# Trig question

1. Jul 29, 2011

### Saitama

1. The problem statement, all variables and given/known data
This is a question from my test paper.

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

(a)x,y $\in$ R+
(b)x,y $\in$ R-
(c)x=y$\neq$0
(d)x$\in$R+, y$\in$R-

2. Relevant equations

3. The attempt at a solution

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

2. Jul 29, 2011

### Staff: Mentor

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. Jul 29, 2011

### Saitama

What i did is $cos \theta=\frac{(x+y)^2}{4xy}$.
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?

4. Jul 29, 2011

### eumyang

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. Jul 29, 2011

### Saitama

What's this "R"?

6. Jul 29, 2011

### eumyang

"R" stands for radius. See the attached diagram, and the website below (look under the heading "TRIGONOMETRIC FUNCTIONS").

#### Attached Files:

• ###### trig_cart.jpg
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Last edited by a moderator: May 5, 2017
7. Jul 29, 2011

### Saitama

Ok, i got it, thanks!!

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

Last edited by a moderator: May 5, 2017
8. Jul 29, 2011

### PeterO

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. Jul 29, 2011

### eumyang

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. Jul 29, 2011

### PeterO

Perhaps a sketch graph may help? [use adding ordinates if necessary]

11. Jul 29, 2011

### Saitama

Range of sin x and cos x is [-1,1].
What next?

12. Jul 29, 2011

### eumyang

Well, what happens when you take the absolute value of any number between -1 and 1 inclusive? What does the range become?

13. Jul 29, 2011

### Saitama

Taking the absolute value the range becomes [0,1].

14. Jul 29, 2011

### eumyang

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. Jul 29, 2011

### PeterO

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. Jul 29, 2011

### Saitama

That only i don't know. I don't understand how to do that?

17. Jul 29, 2011

### eumyang

EDIT: Big mistake...

Last edited: Jul 29, 2011
18. Jul 29, 2011

### Saitama

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

19. Jul 29, 2011

### eumyang

EDIT: Another big mistake...

Last edited: Jul 29, 2011
20. Jul 29, 2011

### PeterO

sketch graph!!! the answer is not (a)