Theta Eridani (Acamar) is a star in the constellation Eridanus.It is about 16 times larger than our Sun.
2*theta
Acamar is spectral class A4, which makes it bluish-white.
Theta
I think you will be out of luck.Firstly, Theta Eridani (Acamar) is a binary - possibly a multiple star systemSecondly, dating stars ages is a difficult if not impossible task.However, what can be deduced is that Theta Eridani brightest stars have a stellar class of A1 and A4 respectively.All A class stars have a short lifespan of about 1 billion years.Because it has been observed since the ancients - that's 2,000 years in it's favour and it's still fusing hydrogen - another one in it's favour. Basing this with other known stars of similar types I would guess the age at around 500 million years old.
Just did this in my trig class yesterday. Arc length = radius * theta(radians) Circumference of Earth = radius of earth * 2pi Note: The arc length is the circumference of the Earth only in this case because theta is equal to 2pi.
The only real solution is theta = 0For theta < 0 square root of 3 theta is not defined.For theta > 0, sin theta increases slower than 3*theta and so the sum is always negative.The only real solution is theta = 0For theta < 0 square root of 3 theta is not defined.For theta > 0, sin theta increases slower than 3*theta and so the sum is always negative.The only real solution is theta = 0For theta < 0 square root of 3 theta is not defined.For theta > 0, sin theta increases slower than 3*theta and so the sum is always negative.The only real solution is theta = 0For theta < 0 square root of 3 theta is not defined.For theta > 0, sin theta increases slower than 3*theta and so the sum is always negative.
tan2(theta) + 5*tan(theta) = 0 => tan(theta)*[tan(theta) + 5] = 0=> tan(theta) = 0 or tan(theta) = -5If tan(theta) = 0 then tan(theta) + cot(theta) is not defined.If tan(theta) = -5 then tan(theta) + cot(theta) = -5 - 1/5 = -5.2
cos2(theta) = 1 cos2(theta) + sin2(theta) = 1 so sin2(theta) = 0 cos(2*theta) = cos2(theta) - sin2(theta) = 1 - 0 = 1
cos2(theta) = 1 so cos(theta) = ±1 cos(theta) = -1 => theta = pi cos(theta) = 1 => theta = 0
There are three of them. Granted this means that there are different variations of all three. I'll show you the variations as well. This is coming straight from my Math 1060 (Trigonometry) notebook. Sorry there is no key to represent the angle; Theta.1. Sin2 (of Theta) + Cos2 (of Theta)= 1Variations: Sin2 (of Theta) = 1- Cos2 (of Theta)AND: Cos2 (of Theta) = 1-Sin2 (of Theta)2. Tan2 (of Theta) + 1 = sec2 (of Theta)Variations: Tan2 (of Theta) = Sec2 (of Theta) -13. 1 + Cot2 (of Theta) = Csc2 (of Theta)Variations: Cot2 (of Theta) = Csc2 (of Theta) -1
-Sin^(2)(Theta) + Cos^(2)Theta => Cos^(2)Theta - Sin^(2)Theta Factor (Cos(Theta) - Sin(Theta))( Cos(Theta) + Sin(Theta)) #Is the Pythagorean factors . Or -Sin^(2)Theta = -(1 - Cos^(2)Theta) = Cos(2)Theta - 1 Substitute Cos^(2)Thetqa - 1 + Cos^(2) Theta = 2Cos^(2)Theta - 1