Calc of trig function (1 Viewer)

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By using the same axes and scales , sketch the curves

y=cosx and y=cos2x for 0<=x<=1/2pi

FOr this interval of values of x, find the maximum distance between the two curves measured parallel to the y-axis


Help on how to approach this question after sketching the graphs

THanks
 

rand_althor

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If you sketch it you should be able to see that the maximum separation occurs at .
 

Paradoxica

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After sketching, it should be clear that cosx > cos2x in this interval.
The length in this interval is thus given by l = cosx - cos2x, as length is positive.
Differentiating l with respect to x, we have dl/dx = -sinx + 2sin2x = -sinx + 4sinxcosx
Factorise to obtain dl/dx = (4cosx -1)sinx
We want the maximum distance, solve for the zeroes of the derivative.
sinx = 0 or cosx = 1/4
x=0 or cosx = 1/4
From the graph, it is clear that x = 0 cannot be the maximum value, as cosx and cos2x take on the same value at x=0, and hence their difference is zero. Therefore, we conclude that if a maximum value exists, it's value of x is given by cosx = 1/4
By substituting values of x around cos^-1 (1/4) into the derivative function, it is clear that this value of x is a maximum.
Rewrite l = cosx - cos2x as cosx + 1 - 2cos^2 (x) using double angle identities.
substituting cosx = 1/4, we have:
l_max = 1/4 + 1 - 2(1/4)^2 = 9/8 units. Therefore, the maximum vertical separation distance between the two curves is 9/8 units.

If you sketch it you should be able to see that the maximum separation occurs at .
LoL, nowhere near correct, as you have not justified it is a maximum value.
 

leehuan

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I don't believe 2U examiners will be that fussy.

However, an algebraic approach will require the need to do a full rigorous proof using calculus.

Also, keep in mind 2U do not do double angles.
 

Paradoxica

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I don't believe 2U examiners will be that fussy.

However, an algebraic approach will require the need to do a full rigorous proof using calculus.

Also, keep in mind 2U do not do double angles.
In that case, the calculator will suffice.
 

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