1.find maxima and minima:? y=cos(2 sinx)

Question

2.cos(pi/19) cos(3pi/19) cos(17pi/19)=?? 

Plz provide step by step solution 
Thnx in advance

Answer 1

The function is y = cos[2 sin(x)].

f^,(x) = - sin[2 sin(x)]* (2cos(x) = - 2 * sin[2sin(x)] * cos(x).

f^,,(x) = - 2 * sin[2sin(x)] * [-sin(x)] - 2 * cos(x) * cos[2sin(x)] * 2 cos(x).

f^,,(x) = 2sin(x) * sin[2sin(x)] - 4cos²(x) * cos[2sin(x)].

To find the critical or key numbers, to make the first derivative equal to zero or f ' (x) = 0.

- 2 * sin[2sin(x)] * cos(x) = 0

sin[2sin(x)] = 0 and cos(x) = 0

sin(x) = 0 and cos(x) = 0

sin(x) = sin(0) and cos(x) = cos(π/2).

General solution x = nπ and x = 2nπ ± π/2, where n is an integer.

The critical numbers are x = nπ and x = 2nπ ± π/2.

If f " (c) > 0 (positive) ------> minimum point,.

If f " (c) < 0 (negative) ------> maximum point.

In the interval [0, 2π], the critical numbers are 0 and π/2.

So, lets plug each critical point in f^,,(x) = 2sin(x) * sin[2sin(x)] - 4cos²(x) * cos[2sin(x)].

If x = 0 then f " (0) = 2sin(0) * sin[2sin(0)] - 4cos²(0) * cos[2sin(0)] = - 4 < 0 (negative).

Therefore maximum points (maximum) at x = nπ.

If x = π/2 then f " (π/2) = 2sin(π/2) * sin[2sin(π/2)] - 4cos²(π/2) * cos[2sin(π/2)] = 1.818 < 0 (positive).

Therefore minimum points (minima) at 2nπ ± π/2.

 

To find the f(x) to each x for local max and local min plugging those values in the original function.

Maximum points (Maxima) :

If x = 0 then, f(0) = cos[2 sin(0)] = 1.

If x = π then, f(π) = cos[2 sin(π)] = 1.

If x = 2π then, f(2π) = cos[2 sin(2π)] = 1.

Therefore, Maxima is (nπ, 1), where n is an integer.

Minimum points (Minima) :

If x = π/2 then, f(π/2) = cos[2 sin(π/2)] = - 0.416.

If x = 3π/2 then, f(3π/2) = cos[2 sin(3π/2)] = - 0.416.

If x = -π/2 then, f(-π/2) = cos[2 sin(-π/2)] = - 0.416.

Therefore, Minima is (2nπ ± π/2, - 0.416), where n is an integer.

 

Graph :

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