Question

Find the maximum value of the given function.
y=sin^3 (x) + cos^3(x)-3sinxcosx.
Let t=sinx+cosx. Express y in terms of t
1) find the range of values of t

Answer 1

do you know calculus?

Answer 2

dear @agent0smith.. need your help:)

Answer 3

@primeralph.. not really.. my i know y?

Answer 4

easier when you know calculus.

Answer 5

try me:)

Answer 6

if you knew how to take derivatives this problem would probably be easier because you would take a derivative and find critical points which is where the slope of the original function is = to 0 then you could test values to the left and right of the critical points to judge whether or not they were local maximums or minimums

Answer 7

f'(x)=3sin(x)^(2)*cos(x)-3cos(x)^(2)sin(x)-3[-sin(x)^(2)+cos(x)cos(x)]
someone please check that

Answer 8

let a=cosx and b=sinx now its like a^3 + b^3 - 3ab = (a+b)(a^2 + b^2 - ab) -3ab

now since cos^2x + sin^2x = 1 we can have

(a+b)(1-ab) - 3ab

Answer 9

Dear all, kindly refer to the attached file for your perusal.
need your help.

Answer 10

actually, since: t = sinx + cosx,
using sum to product formula, I got
\(\large t= 2\sqrt2 sin(x+\frac{\pi}{4})\)

Answer 11

which part is unclear?

Answer 12

@mukushla.. have you refer to attached file?
i dont know how to get...\[ since t=\sqrt{2} \sin (x+\pi/4)\]

Answer 13

\[y = \sin ^{3} x + \cos ^{3} x -3sinxcosx = (sinx + cosx)^{3} - 3cosxsinx(cosx+sinx) -3sinxcosx\]

Answer 14

\[t^{3}-3cosxsinx(1+t) \]

Answer 15

and also t=sinx+cosx = \[\sqrt{2}(1/\sqrt{2}cosx + 1/\sqrt{2}sinx)\] = \[\sqrt{2}\sin(x+45)\] and hence the values of x+45 is between -180 and +180

x is between -225 and +135 degrees