# Homework Help: Two trigometric functions intersect point

1. Mar 20, 2007

### likeachild

1. The problem statement, all variables and given/known data

I am trying to find out how to solve for x without a calculator.
Basically where $$tan({x})$$ and $$sqrt{2}*cos{x}$$ intersect.

2. Relevant equations

Find $$x$$ in the range of $$0 \le {x} \le \frac {pi}{2}$$

3. The attempt at a solution

I couldn't find out how to solve this without a calculator.
I tried fooling around with the trigometric properties like the double argument and pythagorean, but I still couldn't find out.
My teacher doesn't know either. lol.

The answer by looking at it graphically is $$\frac {pi}{4}$$

Last edited: Mar 20, 2007
2. Mar 20, 2007

### christianjb

sin(x)=sqrt(2)cos^2(x)

cos^2(x)=1-sin^2(x)

sin(x)=sqrt(2)[1-sin^2(x)]

Quadratic eqn- solve for sin(x)

3. Jun 13, 2011

### Staff: Mentor

Doing without a calculator hints that the solution might be something like 30 deg, or 45 deg, or 90 deg, etc., an angle whose sin, cos, tan you should have memorized. Let's try that....

There are 2 right-angled triangles you need to be able to sketch without even thinking:-

1) an isosceles triangle with base angles of 45 deg. (label the sides 1,1, and root something)
2) a triangle with angles of 30, 60 and 90 degrees, and sides of 1,2, and root something

Construct them. These allow you to, by inspection, write down equations for sin 45, sin 60, tan 30, tan 45, and so on.

Good luck!

Last edited: Jun 13, 2011
4. Jun 18, 2011

### AC130Nav

Since the two items noted are not formulae, as I understand it they can't intersect. What did you really mean? Are the values equal?

5. Jun 18, 2011

### Staff: Mentor

Question concerns two graphs,
viz., y = tan x
and y = root2 * cos x

for x between 0 and Pi / 2 the curves intersect at one point.

6. Jun 19, 2011

### HallsofIvy

Your equation is
$$\frac{sin(x)}{cos(x)}= \sqrt{2}cos(x)$$

Convert ever thing to sin(x) and you will have quadratic equation in sin(x).

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook