I am trying to submit this homework in but i guess i'm not doing it in exact values because it is not accepting it. I know i'm supposed to be using half angle formulas but maybe the quadrants are messing me up. Please help!
Find sin x/2 , cos x/2 , and tan x/2 from the given information. tan x = square root (15) , 0° < x < 90°
Find sin x/2 , cos x/2 , and tan x/2 from the given information.
cos x = (−12/13) 180° < x < 270°
if tanx = √15 or √15/1 , and x is in quad I
make a sketch of a right-angled triangle with opposite = √15
adjacent = 1
r^2 = (√15)^2 + 1^2 = 16
r = 4
then sinx = √15/4 and cosx = 1/4
if x is in quad I, then certainly x/2 is in quad I and everybody stays nicely positive
we know:
cosx = 1 - 2sin^2 (x/2)
1/4 = 1 - 2sin^2 (x/2)
2sin^2 (x/2) = 3/4
sin^2 x/2 = 3/8 = 6/16
sin (x/2) = √6/4
also sin^2 (x/2) + cos^2 (x/2) = 1
cos^2 (x/2) = 1 - 6/16 = 10/16
cos (x/2) = √10/4
also tan(x/2) = sin(x/2) / cos(x/2)
= (√6/4) / (√10/4)
= √6/√10
or √60/10 = 2√15/10 = √15/5
for the second, since x is in the third,
x/2 is in II
(e.g. if x = 182°, x = 91° ; if x = 278 , x/2 = 139°
given: cosx = - 12/13, other side is 5
sinx = -5/13
same as before ...
cos x = 1 - 2sin^2 (x/2)
-12/13 = 1 - 2sin^2 (x/2)
2sin^2 (x/2) = 1 + 12/13 = 25/13
sin^2 (x/2) = 25/26
sin (x/2) = ± 5/√26 , but x/2 is in II, so
sin (x/2) = 5/√26 or 5√26/26
again, as before:
sin^2 (x/2) + cos^2 (x/2) = 1
25/26 + cos^2(x/2) = 1
cos^2(x/2) = 1-25/26 = 1/26
cos(x/2) = ±1/√26, but in II cos(x/2) = -1/√26
finally, tan(x/2) = (5/√26) / (-1/√26) = -5
Sure, I can help you with that. Let's start with the first problem:
Problem 1:
Given tan(x) = √15 and 0° < x < 90°.
Step 1: Use the half-angle formulas for sine, cosine, and tangent.
sin(x/2) = ±√[(1 - cos(x))/2]
cos(x/2) = ±√[(1 + cos(x))/2]
tan(x/2) = sin(x)/(1 + cos(x))
Step 2: Calculate the values.
Since tan(x) = √15, we can calculate cos(x) and sin(x) using the Pythagorean identity:
cos(x) = 1 / √(1 + tan^2(x))
sin(x) = tan(x) * cos(x)
cos(x) = 1 / √(1 + (√15)^2)
= 1 / √(1 + 15)
= 1 / √16
= 1 / 4
= 0.25
sin(x) = (√15/4) * (1/4)
= √15 / 16
Now we can substitute these values into the half-angle formulas.
sin(x/2) = ±√[(1 - cos(x))/2]
= ±√[(1 - 0.25)/2]
= ±√[0.75/2]
= ±√0.375
= ±(√3) / 2
cos(x/2) = ±√[(1 + cos(x))/2]
= ±√[(1 + 0.25)/2]
= ±√[1.25/2]
= ±√0.625
= ±(√5) / 4
tan(x/2) = sin(x)/(1 + cos(x))
= (√15 / 16) / (1 + 0.25)
= (√15 / 16) / (1.25)
= (√15 / 16) * (4/5)
= 2√15 / 20
= √15 / 10
Therefore, the values of sin(x/2), cos(x/2), and tan(x/2) are:
sin(x/2) = ±(√3) / 2
cos(x/2) = ±(√5) / 4
tan(x/2) = √15 / 10
Now let's move on to the second problem:
Problem 2:
Given cos(x) = -12/13 and 180° < x < 270°.
