Question

Find the maximum and minimum values of sin^4x + cos^2x and hence or otherwise find the maximum value of sin^1000 x + cox^2008 x

Answer 1

sin4x + cos2x  = (1-cos2x )2 + cos2x 
= 1 + cos4x -2cos2x + cos2x
= 1+ cos4x -cos2x
= 1/4  + 3/4  + cos4x - cos2x
=(cos2x −12)2 +34 =(2cos2x −12)2 +34=(cos2x)24+34 As (cos2x)2  will lies between [0,1]So above expression has minimum value when  (cos2x)2  = 0 And maximum value when (cos2x)2 = 1 Hence the range is [34,1]

And sin1000x  + cos2008x

Its maximum value will be 1 , as both sinx and cosx range are less than equal to one , so if a number less than 1 has exponents very high like 1000 or 2008 , then it will eventually lead to very small value and ultimately zero.
So the smallest value will be zero and maximum value will  be 1 , when x  = 0 or 90 degree .

I hope this will help you

Answer 2

Answer:

hope it helps sir

Step-by-step explanation:

f(x)=sin  

2

x+(1−sin  

2

x)  

2

 

=1+sin  

2

x−2sin  

2

x+sin  

4

x

=1−sin  

2

x(1−sin  

2

x)

=1−sin  

2

xcos  

2

x

=1−(  

2

sin2x

​  

)  

2

 

sin2x∈[−1,1]  

⇒sin  

2

2x∈[0,1]  

⇒[  

2

sin2x

​  

]  

2

∈[0,  

4

1

​  

]

⇒0≤[  

2

sin2x

​  

]  

2

≤  

4

1

​  

 

⇒0≥−[  

2

sin2x

​  

]  

2

≥−  

4

1

​  

 

⇒1≥1−[  

2

sin2x

​  

]  

2

≥1−  

4

1

​  

 

⇒1≥f(x)≥  

4

3

​  

 

⇒f(x) lies in [  

4

3

​  

,1]