Find sin x/2, cos x/2, and tan x/2 from the given information. (show work please) thank you!!!
cos x = − 4/5, 180° < x < 270°
To find sin(x/2), cos(x/2), and tan(x/2) from the given information, we can use the half-angle formulas.
1. First, let's find sin(x) using the Pythagorean identity: sin^2(x) + cos^2(x) = 1
Since we are given cos(x) = -4/5, we can substitute this value into the equation:
sin^2(x) + (-4/5)^2 = 1
sin^2(x) + 16/25 = 1
sin^2(x) = 1 - 16/25
sin^2(x) = 9/25
Taking the square root of both sides, we get:
sin(x) = ±√(9/25) = ±3/5
2. Since we know that x is between 180° and 270° (which lies in the third quadrant), sin(x) must be negative. Therefore, sin(x) = -3/5
3. Now let's find cos(x/2) using the half-angle formula: cos(x/2) = ± √((1 + cos(x)) / 2)
Substituting the given value cos(x) = -4/5:
cos(x/2) = ± √((1 + (-4/5)) / 2)
cos(x/2) = ± √(1/5) = ±√5 / 5
4. Finally, let's find tan(x/2) using the half-angle formula: tan(x/2) = sin(x) / (1 + cos(x))
Substituting the known values:
tan(x/2) = (-3/5) / (1 + (-4/5))
tan(x/2) = (-3/5) / (1/5)
tan(x/2) = (-3/5) * (5/1)
tan(x/2) = -3
Therefore, the answers are:
sin(x/2) = -3/5
cos(x/2) = ±√5 / 5
tan(x/2) = -3
To find sin(x/2), cos(x/2), and tan(x/2) from the given information, we first need to determine the value of x/2.
Since cos(x) is given as -4/5 and the given range of x is 180° < x < 270°, we can determine the quadrant in which x lies. In the given range, x is between 180° and 270°, which means it lies in the third quadrant. In the third quadrant, cos(x) is negative. Therefore, we know that cos(x) = -4/5.
To find sin(x), we can use the Pythagorean identity: sin^2(x) + cos^2(x) = 1. We know the value of cos(x) (-4/5), so we can substitute it into the equation and solve for sin(x).
(-4/5)^2 + sin^2(x) = 1
16/25 + sin^2(x) = 1
sin^2(x) = 1 - 16/25 = 9/25
Taking the square root of both sides gives us:
sin(x) = ± √(9/25) = ± 3/5
Now we can find sin(x/2), cos(x/2), and tan(x/2) using the half-angle formulas:
sin(x/2) = ± √[(1 - cos(x))/2]
Since cos(x) = -4/5, we can plug it into the formula:
sin(x/2) = ± √[(1 - (-4/5))/2] = ± √[(1 + 4/5)/2] = ± √(9/10) = ± 3/√10 = ± (3√10) / 10
cos(x/2) = ± √[(1 + cos(x))/2]
Substituting cos(x) = -4/5:
cos(x/2) = ± √[(1 + (-4/5))/2] = ± √[(1 - 4/5)/2] = ± √(1/10) = ± 1/√10 = ± √10 / 10
tan(x/2) = sin(x/2) / cos(x/2)
Substituting the values of sin(x/2) and cos(x/2) we found:
tan(x/2) = ((±3√10)/10) / ((±√10)/10) = ±3/1 = ±3
Therefore, the values of sin(x/2), cos(x/2), and tan(x/2) are:
sin(x/2) = ± (3√10) / 10
cos(x/2) = ± √10 / 10
tan(x/2) = ±3
so the angle x is in III and
cosx = -4/5, sinx = -3/5
we know cos 2A = 2cos^2 A -1
then
cos x = 2cos^2 (x/2) - 1
-4/5 = 2cos^2 (x/2) - 1
1/5 = 2cos^2 (/2)
cos(x/2) = -1/√10
then using Pythagoras and CAST
sin(x/2) = -3/√10
and
tan (x/2) = (-3/√10) / (-1/√10) = 3