'csc' = 1/sin
'tan' = sin/cos
So it must follow that
(cos) (csc) / (tan) = (cos) (1/sin)/(sin/cos) = (cos) (1/sin) (cos/sin) = (cos/sin)2
Remember that tan = sin/cos. So your expression is sin/cos times cos. That's sin(theta).
cosec(q)*cot(q)*cos(q) = 1/sin(q)*cot(q)*cos(q) = cot2(q)
For a start, try converting everything to sines and cosines.
The side adjacent to theta divided by the hypotenuse, or the angle opposite of the right angle
cos2(theta) = 1 so cos(theta) = Â±1 cos(theta) = -1 => theta = pi cos(theta) = 1 => theta = 0
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)
There can be no significant simplicfication if some of the angles are theta and others are x, so assume that all angles are x. [csc(x) - cot(x)]*[cos(x) + 1] =[1/sin(x) - cos(x)/sin(x)]*[cos(x) + 1] =1/sin(x)*[1 - cos(x)]*[cos(x) + 1] =1/sin(x)*[1 - cos2(x)] =1/sin(x)*[sin2(x)] = sin(x)
By converting everything to sines and cosines. Since tan x = sin x / cos x, in the cotangent, which is the reciprocal of the tangent: cot x = cos x / sin x. You can replace any other variable (like thetha) for the angle.
(Sin theta + cos theta)^n= sin n theta + cos n theta
It is cotangent(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
Zero. Anything minus itself is zero.
A - WORKWork = F.s cos (theta)
The question contains an expression but not an equation. An expression cannot be solved.
You can use the Pythagorean identity to solve this:(sin theta) squared + (cos theta) squared = 1.
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
In a right triangle, the right angle is formed by sides a and b. Side c is the hypotenuse.Theta is the interior angle that joins (let's say) sides b and c. The sin of theta is the length of a over the length of c. The cos of theta is the length of b over the length of c. The tan of theta is the length of a over the length of b.Sin theta= opposite divided by hypotenuse. Cos theta=adjacent divided by hypotenuse. Tan theta=opposite over adjacent.(Sin1-Cos1)-Tan1=-1.25623905Sorry, that was a mathematician's joke.
cosine (90- theta) = sine (theta)
For such simplifications, it is usually convenient to convert any trigonometric function that is not sine or cosine, into sine or cosine. In this case, you have: sin theta / sec theta = sin theta / (1/cos theta) = sin theta cos theta.
It's 1/2 of sin(2 theta) .
4*cos2(theta) = 1 cos2(theta) = 1/4 cos(theta) = sqrt(1/4) = Â±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.
The equation cannot be proved because of the scattered parts.
cos2(theta) = 1 cos2(theta) + sin2(theta) = 1 so sin2(theta) = 0 cos(2*theta) = cos2(theta) - sin2(theta) = 1 - 0 = 1
For Single Phase, P = VI cos (theta) therefore cos(theta) = P/VI here P = Power V = voltage I = current theta = phase angle current to voltage cos(theta) = power factor For Three Phase, P = 3VI cos(theta) where V = phase voltage and I = phase current and theta = phase angle
The derivative of (sin (theta))^.5 is (cos(theta))/(2sin(theta))