The cosine is simply the x-coordinate of the unitary circle. It helps to draw the circle, and the sine and cosine (x and y coordinates), to visualize this. The y-coordinate is the same for a positive angle and for the corresponding negative angle.
Remember that tan = sin/cos. So your expression is sin/cos times cos. That's sin(theta).
Let 'theta' = A [as 'A' is easier to type] sec A - 1/(sec A) = 1/(cos A) - cos A = (1 - cos^2 A)/(cos A) = (sin^2 A)/(cos A) = (tan A)*(sin A) Then you can swap back the 'A' with theta
cosine (90- theta) = sine (theta)
It's 1/2 of sin(2 theta) .
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
Cos theta squared
The equation cannot be proved because of the scattered parts.
because sin(2x) = 2sin(x)cos(x)
You must think of the unit circle. negative theta is in either radians or degrees and represents a specific area on the unit circle. Remember the unit circle is also like a coordinate plane and cos is the x while sin is the y coordinate. Here is an example: cos(-45): The cos of negative 45 degrees is pi/4 and cos(45) is also pi/4
- cos theta
No, they cannot all be negative and retain the same value for theta, as is shown with the four quadrants and their trigonemtric properties. For example, in the first quadrant (0
Remember that tan = sin/cos. So your expression is sin/cos times cos. That's sin(theta).
(Sin theta + cos theta)^n= sin n theta + cos n theta
The identity for tan(theta) is sin(theta)/cos(theta).
The solution is found by applying the definition of complementary trig functions: Cos (&Theta) = sin (90°-&Theta) cos (62°) = sin (90°-62°) Therefore the solution is sin 28°.
Let 'theta' = A [as 'A' is easier to type] sec A - 1/(sec A) = 1/(cos A) - cos A = (1 - cos^2 A)/(cos A) = (sin^2 A)/(cos A) = (tan A)*(sin A) Then you can swap back the 'A' with theta