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Find the exact value of each trigonometric function of cos pi over 3?
The exact value of (\cos\left(\frac{\pi}{3}\right)) is (\frac{1}{2}). Consequently, the values of the other trigonometric functions are as follows: (\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}), (\tan\left(\frac{\pi}{3}\right) = \sqrt{3}), (\sec\left(\frac{\pi}{3}\right) = 2), (\csc\left(\frac{\pi}{3}\right) = \frac{2\sqrt{3}}{3}), and (\cot\left(\frac{\pi}{3}\right) = \frac{1}{\sqrt{3}}).
What is the solution to the equation 4 sin x - 3 equals 0 to the nearest degree over the domain 0degrees360degrees?
4 sin(x) - 3 = 0 Therefore sin(x) = 3/4 And so the primary solution is x = sin-1(3/4) = 49 deg The second solution in the domain is 180 - 49 = 131 deg.
Sin3alfa equals to?
The expression (\sin(3\alpha)) can be expanded using the triple angle formula for sine, which is (\sin(3\alpha) = 3\sin(\alpha) - 4\sin^3(\alpha)). This formula allows you to express (\sin(3\alpha)) in terms of (\sin(\alpha)).
How do you solve 3 sin theta 1.5 when 0 isless than or equal to theta which is less than or equal to 4 radians?
I am assuming that the equation is 3*sin(t) = 1.5 even though the equality sign is not visible - due to the browser limitations. Then sin(t) = 1.5/3 = 0.5 So t = sin-1(0.5) which gives the principal value of t = 0.5236. The next value of t, in the domain, is pi - 0.5236 = 2.618 radians. There are no further values in the specified domain.
What is 3pi covert to degree?
To convert (3\pi) radians to degrees, you can use the conversion factor (180^\circ/\pi). Multiply (3\pi) by this factor: [ 3\pi \times \frac{180^\circ}{\pi} = 3 \times 180^\circ = 540^\circ. ] Therefore, (3\pi) radians is equivalent to (540^\circ).
What is the answer to inverse tangent square root 3 over 3?
30 degrees or pi/6
What is the value of sin 3 pi over 2?
sin(3π/2) = -1
Why does sin 2pi over 6 radians equal .5?
Sin(2*pi/6) = sin(pi/3) which, by definition, is 0.5 If you wish, you can calculate y/1! - y^3/3! + y^5/5! - y^7/7! + ... where y = pi/3.
What is sin of 3 pi over 2?
Sin(3pi/2) = Sin(2pi - pi/2) Double angle Trig. Identity. Hence Sin(2pi)Cos(pi/2) - Cos(2pi) Sin(pi/2) Sin(2pi) = 0 Cos(pi/2) = 0 Cos(2pi) = 1 Sin(pi/2) = 1 Substituting 0 x 0 - 1 x 1 = 0 - 1 = -1 The answer!!!!!
What is the exact value using a sum or difference formula of the expression cos 11pi over 12?
11pi/12 = pi - pi/12 cos(11pi/12) = cos(pi - pi/12) cos(a-b) = cos(a)cos(b)+sin(a)sin(b) cos(pi -pi/12) = cos(pi)cos(pi/12) + sin(pi)sin(pi/12) sin(pi)=0 cos(pi)=-1 Therefore, cos(pi -pi/12) = -cos(pi/12) pi/12=pi/3 -pi/4 cos(pi/12) = cos(pi/3 - pi/4) = cos(pi/3)cos(pi/4)+sin(pi/3) sin(pi/4) cos(pi/3)=1/2 sin(pi/3)=sqrt(3)/2 cos(pi/4)= sqrt(2)/2 sin(pi/4) = sqrt(2)/2 cos(pi/3)cos(pi/4)+sin(pi/3) sin(pi/4) = (1/2)(sqrt(2)/2 ) + (sqrt(3)/2)( sqrt(2)/2) = sqrt(2)/4 + sqrt(6) /4 = [sqrt(2)+sqrt(6)] /4 Therefore, cos(pi/12) = (sqrt(2)+sqrt(6))/4 -cos(pi/12) = -(sqrt(2)+sqrt(6))/4 cos(11pi/12) = -(sqrt(2)+sqrt(6))/4
What is the sin of pi divided by 3?
The question is ambiguous and the two possible answers are: sin(pi)/3 = 0 and sin(pi/3) = sqrt(3)/2 It is assumed, of course, that since the angles are given in terms of pi, they are measured in radians and not degrees!
Is sin3A plus sinA equal to 0?
No, for example if A = Pi/3. Then sin3A = sin pi = 0, but sinA = sin Pi/3 = 1/2. So for A = Pi/3, the sum is 1/2, not zero. It can't be proved because the statement is false. For example if A = Pi/3. Then sin3A = sin pi = 0, but sinA = sin Pi/3 = 1/2. So for A = Pi/3, the sum is 1/2, not zero.
What is sin of pi over 12 on a radian circle?
The sine of (\frac{\pi}{12}) radians (which is equivalent to 15 degrees) can be calculated using the sine subtraction formula: (\sin(a - b) = \sin a \cos b - \cos a \sin b). By letting (a = \frac{\pi}{4}) (45 degrees) and (b = \frac{\pi}{3}) (60 degrees), we find that (\sin\left(\frac{\pi}{12}\right) = \sin\left(\frac{\pi}{4} - \frac{\pi}{3}\right) = \sin\frac{\pi}{4} \cos\frac{\pi}{3} - \cos\frac{\pi}{4} \sin\frac{\pi}{3}). This evaluates to (\frac{\sqrt{2}}{2} \cdot \frac{1}{2} - \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{2}}{4} - \frac{\sqrt{6}}{4} = \frac{\sqrt{2} - \sqrt{6}}{4}).
Who can Solve sin(-1)sin((5pi )(7)) with step by Step please )?
Unfortunately, the browser used for posting questions is hopelessly inadequate for mathematics: it strips away most symbols. All that we can see is "sin(-1)sin((5pi )(7))". From that it is not at all clear what the missing symbols (operators) between (5pi ) and (7) might be. There is, therefore no sensible answer. It makes little sense for me to try and guess - I may as well make up my own questions and answer them!All that I can tell you that the principal sin-1 is the inverse for sin over the domain (-pi/2, pi/2). Thus sin-1(sin(x) = x where -pi/2 < x
What is the sin of 3 pi?
The sine of (3\pi) is 0. This is because (3\pi) corresponds to a point on the unit circle where the angle is a multiple of (2\pi) (specifically, (3\pi = 2\pi + \pi)), and the sine of any integer multiple of (2\pi) is always 0. Thus, (\sin(3\pi) = 0).
What is the additive inverse of -2 over -3?
-5
What is the period for y-3 sin x?
y = 3 sin x The period of this function is 2 pi.
