Trig question

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  • #1
Saitama
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Homework Statement


This is a question from my test paper.

If sec [itex]\theta[/itex]=[itex]\frac{4xy}{(x+y)^2}[/itex], where x,y [itex]\in[/itex] R, then

(a)x,y [itex]\in[/itex] R+
(b)x,y [itex]\in[/itex] R-
(c)x=y[itex]\neq[/itex]0
(d)x[itex]\in[/itex]R+, y[itex]\in[/itex]R-

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The Attempt at a Solution



I don't understand what i have to do and where should i start with? :confused:
 

Answers and Replies

  • #2
NascentOxygen
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Homework Statement


This is a question from my test paper.

If sec [itex]\theta[/itex]=[itex]\frac{4xy}{(x+y)^2}[/itex], where x,y [itex]\in[/itex] R, then

(a)x,y [itex]\in[/itex] R+
(b)x,y [itex]\in[/itex] R-
(c)x=y[itex]\neq[/itex]0
(d)x[itex]\in[/itex]R+, y[itex]\in[/itex]R-

Remembering that sec a is by definition 1/(cos a),
then what values for x and y can you throw into that quotient term
and still yield a value that is a valid value for sec ? Do you need to
restrict the permissible values for x and/or y to a narrower domain than all Reals ?
 
  • #3
Saitama
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Remembering that sec a is by definition 1/(cos a),
then what values for x and y can you throw into that quotient term
and still yield a value that is a valid value for sec ? Do you need to
restrict the permissible values for x and/or y to a narrower domain than all Reals ?

What i did is [itex]cos \theta=\frac{(x+y)^2}{4xy}[/itex].
I substituted different values according to the given conditions in my equation.
I found my answer to be (c) option. Is that correct?
But what should be the correct method to find out the answer? :confused:
 
  • #4
eumyang
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What i did is [itex]cos \theta=\frac{(x+y)^2}{4xy}[/itex].
I substituted different values according to the given conditions in my equation.
I found my answer to be (c) option. Is that correct?
But what should be the correct method to find out the answer? :confused:

What's wrong with guess-and-check as a method? All you need is one example that would make an answer choice wrong.

In any event, you could try using the definition sec θ = R/X in the Cartesian plane (I'm using capital letters to distinguish from the lower case x and y that you already used), and that R2 = X2 + Y2. Find Y, and then look at the answer choices again to see what must hold.
 
  • #5
Saitama
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What's wrong with guess-and-check as a method? All you need is one example that would make an answer choice wrong.

In any event, you could try using the definition sec θ = R/X in the Cartesian plane (I'm using capital letters to distinguish from the lower case x and y that you already used), and that R2 = X2 + Y2. Find Y, and then look at the answer choices again to see what must hold.

What's this "R"? :confused:
 
  • #6
eumyang
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What's this "R"? :confused:
"R" stands for radius. See the attached diagram, and the website below (look under the heading "TRIGONOMETRIC FUNCTIONS").
http://www.mathacademy.com/pr/prime/articles/trig_ident/index.asp" [Broken]
 

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  • #7
Saitama
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See the attached diagram, and the website below (look under the heading "TRIGONOMETRIC FUNCTIONS").
http://www.mathacademy.com/pr/prime/articles/trig_ident/index.asp" [Broken]

Ok, i got it, thanks!! :smile:

I have one more question.
If f(x)=|sin x|+|cos x|, x[itex]\in[/itex]R, then
(a)f(x)[itex]\in[/itex][0,2]
(b)f(x)[itex]\in[/itex][0,[itex]\sqrt{2}[/itex]]
(c)f(x)[itex]\in[/itex][0,1]
(d)f(x)[itex]\in[/itex][1,[itex]\sqrt{2}[/itex]]

How should i start with in this question? :confused:
 
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  • #8
PeterO
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What i did is [itex]cos \theta=\frac{(x+y)^2}{4xy}[/itex].
I substituted different values according to the given conditions in my equation.
I found my answer to be (c) option. Is that correct?
But what should be the correct method to find out the answer? :confused:

I like that you have identified that it is unsuitable for both x and y to be 0, but if x and y are alternately positive or negative real numbers, then the possibility of them both being zero is still excluded.

Think about the maximum and minimum value the cosine function can have, and the significance that would have to the relative sizes of the numerator and denominator of the fraction you have.
 
  • #9
eumyang
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Ok, i got it, thanks!! :smile:

I have one more question.
If f(x)=|sin x|+|cos x|, x[itex]\in[/itex]R, then
(a)f(x)[itex]\in[/itex][0,2]
(b)f(x)[itex]\in[/itex][0,[itex]\sqrt{2}[/itex]]
(c)f(x)[itex]\in[/itex][0,1]
(d)f(x)[itex]\in[/itex][1,[itex]\sqrt{2}[/itex]]

How should i start with in this question? :confused:

Start by finding the range of sin x and cos x. In other words, what are all the possible values of y in y = sin x, and what are all the possible values of y in y = cos x?
 
