The value of cos(theta) cannot be infinite. The cosine function, which represents the ratio of the adjacent side to the hypotenuse in a right triangle, has a range of values between -1 and 1. Therefore, it never reaches infinity or negative infinity for any angle theta.
The expression (\cos^2(90^\circ - \theta)) can be simplified using the co-function identity, which states that (\cos(90^\circ - \theta) = \sin(\theta)). Therefore, (\cos^2(90^\circ - \theta) = \sin^2(\theta)). This means that (\cos^2(90^\circ - \theta)) is equal to the square of the sine of (\theta).
The identity for tan(theta) is sin(theta)/cos(theta).
Zero. Anything minus itself is zero.
The question contains an expression but not an equation. An expression cannot be solved.
The fourth Across the quadrants sin theta and cos theta vary: sin theta: + + - - cos theta: + - - + So for sin theta < 0, it's the third or fourth quadrant And for cos theta > 0 , it's the first or fourth quadrant. So for sin theta < 0 and cos theta > 0 it's the fourth quadrant
You can use the Pythagorean identity to solve this:(sin theta) squared + (cos theta) squared = 1.
Cos theta squared
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).
-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
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
(Sin theta + cos theta)^n= sin n theta + cos n theta
The identity for tan(theta) is sin(theta)/cos(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
It is cotangent(theta).
Zero. Anything minus itself is zero.
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°.