If a< or =3sin(A - pie/6) + 2 cos(A + pie/3) < or =b then a + b=?

Asked by rekkakr1095 | 9th Nov, 2010, 09:26: PM

Expert Answer:

Dear Student,

Let us find the minimum and maximum values of sinx ± cosx.

y = sin(x) + cos(x)

dy/dx = cos(x) - sin(x)

Now, you make dy/dx = 0.

0 = cos(x) - sin(x)

Move sin(x) to the left hand side.

sin(x) = cos(x)

Divide both sides by cos(x).

sin(x)/cos(x) = 1
tan(x) = 1
x = { pi/4 + k×pi | k an integer }

So we have an infinite number of critical values. Since this is periodic and repeats every pi times, and pi is half a circle, all we need to use is two consecutive values.
Let's try pi/4, 5pi/4.

If x = pi/4, then
y = sin(pi/4) + cos(pi/4)
= √ (2)/2 + √ (2)/2
= 2√ (2)/2
= √ (2)

If x = 5pi/4, then
y = sin(5pi/4) + cos(5pi/4)
y = -√ (2)/2 + -√ (2)/2
y = -√ (2)

So,

-√2  ≤ sin(x) + cos(x) ≤ √2

In general, if we follow similar steps, it comes out that

-√(p2+q2) ≤ psin(x) ± qcos(x) a ≤ √(p2+q2)

Hence, in the given problem,

a ≤ 3sin(A - Π/6) + 2 cos(A + Π/3) ≤ b

=> a ≤ 3sin[Π - (A - Π/6)] + 2 cos(A + Π/3) ≤ b

=> a ≤ 3sin(A + Π/3) + 2 cos(A + Π/3) ≤ b

=> a = -√(32 + 22) and b = √(32 + 22)

=> a = -√13 and b = √13

=> a+b =0

Regards Topperlearning.

Answered by  | 11th Nov, 2010, 11:11: PM

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