Solve Basic Trigonometry Homework: m^2+n^2=cosec^2θ

[ritwik06]

Homework Statement

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 $$m^{2}+n^{2}=cosec^{2} \theta$$

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 $$cosec ^{2} \theta$$

 

[tiny-tim]

draw the triangle!

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 $$m^{2}+n^{2}=cosec^{2} \theta$$

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 …

 

[ritwik06]

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 …

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 won't solve my purpose, i have tried it.
regards

 

[tiny-tim]

cosine rule for triangles …

Please tell me why am I expected to make a triangle ??

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.

 

[HallsofIvy]

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 won't solve my purpose, i have tried it.
regards

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 [itex]m^2+ m'^2+ 2mm'cos(\theta)= 1[/quote] differs from that only only in having "+" instead of "-". Since $cos(\pi- \theta)= -cos(\theta$), 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.