Tan^2
You can use the Pythagorean identity to solve this:(sin theta) squared + (cos theta) squared = 1.
sin/cos
It depends if 1 plus tan theta is divided or multiplied by 1 minus tan theta.
(/) = theta sin 2(/) = 2sin(/)cos(/)
Zero. Anything minus itself is zero.
cosine (90- theta) = sine (theta)
Tan^2
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)
Cos theta squared
cos(theta) = 0.7902 arcos(0.7902) = theta = 38 degrees you find complimentary angles
108.435 degrees 288.435 degrees (decimal is rounded)
cos2(theta) = 1 so cos(theta) = ±1 cos(theta) = -1 => theta = pi cos(theta) = 1 => theta = 0
Remember that tan = sin/cos. So your expression is sin/cos times cos. That's sin(theta).
You cannot prove it because it is not true! cos(0) = 1 cos(2*pi) = 1 cos(4*pi) = 1 ... cos(2*k*pi) = 1 for all integers k or, if you still work in degrees, cos(0) = 1 cos(360) = 1 cos(720) = 1 ... cos(k*360) = 1 for all integers k
(Sin theta + cos theta)^n= sin n theta + cos n theta
The identity for tan(theta) is sin(theta)/cos(theta).