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Grade: 11
maximum value of the expression 1/sin^2theta+3sinthetacostheta+5cos^2theta 
one year ago

Answers : (2)

24483 Points

For the given expression to be maximum , denominator should be minimum which is sin^2x+3sinxcosx+5 cos^2x should be minimum.

let y= sin^2x+3sinxcosx+5cos^2x

2y= 2sin^2x+6sinxcosx+10cos^2x

(I am multiplying by 2 on both sides for simple calculation . )

2y =sin^2x + 2 (sin x) (3cosx) + (3 cos x )^2 + sin^2x+cos^2x

2y= (sin x + 3 cos x ) ^2 +1

The minimum value of (sin x+3 cos x )^2 is 0

since sin x + 3cos x can be expressed as a single sine function,

sin (x+ n ) where n = arcsin ( 3/ sqrt10)

so the minimum value of 2y is 0+1 =1

and that of y is 1/2

so the minimum value of sin^2 x + 3sinxcosx + cos^2x is 0.5

the maximum value of 1/(sin^2 x + 3sinxcosx + cos^2x) is 2

one year ago
Sai Soumya
99 Points
dear student.
the maximum and minimum value accoding to the above given question is 2 and 0.5 respectively .
one year ago
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