# trigo(2)

• August 13th 2009, 04:20 AM
thereddevils
trigo(2)
Find for all 0<=x<=90 , the least and most value of

$\frac{1+cosx}{1+2sinx+2cosx}$

Am i supposed to do it like this :

For most value :

$\frac{1+1}{1+2(1)+2(1)}=\frac{2}{5}$

and just substitute the minimum value of cos and sin which is -1 into the equation ?

or the question means something else . Thanks .
• August 13th 2009, 04:57 AM
songoku
Hi thereddevils
Quote:

Originally Posted by thereddevils
$\frac{1+cosx}{1+2sinx+2cosx}$

Am i supposed to do it like this :

For most value :

$\frac{1+1}{1+2(1)+2(1)}=\frac{2}{5}$

The value of sin (x) and cos (x) can't be 1 for the same value of x :)

EDIT :
Quote:

and just substitute the minimum value of cos and sin which is -1 into the equation ?
For $0\leq x \leq 90$, the value of sin and cos can't be negative
• August 13th 2009, 04:58 AM
mr fantastic
Quote:

Originally Posted by thereddevils
Find for all 0<=x<=90 , the least and most value of

$\frac{1+cosx}{1+2sinx+2cosx}$

Am i supposed to do it like this :

For most value :

$\frac{1+1}{1+2(1)+2(1)}=\frac{2}{5}$

and just substitute the minimum value of cos and sin which is -1 into the equation ?

or the question means something else . Thanks .

No. For starters it's impossible for sin x and cos x to both equal 1 at the same time (that is, for the same value of x). There are other reasons but one is enough.

For the given domain, you will find that the maximum and minimum values occur at the endpoints, that is, at x = 0 degrees and x = 90 degrees respectively.
• August 13th 2009, 06:23 AM
thereddevils
Quote:

Originally Posted by mr fantastic
No. For starters it's impossible for sin x and cos x to both equal 1 at the same time (that is, for the same value of x). There are other reasons but one is enough.

For the given domain, you will find that the maximum and minimum values occur at the endpoints, that is, at x = 0 degrees and x = 90 degrees respectively.

oh ok .tat's a big mistake ! So do i do it like this :

For minimum :

(1+1)/(1+2(0)+2(1))=2/3 ??
• August 13th 2009, 09:35 AM
songoku
Hi thereddevils

Your answer is maximum value. You optimize the value of the numerator and minimize the value of denominator. That will give you the maximum value :)
• August 14th 2009, 01:59 AM
thereddevils
Quote:

Originally Posted by songoku
Hi thereddevils

Your answer is maximum value. You optimize the value of the numerator and minimize the value of denominator. That will give you the maximum value :)

Thanks . But i still don get it . Maybe is because i don understand the question .i know It asks to find the min and max value but the substitution seems to confuse me .
what is maximum and minimum for sin and cos respectively ? Thanks again .
• August 14th 2009, 05:28 AM
mr fantastic
Quote:

Originally Posted by thereddevils
Thanks . But i still don get it . Maybe is because i don understand the question .i know It asks to find the min and max value but the substitution seems to confuse me .
what is maximum and minimum for sin and cos respectively ? Thanks again .

As I suggested in my previous post: Substitute 0 degrees to get the maximum value. Substitute 90 degrees to get the minimum value.

This works because the function is strictly decreasing over the interval [0, 90] (measured in degrees).
• August 14th 2009, 05:36 AM
thereddevils
Quote:

Originally Posted by mr fantastic
As I suggested in my previous post: Substitute 0 degrees to get the maximum value. Substitute 90 degrees to get the minimum value.

This works because the function is strictly decreasing over the interval [0, 90] (measured in degrees).

