Question

write the max. and min. value of 1)cos(cos x) 2)sin(sin x) 3)sin(cos x)

Answer 1

☆The maximum value of cosx is 1. At x = 0° and x = 360°, cosx = 1. The minimum value of cosx is −1.



☆ a*sinx + b*cosx, maximum & minimum value is (+,-)√[a^2+b^2]. As here a=b=1, maximum and minimum value of sinx + cosx = (+,-)√[1^2+1^2], i.e. +√2 is maximum value and -√2 is the minimum value. Whether these correspond to maximum or minimum, can be found from the sign of second derivative.

Answer 2

Answer:

Step-by-step explanation:

Let y= sin x + cos x

dy/dx=cos x- sin x

For maximum or minimum dy/dx=0

Setting cosx- sin x=0

We get cos x = sin x

x= π/4, 5π/4———-

Whether these correspond to maximum or minimum, can be found from the sign of second derivative.

d^2y/dx^2=-sin x - cos x=-1/√2–1/√2 (for x=π/4) which is negative. Hence x=π/4 corresponds to maximum.For x=5π/4

d^2y/dx^2=-(-1/√2)-(-1/√2)=2/√2 a positive quantity. Hence 5π/4 corresponds to minimum

Maximum value of the function

y= sin π/4 + cos π/4= 2/√2=√2

Minimum value is

Sin(5π/4)+cos (5π/4)=-2/√2=-√2