Find the range of the function f(x)=cos(x)[sin(x)+sqrt(sin^2(x)+1/2)]

Brooke Richard 2022-11-21 Answered
Find the range of the function f ( x ) = cos ( x ) [ sin ( x ) + sin 2 ( x ) + 1 2 ]
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Answers (1)

boursecasa2je
Answered 2022-11-22 Author has 15 answers
f ( x ) cos x sin x = cos x 1 2 + sin 2 x f ( x ) 2 2 f ( x ) cos x sin x = 1 2 cos 2 x 4 f ( x ) cos x sin x = 2 f ( x ) 2 cos 2 x 16 f ( x ) 2 cos 2 x ( 1 cos 2 x ) = ( 2 f ( x ) 2 cos 2 x ) 2
f ( x ) 2 = y ,   cos 2 x = t :
16 y t ( 1 t ) = ( 2 y t ) 2 4 y 2 + ( 16 t 2 20 t ) y + t 2 = 0 8 y y + ( 16 t 2 20 t ) y + ( 32 t 20 ) y + 2 t = 0
Here y = d y d t . In critical points d f ( x ) d x = 0 d y d t = 0
( 32 t 20 ) y + 2 t = 0 y = t 16 t 10 4 t 2 ( 16 t 10 ) 2 t ( 16 t 2 20 t ) 16 t 10 + t 2 = 0 4 t 2 t ( 16 t 2 20 t ) ( 16 t 10 ) + t 2 ( 16 t 10 ) 2 = 0 32 t 2 ( 5 t 3 ) = 0
t = 0 gives minimum y = 0. t = 3 5 gives maximum y = 3 2
Minimum f ( x ) = 3 2 is obtained at cos x = 3 5 , sin x = 2 5
Maximum f ( x ) = 3 2 is obtained at cos x = 3 5 , sin x = 2 5
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