^{1/2}sinx + cox is maximum when x is

^{2}+1

^{2}) sin (x + tan

^{-1}(1/√3))

^{o})

3^{1/2}sinx + cox is maximum when x is

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Chen Jang T. | MaCT (Math and Chemistry Tutor)MaCT (Math and Chemistry Tutor)

Marked as Best Answer

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}+1^{2}) 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})

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.

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, ...

SURENDRA K. | An experienced,patient & hardworking tutorAn experienced,patient & hardworking tut...

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

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