Trigonometric Functions (1 Viewer)

[yeyjasminetime]

2sec^2 (x) -1=0

apparently x= 3pi /2


?????how?????

ty

 

[yeyjasminetime]

You can't cos^-1 (root 2)

 

[Drongoski]

But what was the question? Of course no value of x can satisfy that equation.

 

[yeyjasminetime]

The question was secx(2sec^2 x - 1) = 0

This is actually the derivative of another equation and I have to find values for x that are decreasing.
^^

 

[yeyjasminetime]

in the interval 0 < x < 2pi

 

[Drongoski]

It seems you have completely distorted the intent of the question, by your interpretation. sec x(2sec^2 x -1) = 0 has no solution.

Why not post the original question?

 

[yeyjasminetime]

a) Show that d/dx(secxtanx) = secx (2sec^2 x -1)
b) Hence find the values of x for which the function y = secxtanx is decreasing in the interval 0 < x < 2pi.

THANKYOU!

 

[Drongoski]

Well that makes a world of difference. In this case you are looking for values of x that will satisfy: sec x(2sec^2 x -1) < 0.

Since 2sec^2 x - 1 is always > 0, the only way this inequality can happen is when sec x itself is negative, and subject to the range 0 < x < 2 pi, this can happen only when pi/2 < x < 3pi/2, the required answer.

 

[yeyjasminetime]

I thought i interpreted it correctly though?

 

[yeyjasminetime]

Well that makes a world of difference. In this case you are looking for values of x that will satisfy: sec x(2sec^2 x -1) < 0.

Since 2sec^2 x - 1 is always > 0, the only way this inequality can happen is when sec x itself is negative, and subject to the range 0 < x < 2 pi, this can happen only when pi/2 < x < 3pi/2, the required answer.

Why? I don't understand why "this can happen only when..."

 

[Drongoski]

I thought i interpreted it correctly though?

No. By re-processing the question, you have presented something confusing or misleading. It is dangerous presenting a half-digested question as you can see in this case

 

[yeyjasminetime]

I'm sorry can you please explain it again? I don't understand what 0 < x < 2 pi, has anything to do with pi/2 < x < 3pi/2 just because sec x is negative.

Ty!

 

[Drongoski]

If you are not confined to that range, then you'll have an infinite number of similar ranges of values of x that will satisfy the inequality: e.g. -3pi/2 < x < -pi/2, 5pi/2 < x < 7pi/2, etc if I've worked it out correctly.

 

[yeyjasminetime]

Okay, can I tell you what I did and you tell me where I went wrong?

So to solve b) I wanted to find the stationary points, because I can test around the stationary points to find where y is decreasing.
To do this I need to solve secx (2sec^2 x -1)=0 right?

I then solved (sec x) = 0 first and then (2 sec^2 x -1) = 0

For (sec x) = 0 , I got pi/2 and since that was right I tried to solve (2 sec^2 x -1) = 0 and that's where I got stuck.

 

[Drongoski]

please enter (1/cos 90)) to get sec 90 [pi/2 radian], into your calculator. What do you get?

 

[bokat]

Okay, can I tell you what I did and you tell me where I went wrong?

So to solve b) I wanted to find the stationary points, because I can test around the stationary points to find where y is decreasing.
To do this I need to solve secx (2sec^2 x -1)=0 right?

I then solved (sec x) = 0 first and then (2 sec^2 x -1) = 0
For (sec x) = 0 , I got pi/2 and since that was right I tried to solve (2 sec^2 x -1) = 0 and that's where I got stuck.

secx=0 has no solution. (If you look at the graph of secx it is always greater or equal to one or less than or equal to -1, in other words secx cannot take any values between -1 and 1, so it cannot be zero).
Also 2sec^2x-1=0 has no solution.
Similarly sec^2 x will be always greater than one so 2sec^2 x will be larger than or equal 2 resulting in 2sec^2x-1 being greater or equal to one.
So your derivative will be negative only when secx is negative this is the case when x is in the second or the third quadrant.
So x will be between pi/2 and 3pi/2.
In this case you have no stationary points, but you can determine when your function is decreasing or increasing.
You do not need to look for stationary points all the time (as they do not always exist.) Another example is y=(1/x)^2, it has no stationary points but increases for x<0 and decreases for x>0.

Reza Bokat,
www.cambridgecoaching.com.au