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avatar+1124 

Hi friends,

 

Please kindly explain this to me:

 

For which value of x for the interval X{0; 90} will \(16Sinx.Cos^3x - 8Sinx.Cosx\) have it's minimum value?

 

I have no idea what the meaning of the question is, let alone try to solve it.. Please help...Thank you all very much.

 Mar 4, 2023
 #1
avatar+118617 
+1

Do you want the answer in degrees or radians? Edit:  Sorry I see it is in degrees.

I have been surprised that your work has been in degrees today.

These types of questions are usually in radians.

 

Also,

Normally I would attack this as a calculus problem but i i expect your calculus is not so great so I have attacked it from a non-calculus angle. 

 

I'll answer properly soon.

 Mar 4, 2023
edited by Melody  Mar 4, 2023
 #2
avatar+118617 
+1

\(16sinxcos^3x-8sinxcosx\\ =8sinxcosx(2cos^2x-1)\\ =4*2sinxcosx(cos^2x+cos^2x-1)\\ =4*sin(2x)(cos^2x+1-sin^2x-1)\\ =4sin(2x)(cos^2x-sin^2x)\\ =2*2sin(2x)cos(2x)\\ =2sin(4x)\)

 

This is a sin curve 

it has a amplitude of 2

y=0 is the midline  so     -2<=y<=2

the wavelength will be   360/4 = 90 degrees

90/4= 22.5 degrees

It will pass through

(0,0),    (22.5,2),   (45,0),    (67.5, -2)  and   (90,0)

 

The minimum value is -2  and this is when x = 67.5 degrees

 

Here is the graph.

 

 

 

 

 

LaTex:

16sinxcos^3x-8sinxcosx\\
=8sinxcosx(2cos^2x-1)\\
=4*2sinxcosx(cos^2x+cos^2x-1)\\
=4*sin(2x)(cos^2x+1-sin^2x-1)\\
=4sin(2x)(cos^2x-sin^2x)\\
=2*2sin(2x)cos(2x)\\
=2sin(4x)

 Mar 4, 2023
 #3
avatar+1124 
0

PLEASE COME TO SOUTH AFRICA!!!

 

YOU MAKE THIS SOO VERY EASY TO UNDERSTAND...

 

If you did this using Calculus...would you get the derivative of the \(2Sin(4x)\)?..I tried that...but got stuck?

juriemagic  Mar 4, 2023
edited by juriemagic  Mar 4, 2023
edited by juriemagic  Mar 4, 2023
 #4
avatar+118617 
+1

Thanks, I'd love to come to South Africa one day  :)

 

The actual curve when simplified is   y= 2sin(4x)

the simplified differential is    y'=8cos(4x)

the second differential is y''= -32sin(4x)

 

The minimum will be when  y'=0  and   y">0

 

y'=0

8cos(4x)=0

cos(4x)=0

4x=90, 270

x=22.5  ,    67.5 degrees

 

when x=22.5    y-32*sin90 = -1     Maximum

when x=67.5     y"=-32*sin270 = -32*-1 = 32    Minimum.

Melody  Mar 4, 2023
 #5
avatar+1124 
0

Perfect, I just love this...thank you Melody

juriemagic  Mar 4, 2023
 #6
avatar+118617 
0

You are very welcome :)

Melody  Mar 4, 2023

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