1.25
sec(x) = 5*pi/4 can be solved for x but that is no help at all in finding possible values for 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 =)
sin(0)=0 and sin(very large number) is approximately equal to that same very large number.
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.
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
sin(0)=0 and sin(very large number) is approximately equal to that same very large number.
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.
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.
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.
Yes. (Theta in radians, and then approximately, not exactly.)
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).