The only real solution is theta = 0For theta < 0 square root of 3 theta is not defined.For theta > 0, sin theta increases slower than 3*theta and so the sum is always negative.The only real solution is theta = 0For theta < 0 square root of 3 theta is not defined.For theta > 0, sin theta increases slower than 3*theta and so the sum is always negative.The only real solution is theta = 0For theta < 0 square root of 3 theta is not defined.For theta > 0, sin theta increases slower than 3*theta and so the sum is always negative.The only real solution is theta = 0For theta < 0 square root of 3 theta is not defined.For theta > 0, sin theta increases slower than 3*theta and so the sum is always negative.
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.
If sine theta is 0.28, then theta is 16.26 degrees. Cosine 2 theta, then, is 0.8432
tan (theta x theta) : must square the value of the angle, theta, before applying the trig function, tangent.
It is a non-linear function of theta.
Theta is not a vector quantity because it only represents the magnitude of an angle and does not have a direction associated with it. Scalars, like theta, only have magnitude, while vectors have both magnitude and direction.
Magnitude
What about the two vectors? Are they of same magnitude? If so then the resultant is got by getting the resolved components. Here we need adjacent components. F cos30 + F cos30 = 2 F cos 30 = ./3 F If forces of different magnitude then we use R = ./ (P^2 + Q^2 + 2 P Q cos 60)
The magnitude of force f can be calculated using the equation f = mgsin(theta), where m is the mass of the object, g is the acceleration due to gravity, and theta is the angle of the incline. Given the angle of 30 degrees, the force can be calculated by plugging in the values of mass and acceleration due to gravity.
work = the dot product of the force (F) and displacement vectors (D) = f * d * cos (theta), where 'f' and 'd' the magnitude of F and D, respectively; 'theta' is the angle between the two vectors. If theta = 90o, cos(theta) = 0. No work is done. That is, F is orthogonal to D. If d = 0, no work is done. That is, if the object is returning to the starting point, D = 0. ========================================
The only real solution is theta = 0For theta < 0 square root of 3 theta is not defined.For theta > 0, sin theta increases slower than 3*theta and so the sum is always negative.The only real solution is theta = 0For theta < 0 square root of 3 theta is not defined.For theta > 0, sin theta increases slower than 3*theta and so the sum is always negative.The only real solution is theta = 0For theta < 0 square root of 3 theta is not defined.For theta > 0, sin theta increases slower than 3*theta and so the sum is always negative.The only real solution is theta = 0For theta < 0 square root of 3 theta is not defined.For theta > 0, sin theta increases slower than 3*theta and so the sum is always negative.
tan2(theta) + 5*tan(theta) = 0 => tan(theta)*[tan(theta) + 5] = 0=> tan(theta) = 0 or tan(theta) = -5If tan(theta) = 0 then tan(theta) + cot(theta) is not defined.If tan(theta) = -5 then tan(theta) + cot(theta) = -5 - 1/5 = -5.2
Given two vectors a and b, the area of a parallelogram formed by these vectors is:a x b = a*b * sin(theta) where theta is the angle between a and b, and where x is the norm/length/magnitude of vector x.
cos2(theta) = 1 cos2(theta) + sin2(theta) = 1 so sin2(theta) = 0 cos(2*theta) = cos2(theta) - sin2(theta) = 1 - 0 = 1
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cos2(theta) = 1 so cos(theta) = ±1 cos(theta) = -1 => theta = pi cos(theta) = 1 => theta = 0
Cotan(theta) is the reciprocal of the tan(theta). So, cot(theta) = 1/2.