+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.
+118680 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.
+118680 \(16sinxcos^3x-8sinxcosx\\ =8sinxcosx(2cos^2x-1)\\ =4*2sinxcosx(cos^2x+cos^2x-1)\\ =4*sin(2x)(cos^2x+1-sin^2x-1)\\ =4sin(2x)(cos^2x-sin^2x)\\ =2*2sin(2x)cos(2x)\\ =2sin(4x)\)
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)
+1124 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?
+118680 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.
