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How do solve sin 2 theta - square root of 3 theta- equals 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.
What is the equivalent of sine theta over secant 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.
What is cosine 2 theta when sine theta equals .28?
If sine theta is 0.28, then theta is 16.26 degrees. Cosine 2 theta, then, is 0.8432
When is tan theta theta?
tan (theta x theta) : must square the value of the angle, theta, before applying the trig function, tangent.
What is Theta multiplied by cosine Theta?
It is a non-linear function of theta.
Why theta is a not a vector quantity?
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.
when a complex number z is written in its polar form, z = r (cos(theta) + i * sin(theta)), the nonnegative number r is called the or modulus, or z?
Magnitude
What is the magnitude of the resultant vectors when the angle between them is 60 degrees?
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)
Frictionless 30.0 incline What is the approximate magnitude of force f?
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.
What two condititons must be for work to be done?
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. ========================================
How do solve sin 2 theta - square root of 3 theta- equals 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.
If tansqtheta plus 5tantheta0 find the value of tantheta plus cottheta?
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
How do you find the area of a parallelogram using 2 vectors?
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.
How would you solve and show work for cos2 theta if cos squared theta equals 1 and theta is in the 4th quadrant?
cos2(theta) = 1 cos2(theta) + sin2(theta) = 1 so sin2(theta) = 0 cos(2*theta) = cos2(theta) - sin2(theta) = 1 - 0 = 1
What are the names of the stars in Auriga?
α Aur Capellaβ Aur Menkalinanγ Aur El Nathδ Aur Prajaε Aur Al Mazζ Aur Saclateniζ Aur Hoedus Iθ Aur Manusη Aur Hoedus IIι Aur Hasselehλ Aur Al Hurr
How do you solve theta if cos squared theta equals 1 and 0 is less than or equal to theta which is less than 2pi?
cos2(theta) = 1 so cos(theta) = ±1 cos(theta) = -1 => theta = pi cos(theta) = 1 => theta = 0
If tan Theta equals 2 with Theta in Quadrant 3 find cot Theta?
Cotan(theta) is the reciprocal of the tan(theta). So, cot(theta) = 1/2.
