# Basic Trigonometry

1. Sep 1, 2008

### ritwik06

1. The problem statement, all variables and given/known data
If $$m^{2}+m' ^{2}+2mm' cos \theta=1$$

$$n^{2}+n' ^{2}+2nn' cos \theta=1$$

$$mn+m'n'+(mn'+m'n)cos \theta =0$$,
then prove that $$m^{2}+n^{2}=cosec^{2} \theta$$

3. The attempt at a solution
I have tri a lot but its difficult to eliminate m' or n'

If I add the first to equations and separate m^2+n^2 to left ide, again its difficult to prov the RHS to $$cosec ^{2} \theta$$

2. Sep 2, 2008

### tiny-tim

draw the triangle!

Hi ritwik06!

Well, looking at the the first two equations, you're obviously supposed to draw a triangle with sides m and m', and an angle of π - θ between them.

Give the name φ to the angle opposite m (so the angle opposite m' is … ?).

Then do the same with n n ' and ψ.

And then use the sine rule …

Last edited: Sep 2, 2008
3. Sep 2, 2008

### ritwik06

Re: draw the triangle!

Hi tim

Well I am not familiar with this. though I have used sine rule in physics. Please tell me why am I expected to make a triangle ?? and what shall I do with this thing? Simply applying:

sin a /a=sin b/ b =sin c /c wont solve my purpose, i have tried it.
regards

4. Sep 2, 2008

### tiny-tim

cosine rule for triangles …

ah … perhaps you haven't done the cosine rule for triangles …

a2 = b2 + c2 - 2bc cosA​

so the reason to make this particular triangle is because the first two equations given are the cosine rule for two triangles, both with one side of length a = 1, and both with opposite angle π - θ, one with the other sides of length m and m', and the other n and n'.

(check for yourself that the cosine rule really does give those two equations for those triangles)

Draw those triangles, and apply the sine rule to get m and m' in terms of θ and φ, and n and n' in terms of θ and ψ.

Then plug those values into the third given equation, which should give you a nice relationship between θ φ and ψ.

5. Sep 2, 2008

### HallsofIvy

Staff Emeritus
Re: draw the triangle!

Because the cosine law says that if a triangle has sides of length a, b, c and A is the angle opposite side A, then c2= a2+ b2- 2ab cos(A).

The equation $m^2+ m'^2+ 2mm'cos(\theta)= 1[/quote] differs from that only only in having "+" instead of "-". Since [itex]cos(\pi- \theta)= -cos(\theta$), that should make you think of a triangle having sides of length m, m', and 1 and having angle $\pi- \theta$ opposite the side of length 1.