# How do I prove this seemingly simple trigonometric identity

1. Dec 12, 2014

### Ganesh Ujwal

• 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: Dec 12, 2014
2. Dec 12, 2014

### Staff: Mentor

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

3. Dec 12, 2014

### Ganesh Ujwal

4. Dec 12, 2014

### Staff: 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.

5. Dec 12, 2014

### Ganesh Ujwal

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

6. Dec 12, 2014

### Staff: Mentor

\\ doesn't do anything. At the beginning and end of each line, put # # (no space between).

7. Dec 12, 2014

### Ganesh Ujwal

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

8. Dec 12, 2014

### Staff: 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.

9. Dec 12, 2014

### rcokitsfrollit

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.

10. Dec 12, 2014

### Ganesh Ujwal

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?

11. Dec 12, 2014

### Staff: Mentor

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.

12. Dec 14, 2014

### vela

Staff Emeritus
Aren't you basically done? You have two expressions that equal $16\cos^2\frac{\theta+\phi}2$. Set them equal to each other and simplify.