3 - 3cos (x) = 2sin^2 (x) What is x?
First, recall the pythagorean identity: sin(x)^2 + cos(x)^2 = 1 This means that: cos(x)^2 = 1 - sin(x)^2 sin(x)^2 = 1 - cos(x)^2 3 - 3cos(x) = 2sin(x)^2 3 - 3cos(x) = 2 * (1 - cos(x)^2) 3 - 3cos(x) = 2 - 2cos(x)^2 2cos(x)^2 - 3cos(x) + 3 - 2 = 0 2cos(x)^2 - 3cos(x) + 1 = 0 Now it's just a quadratic equation of the form: am^2 + bm + c = 0 m = (-b +/- sqrt(b^2 - 4ac)) / (2a) m = cos(x) a = 2 b = -3 c = 1 cos(x) = (3 +/- sqrt(9 - 4 * 2 * 1)) / (2 * 2) cos(x) = (3 +/- sqrt(1)) / 4 cos(x) = (3 +/- 1) / 4 cos(x) = 4/4 , 2/4 cos(x) = 1 , 1/2 x = 0 , 60 , 300 , 360
Solve 2 sin x = 3 cos x?
Let 2 sin x - 3 cos x = k cos ( x - α ) for k > 0 2 sin x - 3 cos x = (k cos α) cos x + (k sin α) sin x 2 = k sin α -3 = k cos α Thus α in 2nd quadrant tan α = - 2/3 α = 146.3 ° 4 + 9 = k² k = √13 √13 cos ( x - 146.3 ) = 0 cos ( x - 146.3) = 0 x - 146.3 = 90 , 270 x = 236.3 ° , 56.3 °
What is the range of y=1+2sinx+3cos^2x?
[math]cos^2 x=1-sin^2 x[/math]. So, [math]y=1+2sinx+3–3sin^2 x=4+2sin x-3sin^2 x[/math]. Let sin x=t so that [math]y=4+2t-3t^2, -1<=t<=1[/math]. Differentiate the equation, we get t=1/3, which can be the possible value for t, differentiating again [math]d^2 y/dx^2=-3<0[/math], so maximum value is at t=1/3 and value of y=13/3 and minimum value is at t=-1 (border point), y=-1.Range of y is (-1,13/3)
How do you verify (sin²x+2cosx-1) / (sin²x+3cosx-3) =1/ (1-secx)?
sin^2 x + cos ^ 2 x = 1.Replace 1 in the numerator by sin^2 x + cos ^ 2 x and 3 in the denominator by3(sin^2 x + cos ^ 2 x ).Numerator = sin^2 x + 2 cosx - sin^2x - cos^2x = 2cosx - cos^2x = cosx(2-cosx)Denominator = sin^2x + 3cosx - 3(sin^2 x + cos ^ 2 x )= 3cosx - 3cos^2x - 2 sin^2x= 3cosx -cos^2x - 2cos^2x - 2sin^2x= 3cosx -cos^2x - 2(sin^2 x + cos ^ 2 x )= 3cosx - cos^2x - 2= -(cox^2x + 3cosx-2) = -(cosx-1)(cosx-2) = (cosx -1)(2-cosx)So numerator/denominator = cosx(2-cosx) / (cosx -1)(2-cosx).since cosx -2 can never be equal to 0, we can cancel it.= cosx/(cosx-1).dividing numerator and denominator by cosx , we get the answer= 1/(1-secx).
How do I write x (e^(x^2))-sin^2x+3cosx+5 and 2(x^2) e^(x^2) +e^(x^2)-2cosxsinx-3sinx in C++?
In vanilla C (to make the include file easy), here's a straight transcription without any optimization. It's the same in C++.#include
The function f is such that f(x) = 2sin^2x - 3cos^2x for 0=< x =< pi.?
i) Note that: f(x) = 2sin^2(x) - 3cos^2(x) . . . = [2sin^2(x) + 2cos^2(x)] - 5cos^2(x) . . . = 2 - 5cos^2(x). This implies that a = 2 and b = -5. ii) Since 0 <= cos^2(x) <= 1, we have: 2 - 5(1) <= 2 - 5cos^2(x) <= 2 - 5(0) ==> -3 <= 2 - 5cos^2(x) <= 2. iii) Since f(x) = 2 - 5cos^2(x), we have: f(x) + 1 = 0 ==> 2 - 5cos^2(x) + 1 = 0 ==> 5cos^2(x) = 3 ==> cos^2(x) = 3/5 = 15/25 ==> cos(x) = -√15/5 and cos(x) = √15/5. From here, you can solve these two equations separately. I hope this helps!
What is min & max value of 2 sin^2x + 3 cos^2x?
sin²x = 1 - cos²x => 2(1 - cos²x) + 3cos²x i.e. 2 + cos²x.......(1) cos2x = cos²x - sin²x => cos2x = 2cos²x - 1 so, cos²x = (cos2x + 1)/2 From (1) => 5/2 + (1/2)cos2x Now, cos2x minimum is -1 and cos2x maximum is 1 => max 5/2 + 1/2 = 3 and min 5/2 - 1/2 = 2 :)>
Solve cos2x-3cosx-3-cos^2x = sin^2x for 0 Use cos2x=2cos²x-1, sin²x=1-cos²x You get: 2cos²x-1-3cosx-3-cos²x=1-cos²x 2cos²x-3cosx-5=0 (2cosx-5)(cosx+1)=0 cosx=5/2 not valid cosx=-1 => x=π
What is the integral in (3 sin x cos^2 x - sin^3 x) dx?
First of all, let us simplify the given expression.We have, 3sinx(cosx)^2 -(sinx)^3= 3sinx{1-(sinx)^2} -(sinx)^3= 3sinx -3(sinx)^3 -(sinx)^3= 3sinx -4(sinx)^3= sin3xTherefore, ∫sin3x dx= (-1/3)×cos3x + c; where, c is the constant of integration(independent of x). (Ans)Hope, you'll understand..!Thank You!
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