cos(5π//6) = -(√3)/2 ≈ -0.866
4/5
sin(t) = 2/3 sin2(t) + cos2(t) = 1 so cos(t) = ± sqrt[1 - sin2(t)] but because t is in the first quadrant, cos(t) > 0 so cos(t) = + sqrt[1 - sin2(t)] = sqrt[1 - 4/9] = sqrt[5/9] = sqrt(5)/3 Then sec(t) = 1/cos(t) = 1/sqrt(5)/3 = 3/sqrt(5) = 3*sqrt(5)/5
The integral of cos 5x is 1/5 sin (5x)
the height is 4.33 units cos 30 = .866 15 / 3 = 5 5 x .866 = 4.33
-5
f(-2,3) = 11 f(5,-3) = -5 f(1,4) = 22, maximum
1.6667
The only variable on the right hand side is sin(x). The maximum value of sin(x) is 1. So, the max value of 3sin(x) is 3*1 = 3 and so, the max value of 3sin(x) + 2 is 3+2 = 5.
The face value of 3 is 3: the value of 3 is 3000The face value of 5 is 5: the value of 5 is 500The face value of 3 is 3: the value of 3 is 3000The face value of 5 is 5: the value of 5 is 500The face value of 3 is 3: the value of 3 is 3000The face value of 5 is 5: the value of 5 is 500The face value of 3 is 3: the value of 3 is 3000The face value of 5 is 5: the value of 5 is 500
cos(5π//6) = -(√3)/2 ≈ -0.866
4/5
sin(t) = 2/3 sin2(t) + cos2(t) = 1 so cos(t) = ± sqrt[1 - sin2(t)] but because t is in the first quadrant, cos(t) > 0 so cos(t) = + sqrt[1 - sin2(t)] = sqrt[1 - 4/9] = sqrt[5/9] = sqrt(5)/3 Then sec(t) = 1/cos(t) = 1/sqrt(5)/3 = 3/sqrt(5) = 3*sqrt(5)/5
sin^5 2x = 1/8 sin2x (cos(8x) - 4 cos(4x)+3)
integral sin(3 x) cos(5 x) dx = 1/16 (8 cos^2(x)-cos(8 x))+C
Are you sure sin x = 5/3? The hypotenuse is always longer than the legs, and saying sine x = 5/3 means the leg is 5 units long and the hypotenuse is 3 units long. My best guess would be cos x = 5/4, since 5/3 points toward a 3/4/5 triangle, even if it is set up incorrectly. If the question is indeed correct, this can only be solved using complex numbers: using the identity sin^2(x) + cos^2(x) = 1 it follows that cos^2(x) = -16/9 Therefore cos(x) = + or minus 4/3 i, where i = sqrt(-1)
It would be 1 over square root 5.