Find the minimum value of 9Tan^2x + 4Cot^2x.
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.
set calculator to radians,
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
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)
To find the minimum value of the expression 9Tan^2x + 4Cot^2x, we can use some trigonometric identities to simplify the expression.
First, let's recall the identity: Tan^2x + Cot^2x = Sec^2x.
Using this identity, we can rewrite the expression as follows:
9Tan^2x + 4Cot^2x = 9(Tan^2x + Cot^2x) + (4-9)Cot^2x
= 9Sec^2x - 5Cot^2x
Next, we can use another identity: Sec^2x - Cot^2x = 1.
Substituting this identity into our expression, we get:
9Sec^2x - 5Cot^2x = 9(1 + Cot^2x) - 5Cot^2x
= 9 + 9Cot^2x - 5Cot^2x
= 9 + 4Cot^2x
Now, we want to minimize the expression, 9 + 4Cot^2x.
Since Cot^2x is always non-negative, the minimum value of 4Cot^2x is 0 (when Cot^2x = 0).
Therefore, the minimum value of the expression 9 + 4Cot^2x is 9.
So, the minimum value of 9Tan^2x + 4Cot^2x is 9.