Tuesday
May 21, 2013

# Homework Help: TRIGO

Posted by anandi on Friday, June 3, 2011 at 5:25am.

Find the minimum value of 9Tan^2x + 4Cot^2x.

• TRIGO - Reiny, Friday, June 3, 2011 at 7:29am

take the derivative and set that equal to zero

let y = 9tan^2 x + 4cot^2 x
dy/dx = 18tanx(sec^2x) + 8cotx(-csx^2x)
= 0

18(sinx)/(cosx)(1/cos^2x) + 8(cosx/sinx)(-1/sin^2x) = 0
18sinx/cos^3x) - 8cosx/sin^3x = 0
18sinx/cos^3x) = 8cosx/sin^3x
18sin^4x = 8cos^4x
sin^4x/cos^4x = 8/18
tan^4x = 4/9
tanx = ± (4/9)^.25 = ± .8165 appr.

x = .6847 or π-.6847 or π+.6847 or 2π-.6847

sub each of those into original to see which one gives the smaller value

• TRIGO - Mgraph, Friday, June 3, 2011 at 8:39am

Let z=tan^2(x), z>0
F(z)=9z+4/z,
F'(z)=9-4/z^2, F'(z)=0 if z=2/3
minF(z)=F(2/3)=12

(corresponding value of x exists)

Related Questions

math - I'm having some troubles with these. Thanks in advance. If a&gt;0...
MATH - How do i solve the pair of simultaneous equations y=4-2x y=2x^2-3x+1 ?? ...
Algebra- Maximum and minimum. - I dont understand how to find the maximum and ...
Algebra2 - Complete parts a – c for each quadratic function: a. Find the y-...