prove that the value of cosx + secx can never be 3/2.

1 answer

First make a sketch, here is Wolfram's version
http://www.wolframalpha.com/input/?i=y+%3D+cosx+%2B+1%2Fcosx+for+-%CF%80+to+%2B%CF%80

The vertical lines are asymptotes, and thus will not yield any values. You can see that the min value is
appr 2
Here is a larger picture, showing that it is periodic
http://www.wolframalpha.com/input/?i=y+%3D+cosx+%2B+1%2Fcosx+

let y = cosx + secx
dy/dx = -sinx + secx tanx
= 0 for a max or min

secx tanx = sinx
1/cosx * sinx/cox = sinx
sinx/cos^2 x = sinx
times cos^2 x
sinx = sinx(cos^2 x)
sinx - sinx(cos^2x) = 0
sinx(1 - cos^2 x) = 0
sinx(sin^2 x) = 0
sin^3 x = 0
sinx = 0

sinx x = 0
x = 0, π, 2π, 3π, ...

so let's look at some of the y values
if x = 0,
y = 1 + 1/1 = 2 , ahh, so it was exactly 2
if x = π
y = -1 -1 = -2
So that seems to be our max and min at its corresponding loop
that is, in the main loop between -π and +π, the min is 2

so there are no values of the function that fall between
-2 and +2
and since 3/2 lies in that range, we have proven the case.
Similar Questions
  1. Trigonometric IdentitiesProve: (tanx + secx -1)/(tanx - secx + 1)= tanx + secx My work so far: (sinx/cosx + 1/cosx +
    1. answers icon 0 answers
  2. Prove the following identities.1. 1+cosx/1-cosx = secx + 1/secx -1 2. (tanx + cotx)^2=sec^2x csc^2x 3. cos(x+y) cos(x-y)= cos^2x
    1. answers icon 1 answer
  3. 1/tanx-secx+ 1/tanx+secx=-2tanxso this is what I did: =tanx+secx+tanx-secx =(sinx/cosx)+ (1/cosx)+(sinx/cosx)-(1/cosx)
    1. answers icon 0 answers
  4. Having trouble verifying this identity...cscxtanx + secx= 2cosx This is what I've been trying (1/sinx)(sinx/cosx) + secx = 2cosx
    1. answers icon 2 answers
more similar questions