3

^{1/2}sinx + cox is maximum when x is3^{1/2}sinx + cox is maximum when x is

Tutors, please sign in to answer this question.

Chino Hills, CA

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})

Orlando, FL

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.

Olney, MD

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, ...

San Diego, CA

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

Malik K.

MIT Grad, Math/Physics Tutor, Summer Math/Engineering Camp Instructor

New York, NY

5.0
(17 ratings)

Noah C.

An Expert, Passionate SAT Instructor

New York, NY

5.0
(70 ratings)

Richard B.

Effective and Experienced Math and Science Tutor and Teacher

South Orange, NJ

5.0
(188 ratings)

- Math 6890
- Algebra 2 2745
- Precalculus 1222
- Algebra 1 3131
- Math Help 4061
- Geometry 1397
- Calculus 1726
- Trig 145
- Prealgebra 141
- Trigonometric Functions 169

## Comments