Question

The maximum value of the expression 1/(sin^2 x + 3sinx cox + 5 cos^2 x)

Answer 1

@satellite73 @satellite73 @satellite73 @satellite73 @satellite73 @satellite73 @satellite73

Answer 2

@amistre64 @amistre64 @amistre64 @amistre64 @amistre64 @amistre64 @amistre64 @amistre64

Answer 3

hello?

Answer 4

hi..@satellite73 plzz help

Answer 5

the bottom looks like it could be quadratic withthe proper subs

Answer 6

a^2 + 3ab + 5b^2

Answer 7

or, since cos^2 = 1-sin^2, that might help reduce the clutter

Answer 8

scratch that, it is wrong

Answer 9

what amistre said
same as \(4\cos^2(x)+3\cos(x)\sin(x)+1\) by replacing \(\sin^2(x)+\cos^2(x)=1\)

Answer 10

apparently the minimum of the denominator is \(\frac{1}{2}\) so the max is \(2\) but i do not see why yet
amistre?

Answer 11

yup @satellite73 u r correct

Answer 12

but hw did u get that

Answer 13

no, wolfram is

Answer 14

(sin^2 x + 3sinx cox + 5 cos^2 x)^(-1)

-(sin^2 x + 3sinx cox + 5 cos^2 x)^(-2) * (2sinxcosx + 3cos^2x - 3 sin^2x - 5sinxcosx)

-(sin^2 x + 3sinx cox + 5 cos^2 x)^(-2) * (3cos^2x - 3 sin^2x -3sinxcosx) = 0; when 3cos^2x - 3 sin^2x -3sinxcosx = 0; for extrema

Answer 15

3cos^2x - 3 sin^2x -3sinxcosx

cos^2x - sin^2x -sinxcosx = 0

Answer 16

cos^2x - (1-cos^2x) - sin(2x)/2 = 0

cos^2x - 1 +cos^2x - sin(2x)/2 = 0

2cos^2x - sin(2x)/2 = 1

4cos^2x - sin(2x) = 2

etc ....

Answer 17

unfortunately i seem to be getting the zero of the derivative at \(\tan^{-1}(\frac{1}{3})\)

Answer 18

the wolf says we can simplify to:

sin(2x) = 2cos(2x)

tan(2x) = 2

2x = tan-1(2)

x = tan-1(2)/2 abt:: .5536 rads, or aby 31 degrees

Answer 19

i loathe trigging stuff lol

Answer 20

i was just working with the denominator to find the min
got the derivative via wolf as \(3\cos(2x)-4\sin(2x)\) which is a clue that the problem is cooked somehow, because we can write this as a single function of sine via
\[3\cos(2x)-4\sin(2x)=-5\sin\left(2x-\tan^{-1}(\frac{3}{4})\right)\]

Answer 21

but i am stuck from there
this is zero when \[2x=\tan^{-1}(\frac{3}{4}) \]

Answer 22

i think there must be some simpler way to tackle this