1.25
I'm asuming you meant "ten theta"the square of 5 is 2525 divided by 10 is 2.5so theta equals 2.5there you go =)
Yes, it can. If you plot theta and sin(theta) on the same graph, you will see where they intersect. I do not know of an analytical expression for this point; I think only numerical results are possible.
sin(0)=0 and sin(very large number) is approximately equal to that same very large number.
Using x instead of theta, cos2x/cosec2x + cos4x = cos2x*sin2x + cos4x = cos2x*(sin2x + cos2x) = cos2x*1 = cos2x
Yes. (Theta in radians, and then approximately, not exactly.)
It depends if 1 plus tan theta is divided or multiplied by 1 minus tan theta.
I'm asuming you meant "ten theta"the square of 5 is 2525 divided by 10 is 2.5so theta equals 2.5there you go =)
-0.5736
Yes, it can. If you plot theta and sin(theta) on the same graph, you will see where they intersect. I do not know of an analytical expression for this point; I think only numerical results are possible.
sin(0)=0 and sin(very large number) is approximately equal to that same very large number.
Using x instead of theta, cos2x/cosec2x + cos4x = cos2x*sin2x + cos4x = cos2x*(sin2x + cos2x) = cos2x*1 = cos2x
sin(theta) = 15/17, cosec(theta) = 17/15 cos(theta) = -8/17, sec(theta) = -17/8 cotan(theta) = -8/15 theta = 2.0608 radians.
Yes. (Theta in radians, and then approximately, not exactly.)
Since secant theta is the same as 1 / cosine theta, the answer is any values for which cosine theta is zero, for example, pi/2.
sin (theta) = [13* sin (32o)]/8 = 13*0.529919264/8 = 0.861118804 [theta] = sin-1 (0.861118804) [theta] = 59.44o
theta = arcsin(0.0138) is the principal value.
Since sin(theta) = 1/cosec(theta) the first two terms simply camcel out and you are left with 1 divided by tan(theta), which is cot(theta).