cos 4x=7/18. 4x is on quadrant 4. sin x?
3pi/2 < 4x < 2pi, so
3pi/8 < x < pi/2
sin2x = √((1+cos4x)/2)
sinx = √((1+cos2x)/2)
sin2x = √((1+7/18)/2) = √(25/36) = 5/6
so, cos2x = √11/6
sinx = √((1+√11/6)/2) = 1/2 √(2+√11/3)
oops -- got my formulas mangled:
cos2x = √((1+cos4x)/2)
sinx = √(1-cos2x)/2)
cos2x = 5/6, as above
sinx = √((1 - 5/6)/2) = √(1/12) = 1/(2√3) = √3/6
Now, that's a lot less work!
To find the value of sin(x) when cos(4x) = 7/18 and 4x is in quadrant 4, we can use the identity sin^2(x) + cos^2(x) = 1.
Given that cos(4x) = 7/18, we can substitute this value into the identity:
sin^2(4x) + (7/18)^2 = 1
To solve for sin(4x), we can take the square root of both sides of the equation:
sin(4x) = √(1 - (7/18)^2)
Next, we can find sin(x) by dividing sin(4x) by 2:
sin(x) = sin(4x) / 2
So, the value of sin(x) when cos(4x) = 7/18 and 4x is in quadrant 4 is sin(4x)/2.
To find sin(x), we first need to find the value of x.
From the given equation cos(4x) = 7/18, we can determine that cos(x) = cos(4x) since the value is the same in quadrant 4.
Since cos(x) = cos(4x), we can rewrite the equation as cos(x) = 7/18.
To find sin(x), we can use the Pythagorean identity: sin^2(x) + cos^2(x) = 1.
In this case, we know cos(x) = 7/18, so we can substitute it into the equation:
sin^2(x) + (7/18)^2 = 1
To solve for sin(x), we can rearrange the equation:
sin^2(x) = 1 - (7/18)^2
sin^2(x) = 1 - 49/324
sin^2(x) = 275/324
Taking the square root of both sides gives:
sin(x) = ±√(275/324)
Since we know that x lies in quadrant 4, where sin(x) is positive, we can disregard the negative sign:
sin(x) = √(275/324)
Therefore, sin(x) is approximately equal to √(275/324).