  • #10
PeterO
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Ok, i got it, thanks!! :smile:

I have one more question.
If f(x)=|sin x|+|cos x|, x[itex]\in[/itex]R, then
(a)f(x)[itex]\in[/itex][0,2]
(b)f(x)[itex]\in[/itex][0,[itex]\sqrt{2}[/itex]]
(c)f(x)[itex]\in[/itex][0,1]
(d)f(x)[itex]\in[/itex][1,[itex]\sqrt{2}[/itex]]

How should i start with in this question? :confused:

Perhaps a sketch graph may help? [use adding ordinates if necessary]
 
  • #11
Saitama
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Start by finding the range of sin x and cos x. In other words, what are all the possible values of y in y = sin x, and what are all the possible values of y in y = cos x?

Range of sin x and cos x is [-1,1].
What next?
 
  • #12
eumyang
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Range of sin x and cos x is [-1,1].
What next?
Well, what happens when you take the absolute value of any number between -1 and 1 inclusive? What does the range become?
 
  • #13
Saitama
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Well, what happens when you take the absolute value of any number between -1 and 1 inclusive? What does the range become?

Taking the absolute value the range becomes [0,1].
 
  • #14
eumyang
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Taking the absolute value the range becomes [0,1].
Okay, so the range of |sin x| is [0, 1], and the range of |cos x| is [0, 1]. What would be the range of f(x)=|sin x| + |cos x|, then?
 
  • #15
PeterO
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Taking the absolute value the range becomes [0,1].

That is the separate range of each function, which you have to add together.

Can they both be zero at the same time? can they both be 1 at the same time?

You might need that sketch graph now.
 
  • #16
Saitama
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Okay, so the range of |sin x| is [0, 1], and the range of |cos x| is [0, 1]. What would be the range of f(x)=|sin x| + |cos x|, then?

That only i don't know. I don't understand how to do that? :confused:
 
  • #17
eumyang
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EDIT: Big mistake...
 
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  • #18
Saitama
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It just requires some thinking. You need to add some number (between 0 and 1 inclusive) to another number (between 0 and 1 inclusive). You are looking for the range of numbers that encompasses all of the possible sums.

If |sin x| = 0 and |cos x| = 0, then
|sin x| + |cos x| = ? Wouldn't that number be in the range of f(x)? Try choosing different values of |sin x| and |cos x| and see if you can't come up with the range of f(x).

I add the max and min values of the range of sin x and cos x.
If |sin x| = 0 and |cos x| = 0, then
|sin x| + |cos x| = 0.
If |sin x| = 1 and |cos x| = 1, then
|sin x| + |cos x| = 2

Is the answer (a) option?
 
  • #19
eumyang
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EDIT: Another big mistake...
 
Last edited:
  • #20
PeterO
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that only i don't know. I don't understand how to do that? :confused:

sketch graph!!! the answer is not (a)
 
  • #21
Saitama
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Yes. Just know that looking at the max and min values for each interval isn't always going to work.

Oops, i ticked the wrong answer in my exam!! :redface:

sketch graph

I won't be able to sketch the graph correctly and i am not that comfortable in adding graphs.
 
  • #22
PeterO
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Oops, i ticked the wrong answer in my exam!! :redface:



I won't be able to sketch the graph correctly and i am not that comfortable in adding graphs.

The sketch graph for the sine partial and the cosine partial is just a series of off-set bumps [it would be great if you could draw real sine curves, but even if each bump looks like an inverted parabola will show you. heck, even a bunch of half circles like tracing round a dime].
 
  • #23
eumyang
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Oops, i ticked the wrong answer in my exam!! :redface:
Maybe not. I have looked over this problem again and I had misguided you all this time.

Previously, I had asked,
if |sin x| = 0 and |cos x| = 0, then what is |sin x| + |cos x|?

Well, it's not possible for both to equal 0 at the same time, isn't it?

If |sin x| = 0, then |cos x| must equal 1. And if |cos x| = 0 then |sin x| must equal 1. Then the minimum value for |sin x| + |cos x| is not 0, is it?
 
  • #24
Saitama
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Maybe not. I have looked over this problem again and I had misguided you all this time.

Previously, I had asked,
if |sin x| = 0 and |cos x| = 0, then what is |sin x| + |cos x|?

Well, it's not possible for both to equal 0 at the same time, isn't it?

If |sin x| = 0, then |cos x| must equal 1. And if |cos x| = 0 then |sin x| must equal 1. Then the minimum value for |sin x| + |cos x| is not 0, is it?

Oops sorry, that was my mistake too that i didn't noticed it. :redface:
So If |sin x| = 0, then |cos x|= 1 and |sin x|+|cos x|=1.
But what is the answer then? :confused:
 
  • #25
PeterO
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Taking you through the sketch graphs.

