# To prove : For all values of x , | sin x − cos | ≤ 2 .

### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

#### Solutions

Chapter 4, Problem 2P
To determine

## To prove: For all values of x, |sinx−cos|≤2 .

Expert Solution

### Explanation of Solution

Proof:

Find absolute minimum and absolute maximum of the function f(x)=sinxcosx over the interval [0,2π] .

Observe that sinxandcosx are both periodic functions of period 2π .

Therefore, f(x)=sinxcosx is also a periodic function of period 2π .

So, find the critical points of the function f on the interval [0,2π] .

Differentiate f(x) with respect to x .

f(x)=cosx+sinx .

Set f(x)=0 and obtain the critical points over the interval [0,2π] .

cosx+sinx=0cosx=sinxtanx=1x=3π4or 7π4

Thus, the critical points are 3π4and 7π4 .

Apply the extreme values of the given interval and the critical number in f(x) .

Substitute x=0 in f(x) ,

f(0)=sin0cos0=01=1

Substitute x=2π in f(x) ,

f(2π)=sin2πcos2π=01=1

Substitute x=3π4 in f(x) ,

f(3π4)=sin(3π4)cos(3π4)=12+12=22=2

Substitute x=7π4 in f(x) ,

f(7π4)=sin(7π4)cos(7π4)=1212=22=2

Since the largest functional value is the absolute maximum and the smallest functional value is the absolute minimum, the absolute maximum of f(x) is 2 and the absolute minimum of f(x) is 2 .

Thus, over the interval [0,2π] , 2sinxcosx2 .

Can be expressed as, |sinxcosx|2 .

Therefore, for all values of x, |sinxcos|2 .

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