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31/2sinx + cox is maximum when x is 

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Chen Jang T. | MaCT (Math and Chemistry Tutor)MaCT (Math and Chemistry Tutor)
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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 (=60o)
Stephen K. | Physics PhD experienced in teaching undergraduatesPhysics PhD experienced in teaching unde...
5.0 5.0 (436 lesson ratings) (436)
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 = A + 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.
Philip P. | Effective and Affordable Math TutorEffective and Affordable Math Tutor
5.0 5.0 (438 lesson ratings) (438)
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) =  60o ± 360n, n=1, 2, 3, ...


Hello Philip,
For the general tangent function
We should write it as
  x= pi/3. +- 180n,  n= 1,2,3.......
Since the period of tan function is pi and not 2pi.
Hi Surendra,
The original function is not tan(x) but y = √3sin(x) + cos(x) which has a period of 2pi, so y has a maximum at x= pi/3 and every ±2pi.  The pi/3 ± pi produces a minimum.
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