Step 1: Use the half-angle formulas for sine, cosine, and tangent.
sin(x/2) = ±√[(1 - cos(x))/2]
cos(x/2) = ±√[(1 + cos(x))/2]
tan(x/2) = sin(x)/(1 + cos(x))
Step 2: Calculate the values.
Since cos(x) = -12/13, we can calculate sin(x) using the Pythagorean identity:
sin(x) = ±√(1 - cos^2(x))
sin(x) = ±√(1 - (-12/13)^2)
= ±√(1 - 144/169)
= ±√(169/169 - 144/169)
= ±√(25/169)
= ±5/13
Now we can substitute these values into the half-angle formulas.
sin(x/2) = ±√[(1 - cos(x))/2]
= ±√[(1 - (-12/13))/2]
= ±√[(1 + 12/13)/2]
= ±√[(25/13)/2]
= ±√(25/26)
= ±5 / (13√2)
cos(x/2) = ±√[(1 + cos(x))/2]
= ±√[(1 + (-12/13))/2]
= ±√[(1 -12/13)/2]
= ±√[(1/13)/2]
= ±√(1/26)
= ±1 / (√26)
tan(x/2) = sin(x)/(1 + cos(x))
= (5/13) / (1 - 12/13)
= (5/13) / (1/13)
= 5
Therefore, the values of sin(x/2), cos(x/2), and tan(x/2) are:
sin(x/2) = ±5 / (13√2)
cos(x/2) = ±1 / (√26)
tan(x/2) = 5
To find the values of sin(x/2), cos(x/2), and tan(x/2), we can use the half-angle formulas. These formulas are derived from the double-angle formulas using some trigonometric identities.
1. For the first problem, where tan(x) = sqrt(15) and 0° < x < 90°:
We know that tan(x) = sin(x) / cos(x). So, we can start by finding sin(x) and cos(x) using the given information.
Given:
tan(x) = sqrt(15)
From the identity tan(x) = sin(x) / cos(x), we have:
sqrt(15) = sin(x) / cos(x)
Cross-multiplying, we get:
sqrt(15) * cos(x) = sin(x)
Now, we can utilize the half-angle formulas for sine and cosine, which are:
sin(x/2) = sqrt((1 - cos(x)) / 2)
cos(x/2) = sqrt((1 + cos(x)) / 2)
Plugging in the values we already have:
sin(x/2) = sqrt((1 - cos(x)) / 2)
= sqrt((1 - sqrt(15) * cos(x)) / 2)
cos(x/2) = sqrt((1 + cos(x)) / 2)
= sqrt((1 + sqrt(15) * cos(x)) / 2)
Finally, we can find tan(x/2) using the formula:
tan(x/2) = sin(x/2) / cos(x/2)
2. For the second problem, where cos(x) = (-12/13) and 180° < x < 270°:
We can start by finding sin(x) using the Pythagorean identity sin^2(x) + cos^2(x) = 1.
Given:
cos(x) = -12/13
Using the Pythagorean identity:
sin^2(x) + cos^2(x) = 1
sin^2(x) + (-12/13)^2 = 1
sin^2(x) + 144/169 = 1
sin^2(x) = 1 - 144/169
sin^2(x) = 25/169
sin(x) = ±sqrt(25/169)
Since 0° < x < 90°, sin(x) is positive. Therefore, sin(x) = sqrt(25/169) = 5/13.
Now, we can apply the half-angle formulas to find sin(x/2) and cos(x/2) as follows:
sin(x/2) = ±sqrt((1 - cos(x)) / 2)
= ±sqrt((1 - (-12/13)) / 2)
= ±sqrt(25/26)
= ±5/√26
cos(x/2) = ±sqrt((1 + cos(x)) / 2)
= ±sqrt((1 + (-12/13)) / 2)
= ±sqrt(1/26)
= ±1/√26
Finally, we can find tan(x/2) using the formula:
tan(x/2) = sin(x/2) / cos(x/2)
= (±5/√26) / (±1/√26)
= ±5/1
= ±5
Remember to adjust the signs based on the given range of x to get the appropriate values for sin(x/2), cos(x/2), and tan(x/2).