Verified answer

sin[x - (3π/14)]

= - cos[π/2 + (x - 3π/14)]

= - cos(x + 2π/7)

=> 6sin(x + 2π/7) - 8sin(x - 3π/14)

= 6sin(x + 2π/7) + 8cos(x + 2π/7)

= 6sinθ + 8cosθ

The maximum value is √(6^2 + 8^2) = 10

[For any trigo function of the type asinθ + bcosθ, the range is

[- √(a^2 + b^2), √(a^2 + b^2)]. This is proved in any standard book of trigo without the need for using calculus. This is a problem of trigo and no calculus is needed to prove this.]

If you are too particular about using calculus, here is the solution.

f(θ) = 6sinθ + 8cosθ

For f(θ) to be maximum, f '(θ) = 0 and f "(θ) < 0

f '(θ) = 0 => 6cosθ - 8sinθ = 0 => θ = kπ + arctan(3/4)

f "(θ) = - (6sinθ + 8cosθ) < 0 for θ = arctan(3/4)

=> sinθ = 3/5 and cosθ = 4/5

=> maximum value of f(θ) = 6*(3/5) + 8*(4/5) = 10.

Edit: You have asked a new more complex problem now. I will definitely attempt it and solve it too, but bear with me till my grandkids spare me to attend to you.

5sin(x + π/30) + 2√2 sin(x - π/20) + √3 sin(x + π/5) + sin(x + 11π/30) + 2sin(x + 8π/15) + 2√3 sin(x + 7π/10)

= 5sin(x + 6°) + 2√2 sin(x - 9°) + √3 sin(x + 36°) + sin(x + 66°) + 2sin(x + 96°) + 2√3 sin(x + 126°)

= 5siny + 2√2 sin(y -15°) + √3 sin(y + 30°) + sin(y + 60°) + 2sin(y + 90°) + 2√3 sin(y + 90° + 30°) [Taking x+6°=y]

= 5siny + 2√2(sinycos15° - cosysin15°) + √3(sinycos30° + cosysin30°) + (sinycos60° + cosysin60°) + 2cosy + 2√3(cosycos30° - sinysin30°)

= 5siny + 2√2[(√6+√2)/4*siny - (√6-√2)/4*cosy] + √3(√3/2*siny + 1/2*cosy) + (1/2*siny + √3/2*cosy) + 2cosy + 2√3(√3/2*cosy - 1/2*siny)

= (5+√3+1+3/2+1/2-√3)siny + (1-√3+√3/2+√3/2+2+3)cosy

= 8siny + 6cosy

Maximum value of this is 10.

This is an exteme value problem. First u must find the derivative of the function. You do that by simply taking the derivative f(x) = 2x^3-9x^2+12x-1 f ' (x) = 6x^2 - 18x + 12 factor out a 6 f ' (x) = 6 (x^2 - 3x + 2) further factor the (x^2 - 3x + 2) into x - 2 and x - 1 f ' (x) = 6 (x - 2) (x - 1) To find the relative maximum we must find the critical points first, now the critical points are the points where the derivative equals 0 so to find the critical points we must set the derivative equal to 0 6 (x - 2) (x - 1) = 0 x = 1, x = 2 now that we found the critical points, we must do a sign analysis to test for a relative maximum. When you do the sign analysis, u find that x = 1, you find that it is the relative max. plug x = 1 into f(x) to find the value of the relative max. f(x) = 2x^3-9x^2+12x-1 f(1) = 2(1)^3 - 9(1)^2 + 12(1) - 1 f(1) = 2 - 9 + 12 -1 f(1) = 4 but we're not done here. we also have to test the endpoints. plug the values of the endpoints into the function f(-1) = 2(-1)^3 - 9(-1)^2 + 12(-1) - 1 f(-1) = -2 -9 -12 -1 f(-1) = -24 f(2) = 2(2)^3 - 9(2)^2 + 12(2) - 1 f(2) = 2(8) - 9(4) + 24 - 1 f(2) = 16 -36 + 24 - 1 f(2) = 3 so f(1) = 4, f(2) = 3, and f(-1) = -24. remember when we are looking for the maximum we are looking for the highest value so the maximum is at x = 1 and the maximum value is 24. When you do these problems remember to 1) find the derivative, 2) find the critical points, 3) perform a sign analysis to find the relative max/mins, 4) plug the relative max/mins into the function, 5) plug the endpoints into the function, 6) compare the values of the max/mins/endpoints to find the absolute max/mins. hope that helped =)

We can use the trig identities for linear combination of sines in [1].

a sin(x) + b sin(x + α) = c sin(x + β)

c² = a² + b² + 2abcos(α)

In our case, α = (2π/7) + (3π/14) = π/2

c = sqrt(6² + 8² + 2*6*8*cos(π/2)) = 10

That is the amplitude of the given sinusoid. Still wondering about the geometrical implications of this. I know you are up to something, because you were talking about a 7-gon earlier.

*****

Well, we can use calculus but I don't know if it will make life easy :)

f(x) = 6 sin[x + (2π/7)] - 8 sin[x - (3π/14)]

. . .= 6 cos[(π/2) - x - (2π/7)] - 8 sin[x - (3π/14)]

. . .= 6 cos[-x + (3π/14)] - 8 sin[x - (3π/14)]

. . .= 6 cos[x - (3π/14)] - 8 sin[x - (3π/14)]

. . .= 6 {cos[x - (3π/14)] - (4/3) sin[x - (3π/14)]}

Let 4/3 = tanφ

f(x) = 6 (cos[x - (3π/14)] - tanφ sin[x - (3π/14)]}

. . .= (6/cosφ) {cosφ.cos[x - (3π/14)] - sinφ.sin[x - (3π/14)]}

. . .= (6/cosφ) cos[x - (3π/14) + φ]

Since tan²φ = 4/3, cosφ = 1/√(1 + tan²φ) = 3/5

f(x) = 6*(5/3) cos[x - (3π/14) + φ]

. . .=10 cos[x - (3π/14) + φ]

The maximum value of f(x) is +10, the minimum value is -10

( 2 pi/7 ) + ( 3 pi/14 ) = pi / 2.

Hence,

sin [ x - ( 3 pi/14 ) ] = sin [ x - ( pi/2) + ( 2 pi/ 7 ) ] = - cos [ x + ( 2 pi/7 ) ].

Hence,

y = 6 sin A - ( - 8 cos A ), where A = x + ( 2 pi/7)

y = 6 sin A + 8 cos A

so that

maximum of y is = sqrt ( 6^2 + 8^2 ) = 10. ............Ans.

θ = 2π/7

α = arccos(0.6)

6Sin(x + (2π/7)) - 8Sin(x - (3π/14))

= 6sin(x+θ) - 8sin(x - π/2 + θ)

= 6sin(x+θ) + 8cos(x + θ)

= 10* [ cosα sin(x+θ) + sinα cos(x + θ)]

= 10 * sin(x + θ + α)

Max = 10