Given that cos2x=7/12 and "270 equal or < 2x equal or < 360", find sinx.
Please help and Thank you
In these type of quesstions it usually asks for the "exact" value of ....
cos 2x = 7/12
cos 2x = 1 - 2sin^2 x
2 sin^2 x = 1 - cos 2x = 1 - 7/12 = 5/12
sin^2 x = 5/24
sin x = ± √(5/24)
but 270 < 2x ≤ 360
135 ≤ x ≤ 180 ---> x in in I or II, so
sinx = +√5/√24 = √30/12
how did you get the �ã30/12?
sqrt(5)/2sqrt(6)
when you rationalize the denominator you times both top and bottom by sqrt(6) only not the 2 right?
To find sin(x), we can use the Pythagorean identity: sin^2(x) + cos^2(x) = 1.
First, let's solve for sin(2x) using the given information:
cos(2x) = 7/12
Using the double-angle formula for cosine, we have:
cos(2x) = 2 * cos^2(x) - 1
7/12 = 2 * cos^2(x) - 1
Rearranging the equation, we get:
2 * cos^2(x) = 1 + 7/12
2 * cos^2(x) = 19/12
Dividing both sides by 2, we have:
cos^2(x) = 19/24
Now, let's solve for sin(x):
sin^2(x) + cos^2(x) = 1
sin^2(x) = 1 - cos^2(x)
sin^2(x) = 1 - 19/24
sin^2(x) = 5/24
Taking the square root of both sides, we get:
sin(x) = ±√(5/24)
Since we are given that "270 ≤ 2x ≤ 360", we know that x is in the fourth quadrant (between 180° and 270°), where sin(x) is negative.
Therefore, sin(x) = -√(5/24), which simplifies to:
sin(x) = -√5/2√6
So, sin(x) is equal to -√5/2√6.
Note: In this explanation, we assumed that 2x refers to an angle measurement in degrees. If it represents radians, the calculations would be slightly different.