A normal sine function is a bump beteen 0 and pi [180degrees if you like], then a dip under the axis between pi and 2xpi.
The absolute value of the sine function is just two bumps beside each other.

Please acknowlege you understand that.
 
  • #26
Saitama
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Taking you through the sketch graphs.

A normal sine function is a bump beteen 0 and pi [180degrees if you like], then a dip under the axis between pi and 2xpi.
The absolute value of the sine function is just two bumps beside each other.

Please acknowlege you understand that.

I knew that. :smile:
And i know the graph of |cos x| too but i don't know to add |sin x|+|cos x|. :frown:
 
  • #27
eumyang
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Oops sorry, that was my mistake too that i didn't noticed it. :redface:
So If |sin x| = 0, then |cos x|= 1 and |sin x|+|cos x|=1.
But what is the answer then? :confused:

If you look at the |sin x| values, as they go from 0 to 1,
the corresponding |cos x| values must go from 1 to 0.

Consider that |sin x| + |cos x| would reach the maximum if |sin x| EQUALED |cos x|. And you know the value of x that makes |sin x| = |cos x| true, don't you?
 
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  • #28
Saitama
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If you look at the |sin x| values, as they go from 0 to 1,
the corresponding |cos x| values must go from 1 to 0.

Consider that |sin x| + |cos x| would reach the maximum if |sin x| EQUALED |cos x|. And you know the value of x that makes |sin x| = |cos x| true, don't you?

Ya, i know. For |sin x|=|cos x|, x=(pi/4).
Right..?
 
  • #29
PeterO
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I knew that. :smile:
And i know the graph of |cos x| too but i don't know to add |sin x|+|cos x|. :frown:

i am glad you know that - I thought you would.

As you know the graph of |cos x| , over the 0 to 2pi [0 to 360] range starts at 1, loops down to zero, just as the sine reaches it peak, loops back up to 1 just as the first sine loop drops to zero, then repeats.

To add the sketches together you draw a series of fine , faint, vertical lines across the sketch of the two graphs. The first is at x = 0, the last is at x = 2 x pi and you want about 12 to 16 of them evenly spaced.

You then measure/estimate how far above the axis the lower graph is, and mark a point that far above where the vertical cuts the upper graph. You then join the dots.

Importantly, you will see that when ever one graph has a zero value, the other one has a value of 1. In fact there is no point where both graphs are at zero, so when you add them you always get a total more than zero.
The answewr you want at the end, is the one that does NOT have a range starting at zero - option (d).
 
  • #30
eumyang
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Ya, i know. For |sin x|=|cos x|, x=(pi/4).
Right..?
Yes. Now find |sin x| + |cos x| if x = π/4.
 
  • #31
PeterO
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Ya, i know. For |sin x|=|cos x|, x=(pi/4).
Right..?

That is correct!!, and the maximum value the sum reaches at that point is Square root 2.

See my other answer as to why the minimum is 1 and the maximum shown here is root 2.

whoops - two overlapping responses too.
 
  • #32
Saitama
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i am glad you know that - I thought you would.

As you know the graph of |cos x| , over the 0 to 2pi [0 to 360] range starts at 1, loops down to zero, just as the sine reaches it peak, loops back up to 1 just as the first sine loop drops to zero, then repeats.

To add the sketches together you draw a series of fine , faint, vertical lines across the sketch of the two graphs. The first is at x = 0, the last is at x = 2 x pi and you want about 12 to 16 of them evenly spaced.

You then measure/estimate how far above the axis the lower graph is, and mark a point that far above where the vertical cuts the upper graph. You then join the dots.

Importantly, you will see that when ever one graph has a zero value, the other one has a value of 1. In fact there is no point where both graphs are at zero, so when you add them you always get a total more than zero.
The answewr you want at the end, is the one that does NOT have a range starting at zero - option (d).

The same way is followed by my teacher but in the exam, graphs go out of my mind and i don't like adding them. Yet they are really helpful when finding out the number of solutions for a equation. :smile:

Yes. Now find |sin x| + |cos x| if x = π/4.
|sin x| + |cos x|=[itex]\sqrt{2}[/itex] if x=pi/4.
 
  • #33
eumyang
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Well, no thanks to PeterO, the answer was already out of the bag. Hopefully, that answer was the one you marked.

And I really shouldn't be posting late at night (it's 1:30am where I currently am) and a little drunk. :redface: Sorry for the mistakes earlier.
 
  • #34
Saitama
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Well, no thanks to PeterO, the answer was already out of the bag. Hopefully, that answer was the one you marked.

And I really shouldn't be posting late at night (it's 1:30am where I currently am) and a little drunk. :redface: Sorry for the mistakes earlier.

Yes, i ticked the (d) option. :smile:

WTH!! You are still awake at 1:30am. Here it's 10:30PM.
 

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