## math>philosophy 3 years ago Prove that the range of $1 + \sin ^{2} x$ is between 1 and 2, inclusive. How do I show this?

1. AravindG

sin x maximum vallue is 1

2. AravindG

so max value of exp will be 2

3. AravindG

now minimum value when x=0 sin x=0

4. AravindG

so minimum value =1

5. AravindG

so 1+x^2 has values [1,2]

6. AravindG

gt tht?

7. AravindG

becaus sin x max value is 1

8. akash_809

sinx , minimum value =-1, sin^2x , minimum value =0 ( sin^2x) cant be negative and maximum value of sin^2x=1....put sin^2x=1 for maximum value and put sin^2x=0 for minimum value

9. akash_809

yes maximum value is 2

10. AravindG

so to maximise sin^x+1 ,,,,sin x maximum should be there

11. AravindG

ie 1+1=2