x/2 is in quadrant I)
similarly, you can find cos(x/2) and from there tan(x/2)
To find sin(x/2), cos(x/2), and tan(x/2), we can use the following trigonometric identities:
1. sin(x/2) = ± √((1 - cos(x)) / 2)
2. cos(x/2) = ± √((1 + cos(x)) / 2)
3. tan(x/2) = sin(x) / (1 + cos(x))
Given that tan(x) = -3/4 and x is in the 2nd quadrant, we can determine the values of cos(x) and sin(x) as follows:
1. tan(x) = sin(x) / cos(x)
-3/4 = sin(x) / cos(x)
sin(x) = -3/4 * cos(x)
2. By using the Pythagorean identity: sin^2(x) + cos^2(x) = 1, we can find cos(x) as follows:
(-3/4 * cos(x))^2 + cos^2(x) = 1
9/16 * cos^2(x) + cos^2(x) = 1
(9/16 + 1) * cos^2(x) = 1
(25/16) * cos^2(x) = 1
cos^2(x) = 16/25
cos(x) = ± √(16/25)
cos(x) = ± 4/5
Since x is in the 2nd quadrant, where cosine is negative, cos(x) = -4/5.
Now, substituting the value of cos(x) into the equation sin(x) = -3/4 * cos(x):
sin(x) = -3/4 * (-4/5) = 3/5
Now, we can find sin(x/2), cos(x/2), and tan(x/2) using the identities mentioned earlier:
1. sin(x/2) = ± √((1 - cos(x)) / 2)
sin(x/2) = ± √((1 - (-4/5)) / 2)
sin(x/2) = ± √((1 + 4/5) / 2)
sin(x/2) = ± √(9/10) = ± 3/√10
2. cos(x/2) = ± √((1 + cos(x)) / 2)
cos(x/2) = ± √((1 + (-4/5)) / 2)
cos(x/2) = ± √((1 - 4/5) / 2)
cos(x/2) = ± √(1/10) = ± 1/√10
3. tan(x/2) = sin(x) / (1 + cos(x))
tan(x/2) = (3/5) / (1 + (-4/5))
tan(x/2) = (3/5) / (1 - 4/5)
tan(x/2) = (3/5) / (1/5)
tan(x/2) = 3
Therefore, in the 2nd quadrant, sin(x/2) = ± 3/√10, cos(x/2) = ± 1/√10, and tan(x/2) = 3.
To find the values of sin(x/2), cos(x/2), and tan(x/2), we can use the double angle formulas for sine, cosine, and tangent.
Given that tan(x) = -3/4 and x is in the 2nd quadrant, we can deduce that x lies between π and (3π/2). Since tan(x) is negative, both sin(x) and cos(x) are negative in the 2nd quadrant.
First, let's find sin(x) and cos(x) using the given information about tan(x).
We know that tan(x) = sin(x)/cos(x), so we can express sin(x) and cos(x) in terms of tan(x) using the trigonometric identity:
tan(x) = sin(x)/cos(x)
(-3/4) = sin(x)/cos(x)
sin(x) = (4/3)(-3) = -4/3
Now, to find cos(x), we'll use the Pythagorean identity:
cos^2(x) = 1 - sin^2(x)
cos^2(x) = 1 - (-4/3)^2
cos^2(x) = 1 - 16/9
cos^2(x) = 9/9 - 16/9
cos^2(x) = -7/9
Since cos(x) is negative in the 2nd quadrant, we'll take the negative square root of -7/9:
cos(x) = -√(-7/9) = -√7/3
Now that we have sin(x) and cos(x), we can use the double angle formulas to find sin(x/2), cos(x/2), and tan(x/2).
sin(x/2) = ±√((1 - cos(x))/2)
cos(x/2) = ±√((1 + cos(x))/2)
tan(x/2) = sin(x/2)/cos(x/2)
Since x is in the 2nd quadrant, we can conclude that sin(x/2) and cos(x/2) are negative, and tan(x/2) is positive.
Now, let's substitute the values we obtained for sin(x) and cos(x) into these formulas:
sin(x/2) = -√((1 - (-√7/3))/2) = -√((1 + √7)/6)
cos(x/2) = -√((1 + (-√7/3))/2) = -√((1 - √7)/6)
tan(x/2) = sin(x/2)/cos(x/2) = -√((1 + √7)/(1 - √7))
Therefore, sin(x/2) = -√((1 + √7)/6), cos(x/2) = -√((1 - √7)/6), and tan(x/2) = -√((1 + √7)/(1 - √7)).