2) prave that Sine Jtcase) hasmaximum qlve atme3

1.	2) prave that Sine Jtcase) hasmaximum qlve atme3
Answer» y=sinx.(1+cosx) dy/dx=sinx.(-sinx) +cosx.(1+cosx) = -sin^2x+cosx+cos^2x =-1+cos^2 x+cosx+cos^2x =2cos^2x +cosx -1 =2cos^2x+2cosx-cosx-1 =2cosx(cosx+1)-1(cosx+1) (cosx+1)(2cosx-1) For maxima or minima put dy/dx=0 ( cosx+1)(2cosx-1)=0 cosx+1=0 => x=180° 2cosx-1=0 =>cosx=1/2 or x=60° dy/dx=-sin^2x+cosx+cos^2x=cosx+ cos^2x-sin^2x dy/dx=cosx+cos2x d2ydx^2= -sinx-2.sin2x d2y/dx^2 (at x=180° )=0 d2y/dx^2 ( at x=60°) -ve ,there exist maxima at x=60°. Maximum value =sin60°(1+cos60°). = (3^1/2)/2 (1+1/2) =(1/4).(3)^3/2 , Like my answer if you find it useful!

Answer»

y=sinx.(1+cosx)

dy/dx=sinx.(-sinx) +cosx.(1+cosx)

= -sin^2x+cosx+cos^2x

=-1+cos^2 x+cosx+cos^2x

=2cos^2x +cosx -1

=2cos^2x+2cosx-cosx-1

=2cosx(cosx+1)-1(cosx+1)

(cosx+1)(2cosx-1)

For maxima or minima put dy/dx=0

( cosx+1)(2cosx-1)=0

cosx+1=0 => x=180°

2cosx-1=0 =>cosx=1/2 or x=60°

dy/dx=-sin^2x+cosx+cos^2x=cosx+ cos^2x-sin^2x

dy/dx=cosx+cos2x

d2ydx^2= -sinx-2.sin2x

d2y/dx^2 (at x=180° )=0

d2y/dx^2 ( at x=60°) -ve ,there exist maxima

at x=60°.

Maximum value =sin60°(1+cos60°).

= (3^1/2)/2 (1+1/2)

=(1/4).(3)^3/2 ,

Like my answer if you find it useful!

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