# How do I prove this seemingly simple trigonometric identity

Originally posted in a technical section, so missing the template
Mod note: Fixed the LaTeX.
##a=sinθ+sinϕ##

##b=tanθ+tanϕ##

##c=secθ+secϕ##

Show that,

##8bc=a[4b^2 + (b^2-c^2)^2]##

I tried to solve this for hours and have gotten no-where. Here's what I've got so far :

##a= 2\sin(\frac{\theta+\phi}{2})\cos(\frac{\theta-\phi}{2}) ##

## b = \frac{2\sin(\theta+\phi)}{\cos(\theta+\phi)+\cos(\theta-\phi)}##

##c=\frac{2(\cos\theta+\cos\phi)}{\cos(\theta+\phi)+\cos(\theta-\phi)}##

##a^2 = \frac{\sin^2(\theta+\phi)[\cos(\theta+\phi)+1]}{\cos(\theta+\phi)+1}##

##\\cos(\theta-\phi)=\frac{ca}{b}-1##

##sin^2(\frac{\theta+\phi}{2})=\frac{2a^2b}{4(ca+b)}##

Last edited:

jedishrfu
Mentor
There's something wrong with your latex as it doesn't render at all.

There's something wrong with your latex as it doesn't render at all.

jedishrfu
Mentor
You need to edit your post to fix it. I don't have the authority to do that.

You're probably missing the tags for embedding latex in your post.

how to embed latex? please show to me.
i am waiting

Mark44
Mentor
how to embed latex? please show to me.
\\ doesn't do anything. At the beginning and end of each line, put # # (no space between).

alright then please anybody solve my problem, i am stuck.

jedishrfu
Mentor
Did you try simply subbing a,b and c expressions into the right hand side to see what you get.

Also you didn't show us what identities you know that would be relevant to this problem.

The ratio, a/b and a/c are very useful in this problem. Simplify your trig expressions for those a= , b= , and c= , and then using trig identities for sin(x+y) and cos(x+y), and cos(x-y), the algebra will simplify greatly. Do not immediately substitute those more complex trig expressions into your equation in a, b, and c.

i tried like this:
\begin{align*}4\cos^2(\theta)\cos^2(\phi)b^2 &= 4\cos^2(\theta)\cos^2(\phi)(\tan(\theta) + \tan(\phi))^2 \\
&= 4(\sin(\theta)\cos(\phi) + \cos(\theta)\sin(\phi))^2 \\
&= 4\sin^2(\theta + \phi) \\
&= 16\sin^2((\theta + \phi)/2)\cos^2((\theta + \phi)/2) \\
&= 16(1 - \cos^2((\theta + \phi)/2))\cos^2((\theta + \phi)/2) \\
&= 16\cos^2((\theta + \phi)/2) - 16\cos^4((\theta + \phi)/2)\end{align*}
\begin{align*}c^2 - b^2 &= \sec^2\theta + \sec^2\phi + 2\sec(\theta)\sec(\phi) - (\tan^2\theta + \tan^2\phi + 2\tan(\theta)\tan(\phi)) \\
&= 2 + 2\sec(\theta)\sec(\phi) - 2\tan(\theta)\tan(\phi)\end{align*}
\begin{align*}\cos(\theta)\cos(\phi)(c^2 - b^2) &= 2(\cos(\theta)\cos(\phi) + 1 - \sin(\theta)\sin(\phi)) \\
&= 2(1 + \cos(\theta + \phi)) \\
&= 4\cos^2((\theta + \phi)/2)\end{align*}
\begin{align*}cos^2(\theta)\cos^2(\phi)(b^2 - c^2)^2 = 16\cos^4((\theta + \phi)/2\end{align*}
\begin{align*}cos^2(\theta)\cos^2(\phi)(4b^2 + (b^2 - c^2)^2) = 16\cos^2((\theta + \phi)/2\end{align*}
\begin{align*}8\cos^2(\theta)\cos^2(\phi)bc/a &= 8\cos^2(\theta)\cos^2(\phi)(\tan(\theta) + \tan(\phi))(\sec(\theta) + \sec(\phi))/(\sin(\theta) + \sin(\phi)) \\
&= 8(\sin(\theta)\cos(\phi) + \cos(\theta)\sin(\phi))(\cos(\theta) + \cos(\phi))/(\sin(\theta) + \sin(\phi)) \\
&= 8\sin(\theta + \phi)(\cos(\theta) + \cos(\phi))/(\sin(\theta) + \sin(\phi)) \\
&= \frac{8(2\sin((\theta + \phi)/2)\cos((\theta + \phi)/2))(2\cos((\theta + \phi)/2)\cos((\theta - \phi)/2))}{2\sin((\theta + \phi)/2)\cos((\theta - \phi)/2)} \\
&= 16\cos^2((\theta+\phi)/2)\end{align*}
now how should i proceed it?

Mark44
Mentor
alright then please anybody solve my problem, i am stuck.
The rules in this forum (under Homework Guidelines at https://www.physicsforums.com/threads/physics-forums-global-guidelines.414380/) don't permit solving your problem for you.
Giving Full Answers: On helping with questions: Any and all assistance given to homework assignments or textbook style exercises should be given only after the questioner has shown some effort in solving the problem. If no attempt is made then the questioner should be asked to provide one before any assistance is given. Under no circumstances should complete solutions be provided to a questioner, whether or not an attempt has been made.

vela
Staff Emeritus