sin(3x) = 3sin(x) - 4sin^3(x)
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.
The best way to answer this question is with the angle addition formulas. Sin(a + b) = sin(a)cos(b) + cos(a)sin(b) and cos(a + b) = cos(a)cos(b) - sin(a)sin(b). If you compute this repeatedly until you get sin(3x)cos(4x) = 3sin(x) - 28sin^3(x) + 56sin^5(x) - 32sin^7(x).
Do you mean sin(x)=sqrt(3)/2? IF so, look at at 30/60/90 triangle. We see the sin 60 degrees is square (root of 3)/2
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.
amplitude of the function y =-3 sin 3x
I think you mean to solve:(-2 cos23x) - (3 sin 3x) = 0cos2 x + sin2 x = 1⇒ cos2x = 1 - sin2 x⇒ -2 cos2 3x - 3 sin 3x = -2(1 - sin2 3x) - 3 sin 3x = 0⇒ 2 sin2 3x - 3 sin 3x - 2 = 0⇒ (2 sin 3x + 1)(sin 3x - 2) = 0⇒ sin 3x = -1/2 or 2sin 3x = 2 is impossible as the range of sine is -1 ≤ sine ≤ 1Thus:sin 3x = -1/2⇒ 3x = 2nπ - π/6 or 2nπ - 5π/6⇒ x = 2/3nπ -π/18 or 2/3nπ -5π/18
Using Euler's Formula, you use (cos(x) + i sin(x))^n = cos (nx) + i sin(nx) Now you let n=3 (cos(x) + i sin (x))3 = cos(3x) + i sin (3x) (cos(x))3 + 3(cos(x))2 * i sin(x) + 3cos(x) * i2 (sin(x))3 = cos(3x)+ i sin(3x) (cos(x))3 + i(3sin(x)(cos (x))2) - 3cos(x)(sin(x)2) - i(sin(x))3 = cos (3x) + i sin(3x) Now only use the terms with i in them to figure out what sin(3x) is... 3sin(x)(cos(x))2 - (sin(x))3 = sin(3x) Hope this helps! :D
y = 2 sin 3x y' = 2(sin 3x)'(3x)' y' = 2(cos 3x)(3) y' = 6 cos 3x
Integrate by parts: ∫ uv dx = u ∫ v dx - ∫ (u' ∫ v dx) dx Let u = -2x Let v = cos 3x → u' = d/dx -2x = -2 → ∫ -2x cos 3x dx = -2x ∫ cos 3x dx - ∫ (-2 ∫ cos 3x dx) dx = -2x/3 sin 3x - ∫ -2/3 sin 3x dx = -2x/3 sin 3x - 2/9 cos 3x + c
y=2 sin(3x) dy/dx = 2 cos(3x) (3) dy/dx = 6 cos(3x)
6
y=3x=3
y = -1 + 3 sin 4xLet's look at the equation of y = 3 sin 4x, which is of the form y = A sin Bx, wherethe amplitude = |A|, and the period = (2pi)/B.So that the amplitude of the graph of y = 3 sin 4x is |3| = 3, which tell us that the maximum value of y is 3 and the minimum value is -3, and the period is (2pi)/4 = pi/2, which tell us that each cycle is completed in pi/2 radians.The graph of y = -1 + 3 sin 4x has the same amplitude and period as y = 3 sin 4x, and translates the graph of y = 3 sin 4x one unit down, so that the maximum value of y becomes 2 and the minimum value becomes -4.
Take f(x) = cos(3x) ∫ f(x) dx = ∫ cos(3x) dx Take u=3x → du = 3dx = ∫ 1/3*cos(u) du = 1/3*∫ cos(u) du = 1/3*sin(u) + C, C ∈ ℝ = 1/3*sin(3x) + C
3x = 2y + 3 3x -3 = 2y +3 -3 3x - 3 = 2y (3x -3)/2 = 2y / 2 y = 1/2 x (3x - 3) 3x = 2y +3 3x - 3x = -3x +2y +3 0 = -3x +2y +3 -1 x (0) = -1 x (-3x +2y +3) 0 = 3x - 2y -3
3x = 2y + 3 3x -3 = 2y +3 -3 3x - 3 = 2y (3x -3)/2 = 2y / 2 y = 1/2 x (3x - 3) 3x = 2y +3 3x - 3x = -3x +2y +3 0 = -3x +2y +3 -1 x (0) = -1 x (-3x +2y +3) 0 = 3x - 2y -3