How to solve this

Question

3^1/2sinx + cox is maximum when x is 

Answer 1

Let me see if I can help clarify how to get to the correct answer.

 

First, draw a circle of radius R on a piece of paper with x & y axes through the center.  Pick some arbitrary angle in the 1st quadrant and label it Θ.  Next using this angle construct a right triangle from the x-axis, such that x = A and y = B.

 

Your problem asks to find the maximum value of √3 sin(x) + cos(x) = ? maximum

 

Let's put this equation in the general form of A sin(x) + B cos(x) = ?

 

Multiply top and bottom by R so that R/R [A sin(x) + B cos(x)] = ?

 

We get away with this because R/R = 1.

 

Now bring the R in the denominator inside and write:  R*[A/R sin(x) + B/R cos(x)] = ?

 

From the picture that we drew with the circle and the angle we can see that:

 

sin(θ) = B/R and cos(Θ) = A/R, where R ² = A ² + B ²

 

Substituting into the equation above we can now write R*[cos(θ)*sin(x) + sin(Θ)cos(x)] = R*[sin(x + Θ)]

 

This takes on a maximum value when sin(x+ Θ)] = 1,

or when x + Θ = pi/2.  then x = pi/2 - Θ.

 

Θ = sin ^-1 (B/R) 

 

For your problem A = √3, B = 1, and R = √(A ² + B ²⁾ 

so that R = 2 and B/R = ½

 

sin(Θ) = 1/2 when Θ = pi/6

 

Finally, x = pi/2 - pi/6 = pi/3.  This is your maximum value for x.

 

Answer 2

Guys, please make sure the math level the students taking. I am sure the student has not taken calculus, so it does not make sense to answer trigonometry question using calculus.

 

The plausible answer is below:

 

√3 sin(x) + 1 cos(x) = √((√3)²+1²) sin (x + tan^-1(1/√3))

                               = 2 sin (x+pi/6)

So the maximum x at sin function above, x = pi/2-pi/6 = pi/3 (=60^o)

Answer 3

y = √3 sinx + cosx

 

Take the derivative of y wrt x, set it 0, solve for x:

 

dy/dx = √3 cosx - sinx

 

0 = √3 cosx - sinx

sin x = √3 cos x

 

tan x = √3

 

x = tan^-1(√3) =  60^o ± 360n, n=1, 2, 3, ...

Answer 4

We take the derivative of the given equation.

Say left hand side is = y

 

So dy/dx = sqrt3 cosx-sinx =0

 

Tanx= sqrt3

 

x= pi/3

 

Calculate second derivative

 

=- sqrt3 Sinx -cosx

= -2.  at x= pi/3

Which is negative , means it is maxima

 

So,  x= pi/3