Using x instead of theta, cos2x/cosec2x + cos4x = cos2x*sin2x + cos4x
= cos2x*(sin2x + cos2x)
= cos2x*1 = cos2x
4*cos2(theta) = 1 cos2(theta) = 1/4 cos(theta) = sqrt(1/4) = ±1/2 Now cos(theta) = 1/2 => theta = 60 + 360k or theta = 300 + 360k while Now cos(theta) = -1/2 => theta = 120 + 360k or theta = 240 + 360k where k is an integer.
Cos2 doesn't equal pi; Cos2 equals roughly -0.416 (with radians).
sec2(x) - tan2(x)= 1/cos2(x) - sin2(x)/cos2(x)= (1 - sin2(x)) / cos2(x)= cos2(x) / cos2(x)= 1
cos2x/cosx = 2cosx - 1/cosx
Given that theta is the angle with respect to the positive X axis of a line of length 1, then sin(theta) = Y and cos(theta) is X, with (X,Y) being the point at the end of the line. As theta sweeps from 0 to 360 degrees, or 0 to 2 pi radians, that point draws a circle of radius 1, with center at (0,0).Since X, Y, and 1 form the sides of a right triangle, where 1 is the hypotenuse, then the pythagorean theorem states that X2 + Y2 = 12. This means that sin2(theta) + cos2(theta) = 1.Tan(theta) is defined as sin(theta) divided by cos(theta), or Y / X. Since division by zero is a limiting invalidity, then tan(theta) is asymptotic to Y=0, having value of +infinity at theta = 90 or pi / 4, and -infinity at 270 or 3 pi / 4.
cos2(theta) = 1 cos2(theta) + sin2(theta) = 1 so sin2(theta) = 0 cos(2*theta) = cos2(theta) - sin2(theta) = 1 - 0 = 1
To determine what negative sine squared plus cosine squared is equal to, start with the primary trigonometric identity, which is based on the pythagorean theorem...sin2(theta) + cos2(theta) = 1... and then solve for the question...cos2(theta) = 1 - sin2(theta)2 cos2(theta) = 1 - sin2(theta) + cos2(theta)2 cos2(theta) - 1 = - sin2(theta) + cos2(theta)
4*cos2(theta) = 1 cos2(theta) = 1/4 cos(theta) = sqrt(1/4) = ±1/2 Now cos(theta) = 1/2 => theta = 60 + 360k or theta = 300 + 360k while Now cos(theta) = -1/2 => theta = 120 + 360k or theta = 240 + 360k where k is an integer.
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
Cos2 doesn't equal pi; Cos2 equals roughly -0.416 (with radians).
1 - sin2(q) = cos2(q)dividing through by cos2(q),sec2(q) - tan2(q) = 1
Since theta is in the second quadrant, sin(theta) is positive. sin2(theta) = 1 - cos2(theta) = 0.803 So sin(theta) = +sqrt(0.803) = 0.896.
cos2 + cos2tan2 = cos2 + cos2*sin2/cos2 = cos2 + sin2 which is identically equal to 1. So the solution is all angles.
cos(t) - cos(t)*sin2(t) = cos(t)*[1 - sin2(t)] But [1 - sin2(t)] = cos2(t) So, the expression = cos(t)*cos2(t) = cos3(t)
sec2(x) - tan2(x)= 1/cos2(x) - sin2(x)/cos2(x)= (1 - sin2(x)) / cos2(x)= cos2(x) / cos2(x)= 1
Use these identities: sin2(x) + cos2(x) = 1, and tan(x) = sin(x)/cos(x) For clarity, the functions are written here without their arguments (the "of x" part). (1 - sin2) = cos2 (1 + tan2) = (1 + sin2/cos2) = (cos2+sin2) / cos2 = 1/cos2 Multiply them: (cos2) times (1/cos2) = 1'QED'