Trig question

  • Thread starter Saitama
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  • #26
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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
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92
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
3,812
92
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
3,812
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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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