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How do you find the value of alpha when the value of sin alpha is 1 divided by 2?
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What is the sin 29 equals x divided by 10?
2.9
Need help with working this Trig problem out. A sin alpha plus cos alpha equals 1 B sin alpha - cos alpha equals 1 Solution is AB equals 1?
A*sin(x) + cos(x) = 1B*sin(x) - cos(x) = 1Add the two equations: A*sin(x) + B*sin(x) = 2(A+B)*sin(x) = 2sin(x) = 2/(A+B)x = arcsin{2/(A+B)}That is the main solution. There may be others: depending on the range for x.
What is the sin of pi divided by 2?
sin (pi/2) = 1
Find the value of 3 sin 40 4 cos 20 - tan 45?
6.25
What is 55 times sin43 divided by sin 23?
55 × sin 43 ÷ sin 23 ≈ 96
What is the sin 29 equals x divided by 10?
2.9
How do you find the value of x with 4 sin 2x tan 5x sin 7x equals sin 4x?
Once way is to plot it out, and note the intersection points. One spot is 0.2225400023465516 Follow the Wolfram|Alpha link for more answers.
Identities to find the exact value of sin 75?
what is the value of sin 75 degree
What is the exact value of sin 405?
sin(405) = square root of 2 divided by 2 which is about 0.7071067812
Find the value of sin 105?
0.96593 (rounded)
Find the value of a if tan 3a is equal to sin cos 45 plus sin 30?
If tan 3a is equal to sin cos 45 plus sin 30, then the value of a = 0.4.
Need help with working this Trig problem out. A sin alpha plus cos alpha equals 1 B sin alpha - cos alpha equals 1 Solution is AB equals 1?
A*sin(x) + cos(x) = 1B*sin(x) - cos(x) = 1Add the two equations: A*sin(x) + B*sin(x) = 2(A+B)*sin(x) = 2sin(x) = 2/(A+B)x = arcsin{2/(A+B)}That is the main solution. There may be others: depending on the range for x.
If sin y and deg a divided by 6 and tan y and deg a divided by b what is the value of cos y and deg (6 points)?
1
Find the value of sin 300 degrees?
sin 300 = -sin 60 = -sqrt(3)/2 you can get this because using the unit circle.
By using trigonometric identities find the value of sin A if tan A a half?
The value of tan A is not clear from the question.However, sin A = sqrt[tan^2 A /(tan^2 A + 1)]
The coefficient of static friction along an inclined plane is equal to tan alpha. What about the coefficient of kinetic friction?
It's not. The coefficient of static friction is only equal to the tangent of the angle of incline at the maximum angle before the object begins to slide. At this point static friction equals the component of the weight along the incline (weight X sin alpha). Static friction is given by the coefficient of static friction times the normal force (weight X cos alpha) fs = us N = us mg cos(alpha) Wx =mg sin(alpha) fs = Wx us mg cos(alpha) = mg sin(alpha) us = [sin(alpha)] / [cos(alpha)] = tan(alpha) Similarly, the coefficient of kinetic friction equals the tangent of the angle of incline only if the object is sliding down the incline at constant velocity (net force equals zero). If the object is accelerating along the incline (make this the x axis): Fnet, x = Wx - f max = mg sin(alpha) - uk mg cos(alpha) uk = [g sin(alpha) - ax] / [g cos(alpha)]
Find value of trigonometric functions of quadrantal angle?
sin 0=13/85
