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What kind of angle is 2 pi over 3?
It's an obtuse angle.
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.
How do you prove that 2 sin 3x divided by sin x plus 2 cos 3x divided by cos x equals 8 cos 2x?
You need to know the trigonometric formulae for sin and cos of compound angles. sin(x+y) = sin(x)*cos(y)+cos(x)*sin(y) and cos(x+y) = cos(x)*cos(y) - sin(x)*sin(y) Using these, y = x implies that sin(2x) = sin(x+x) = 2*sin(x)cos(x) and cos(2x) = cos(x+x) = cos^2(x) - sin^2(x) Next, the triple angle formulae are: sin(3x) = sin(2x + x) = 3*sin(x) - 4*sin^3(x) and cos(3x) = 4*cos^3(x) - 3*cos(x) Then the left hand side = 2*[3*sin(x) - 4*sin^3(x)]/sin(x) + 2*[4*cos^3(x) - 3*cos(x)]/cos(x) = 6 - 8*sin^2(x) + 8cos^2(x) - 6 = 8*[cos^2(x) - sin^2(x)] = 8*cos(2x) = right hand side.
