Find sin x/2, cos x/2, and tan x/2 from the given information.
[1]. sin x = 5/13, 0° < x < 90°
1). sin x/2 =________.
2). cos x/2 =________.
3). tan x/2 =________.
[Note: Plz help with these 3]
Use Trigonometric Identities :
sin ( x / 2 ) = + OR - sqroot [ ( 1 - cos x ) / 2 ]
cos ( x / 2 ) = + OR - sqroot [ ( 1 + cos x ) / 2 ]
0° < x < 90° meen x lies in Quadrant I
In Quadrant I sin x and cos x are positive so :
sin ( x / 2 ) = sqroot [ ( 1 - cos x ) / 2 ]
cos ( x / 2 ) = sqroot [ ( 1 + cos x ) / 2 ]
If sin x = 5 / 13 then :
cos x = sqroot [ 1 - ( sin x ) ^ 2 ]
cos x = sqroot [ 1 - ( 5 / 13 ) ^ 2 ] =
sqroot [ 1 - 25 / 169 ] =
sqroot [ 169 / 169 - 25 / 169 ] =
sqroot [ 144 / 169 ] = 12 / 13
cos x = 12 / 13
Now :
sin ( x / 2 ) = sqroot [ ( 1 - cos x ) / 2 ]
sin ( x / 2 ) = sqroot [ ( 1 - 12 / 13 ) / 2 ] =
sqroot [ ( 13 / 13 - 12 / 13 ) / 2 ] =
sqroot [ ( 1 / 13 ) / 2 ] =
sqroot [ 1 / ( 2 * 13 ) ] =
sqroot ( 1 / 26 ) = 1 / sqroot ( 26 )
sin ( x / 2 ) = 1 / sqroot ( 26 )
cos ( x / 2 ) = sqroot [ ( 1 + cos x ) / 2 ]
cos ( x / 2 ) = sqroot [ ( 1 + 12 / 13 ) / 2 ] =
sqroot [ ( 13 / 13 + 12 / 13 ) / 2 ] =
sqroot [ ( 25 / 13 ) / 2 ] =
sqroot [ 25 / ( 2 * 13 ) ] =
sqroot ( 25 / 26 ) = 5 / sqroot ( 26 )
cos ( x / 2 ) = 5 / sqroot ( 26 )
tan ( x / 2 ) = sin ( x / 2 ) / cos ( x / 2 )
tan ( x / 2 ) = [ 1 / sqroot ( 26 ) ] / [ 5 / sqroot ( 26 ) ] =
[ 1 * sqroot ( 26 ) ] / [ 5 * sqroot ( 26 ) ] = 1 / 5
tan ( x / 2 ) = 1 / 5
1). sin x/2 = I would tell you the answer, but I'm afraid it would be a sine of disrespect. Just kidding! To find sin x/2, you can use the half-angle formula for sine:
sin (x/2) = ±√[(1 - cos x)/2]
Considering that x is in the first quadrant (0° < x < 90°) and sin x = 5/13, we know that sin x/2 is positive. Using the half-angle formula, we get:
sin (x/2) = √[(1 - cos x)/2]
Now, let's plug in the given value for sin x:
sin (x/2) = √[(1 - cos x)/2] = √[(1 - (12/13))/2] = √[(1/13)/2] = √[1/26] = 1/√26
So, sin x/2 = 1/√26.
2). cos x/2 = You'll be happy to know that I won't "cos" you any trouble with this one! We can find cos x/2 using the half-angle formula for cosine:
cos (x/2) = ±√[(1 + cos x)/2]
Considering that x is in the first quadrant (0° < x < 90°) and sin x = 5/13, we know that cos x/2 is positive. Using the half-angle formula, we get:
cos (x/2) = √[(1 + cos x)/2]
Now, let's plug in the given value for sin x:
cos (x/2) = √[(1 + cos x)/2] = √[(1 + (12/13))/2] = √[(25/13)/2] = √[25/26] = √25/√26 = 5/√26
So, cos x/2 = 5/√26.
3). tan x/2 = Oh, tan x/2... it's a fantastic ratio and a half! To find tan x/2, we can use the formula:
tan (x/2) = sin(x) / (1 + cos(x))
Now, let's substitute the given values:
tan (x/2) = (sin(x)) / (1 + cos(x)) = (5/13) / (1 + (12/13)) = (5/13) / (25/13) = 5/25 = 1/5
So, tan x/2 = 1/5.
That's it! Hope the "sinister" math wasn't too scary!
To find sin x/2, cos x/2, and tan x/2 from the given information, we can use the half-angle formulas.
The half-angle formulas are:
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 sin x = 5/13 and 0° < x < 90°, we can use the Pythagorean identity sin^2(x) + cos^2(x) = 1 to find cos x.
Using sin x = 5/13, we can find cos x:
cos x = √(1 - sin^2(x)) = √(1 - (5/13)^2) = √(1 - 25/169) = √(144/169) = 12/13 (positive because it's in the first quadrant)
Now we can substitute cos x = 12/13 into the half-angle formulas:
1). sin x/2 = ±√[(1 - cos x)/2]
sin x/2 = ±√[(1 - 12/13)/2] = ±√[(1/13)/2] = ±√(1/26) = ±(1/√26)
Since 0° < x < 90°, x/2 will also be in the first quadrant, so we take the positive value:
sin x/2 = 1/√26
2). cos x/2 = ±√[(1 + cos x)/2]
cos x/2 = ±√[(1 + 12/13)/2] = ±√[(25/13)/2] = ±√(25/26) = ±(5/√26)
Again, x/2 is in the first quadrant, so we take the positive value:
cos x/2 = 5/√26
3). tan x/2 = sin x / (1 + cos x)
tan x/2 = (5/13) / (1 + 12/13) = (5/13) / (25/13) = 5/25 = 1/5
So, the final answers are:
1). sin x/2 = 1/√26
2). cos x/2 = 5/√26
3). tan x/2 = 1/5
To find sin x/2, cos x/2, and tan x/2, we can use the half-angle identities.
1) First, let's find sin x/2. The half-angle identity for sine is:
sin(x/2) = ±√((1 - cos x)/2)
Since 0° < x < 90°, we are in the first quadrant. In the first quadrant, sin x/2 is always positive. So we can ignore the negative value.
sin(x/2) = √((1 - cos x)/2)
Now, let's substitute the given value of sin x:
sin(x/2) = √((1 - 5/13)/2)
= √(8/26)
= √(4/13)
= 2/√13
Therefore, sin x/2 = 2/√13.
2) Next, let's find cos x/2. The half-angle identity for cosine is:
cos(x/2) = ±√((1 + cos x)/2)
Since 0° < x < 90°, we are in the first quadrant. In the first quadrant, cos x/2 is always positive. So we can ignore the negative value.
cos(x/2) = √((1 + cos x)/2)
Now, let's substitute the given value of sin x:
cos(x/2) = √((1 + 5/13)/2)
= √(18/26)
= √(9/13)
= 3/√13
Therefore, cos x/2 = 3/√13.
3) Finally, let's find tan x/2. The half-angle identity for tangent is:
tan(x/2) = sin x / (1 + cos x)
Now, let's substitute the given values of sin x and cos x:
tan(x/2) = (5/13) / (1 + (3/√13))
= (5/13) / (1 + (3/√13)) * (√13/√13)
= (5/13) * (√13/√13) / ((√13 + 3)/√13)
= (5√13)/(13(√13 + 3))
Therefore, tan x/2 = (5√13)/(13(√13 + 3)).