THanks but sorry for asking so many things . How do i know or how can i see that the function is decreasing from 0 to 90 degrees.
• August 15th 2009, 04:34 AM
songoku
Hi thereddevils
Quote:

Originally Posted by thereddevils
Thanks . But i still don get it . Maybe is because i don understand the question .i know It asks to find the min and max value but the substitution seems to confuse me .
what is maximum and minimum for sin and cos respectively ? Thanks again .

Maximum for sin and cos is 1 and minimum is -1. In this case, minimum value for sin and cos can't be -1. Do you know why? Can you find the minimum value for sin and cos in this case? :)
• August 15th 2009, 06:09 AM
luobo
Quote:

Originally Posted by thereddevils
Find for all 0<=x<=90 , the least and most value of

$\frac{1+cosx}{1+2sinx+2cosx}$

Am i supposed to do it like this :

For most value :

$\frac{1+1}{1+2(1)+2(1)}=\frac{2}{5}$

and just substitute the minimum value of cos and sin which is -1 into the equation ?

or the question means something else . Thanks .

Let $t=\tan\frac{x}{2}$
$0^\circ\leq x \leq 90^\circ \Rightarrow 0 \leq t \leq 1$

Then $\cos x = \frac{1-t^2}{1+t^2}, \sin x=\frac{2t}{1+t^2}$

$\frac{1+\cos x}{1+2\sin x+2\cos x}=\frac{2}{7-(t-2)^2}$
• August 15th 2009, 08:50 AM
thereddevils
Quote:

Originally Posted by songoku
Hi thereddevils

Maximum for sin and cos is 1 and minimum is -1. In this case, minimumvalue for sin and cos can't be -1. Do you know why?(No) you find the minimum value for sin and cos in this case? :)

thanks songoku that's exactly my question . And what is the min and max for sin and cos in this case ?
• August 15th 2009, 09:08 AM
songoku
Hi thereddevils

As Mr. F has said : subs. the endpoints of the domain

When x = 0 --------------------> cos x = 1 and sin x = 0
When x = 90 degrees -------> cos x = 0 and sin x = 1

So, in this case, the maximum value of sin and cos is 1 and minimum value is 0.

When you want to find the maximum value of $\frac{1+cosx}{1+2sinx+2cosx}$, you make the value of the numerator as big as possible and the denominator as small as possible.

Because the max. value of cos is 1, then you take x = 0 to optimize the numerator. For x = 0, the value of sin = 0, so you'll get :

$\frac{1+cosx}{1+2sinx+2cosx}$

$=\frac{1+\cos (0)}{1+2\sin(0)+2\cos(0)}$

$=\frac{1+1}{1+2*0+2*1}$

$=\frac{2}{3}$

To find the min. value, you make the numerator as small as possible and the denominator as big as possible. Give it a try (Sun)
• August 15th 2009, 09:20 AM
thereddevils
Quote:

Originally Posted by songoku
Hi thereddevils

As Mr. F has said : subs. the endpoints of the domain

When x = 0 --------------------> cos x = 1 and sin x = 0
When x = 90 degrees -------> cos x = 0 and sin x = 1

So, in this case, the maximum value of sin and cos is 1 and minimum value is 0.

When you want to find the maximum value of $\frac{1+cosx}{1+2sinx+2cosx}$, you make the value of the numerator as big as possible and the denominator as small as possible.

Because the max. value of cos is 1, then you take x = 0 to optimize the numerator. For x = 0, the value of sin = 0, so you'll get :

$\frac{1+cosx}{1+2sinx+2cosx}$

$=\frac{1+\cos (0)}{1+2\sin(0)+2\cos(0)}$

$=\frac{1+1}{1+2*0+2*1}$

$=\frac{2}{3}$

To find the min. value, you make the numerator as small as possible and the denominator as big as possible. Give it a try (Sun)

Thanks a lot songoku , finally i understood .

Well for the minimum ,

(1+ cos 90)/ (1+2sin 90+2 cos 90) = (1+0)/(1+2(1)+0)=1/3
• August 15th 2009, 09:28 AM
songoku
You're welcome :)