What is the expansion of sin(3x) and cos (3x)?
There are several methods to show this. One of the basic methods is by using some of the simplest formulae of trigonometry. So let's start.You must be aware of the formula[math]\sin(A+B)=\sin A\cos B+\cos A\sin B[/math][math]\cos(A+B)=\cos A\cos B-\sin A\sin B[/math]Now first find [math]\sin 2x[/math] using the above formula.Replace both A and B with [math]x[/math] in 1st formulaWe get [math]\sin(x+x)=\sin x\cos x+\cos x\sin x.[/math]Therefore [math]\sin 2x=2\sin x\cos x \ldots(1)[/math]Similarly[math]\cos 2x= \cos^2x-\sin^2x[/math]Using [math]\sin^2x+\cos^2x=1 \ldots(2)[/math][math]\cos 2x=2\cos^2x-1[/math]Again using eqn [math](2)[/math][math]\cos 2x=1-2\sin^2x \ldots(3)[/math]Now for [math]\sin 3x[/math] replace A and B with [math]x[/math] and [math]2x[/math] or vice versa in the first formula.We get [math]\sin(x+2x)= \sin x\cos 2x+\cos x\sin 2x[/math]Subtituting the values of [math]\sin 2x[/math] and [math]\cos 2x[/math] we get[math]\sin 3x=(\sin x)(1-2\sin^2x)+(\cos x)(2\sin x\cos x)[/math]Using eqn [math](2)[/math] and rearranging terms we get[math]\boxed{\sin 3x=3\sin x-4\sin^3x}[/math]Similarly[math]\boxed{\cos 3x=4\cos^3x-3\cos x}[/math]There's a second interesting Method and it involves complex number. For that you must know De Moivre's theorem which can be stated as[math](\cos θ+i\sin θ)^n=(\cos nθ+i\sin nθ)[/math]So [math](\cos θ+i\sin θ)^3=\cos^3θ-i\sin^3θ+3i\cos^2θ\sin θ-3\sin^2θ\cos θ[/math]Now using eqn [math](2)[/math] we get[math](\cos 3θ+i\sin 3θ)=(4\cos^3θ-3\cos θ)+i(3\sin θ-4\sin^3θ) [/math]When two complex number are equal then their real and imaginary part are also equalHence we get[math]\boxed{\cos 3θ=4\cos^3θ-3\cos θ}[/math][math]\boxed{\sin 3θ=3\sin θ-4\sin^3θ}[/math]I hope it helps
Have I got the right answer to my maths homework?
Given acos(x) + bsin(x) you can rewrite it as: √(a^2+b^2)cos(x - θ), where θ = tan^-1(b/a) This means that 3cos(x) + 5sin(x) becomes: √34cos(x - 1.030376827) in radians or √34cos(x - 59.03624347) in degrees. Now to solve for 3cos(x) + 5sin(x) = 4 (I'll do this on the range [0,360)) √34cos(x - 59.03624347) = 4 cos(x - 59.03624347) = 4/√34 x - 59.03624347 = 46.68614334 or x - 59.03624347 = 313.3138567 x = 105.7223868 or x = 372.3501001 Since the desired range is [0,360) subtract 360 from 372.3501001 So 3cos(x) + 5sin(x) = 4 when x = x = 105.7223868 or x = 12.3501001
3cosx = 4sinx -2 (find the angle between 0 to 360 )?
3 cos x = 4 sin x - 2. [Square.] 9 cos² x = 16 sin² x - 16 sin x + 4. 9 - 9 sin² x = 16 sin² x - 16 sin x + 4. 0 = 25 sin² x - 16 sin x - 5. 0 = 25 sin² x - 16 sin x + 2.56 - 7.56. 7.56 = (5 sin x - 1.6)². ±√7.56 = 5 sin x - 1.6. 1.6 ±√7.56 = 5 sin x. (1.6 ±√7.56) / 5 = sin x. After computation arcsines are about 60.4480761° and -13.291719° BUT examinations of cosine values show that while the first works only in quadrant I, the second only works in quadrant III as 193.291719°.
Trig: How to solve cos^2x-sin^2x-3cosx-1?
cos^2 x - sin^2 x - 3cosx - 1 in order to "solve" there must be an equal sign. May I assume that this is equal to zero? if so, first substitute (1 - cos^2 x) for the sin^2 x: cos^2 x - (1 - cos^2 x) - 3cosx - 1 = 0 simplify: 2cos^2x - 3cosx - 2 = 0 this factors -- you can see it more clearly if you substitute a variable for cosx, let's say "y", and you have: 2y^2 - 3y - 2 = 0 which factors to: (2y + 1)(y - 2) = 0 and set each factor equal to zero: 2y + 1 = 0 or y - 2 = 0 when you substitute back and solve for cosx, throw out cosx=2, since it is out of the possible range for cosine, and you are left with: cosx = -1/2 this is "special" as it's reference angle is 60º. Since the cosine is negative, the angle is in either the 2nd or 3rd quadrants. As such, x = 120º or x = 240º or, if prefer radians: 2π/3 and 4π/3 that's it! ;)
How to solve: (3sinx)^2+(3cosx)^2?
(3sinx)² + (3cosx)² = 3² sin²x + 3² cos²x = 9(sin²x + cos²x) = 9*1 = 9
The value of x which satisfies the eq:( cos^2x-3cosx+2)/sin^2x=1 , lies in the open interval?
ah such a cuite name you have nishi .well dear look what am doing ; as you know sin^2x + cos^2 = +1 so sin^2x = 1-cos^2x formula one; (cos^2x-3cosx+2) / sin^2x=1 cos^2x-3cosx+2 = (sin^2x)*1 cos^2x-3cosx+2 = sin^2x { here put formual (a) instead of sin^2x } cos^2x-3cosx+2 = 1- (cos^2x) cos^2x-3cosx+ cos^2x = 1-2 2cos^2x - 3cosx =-1 factor ( cosx) cosx(2cosx - 3) = -1 cosx= -1 as you know in 0, 2pi cosx = +1 & in pi cosx = -1 Or 2 cosx -3 = -1 ; 2cosx = -1+3 ; 2cosx = 2 ; cosx = +1 x = 0 , x = 180 or pi ; x = 360 or 2pi Good Luck sweetheart.