How do I prove this seemingly simple trigonometric identity

$$a = \sin\theta+\sin\phi\\b=\tan\theta+\tan\phi\\c=\sec\theta+\sec\phi$$ 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 : $$\\a= 2\sin(\frac{\theta+\phi}{2})\cos(\frac{\theta-\phi}{2}) \\ b = \frac{2\sin(\theta+\phi)}{\cos(\theta+\phi)+\cos(\theta-\phi)} \\c=\frac{2(\cos\theta+\cos\phi)}{\cos(\theta+\phi)+\cos(\theta-\phi)} \\a^2 = \frac{\sin^2(\theta+\phi)[\cos(\theta+\phi)+1]}{\cos(\theta+\phi)+1}\\\cos(\theta-\phi)=\frac{ca}{b}-1\\\sin^2(\frac{\theta+\phi}{2})=\frac{2a^2b}{4(ca+b)}$$

• Have you tried substituting the trigonometric forms for each variable into the final equation and seeing if the two sides are equivalent? – Ari Dec 2 '14 at 11:42
• @Ari What do you mean exactly ? – David Sebastian Dec 2 '14 at 13:20

Hint: Divide both sides of $8bc=a[4b^2 + (b^2-c^2)^2]$ by $bc$. You will end up with this: $$8 = a[4\frac{b}{c}+bc(\frac{b}{c}-\frac{c}{b})^2]$$ Take the RHS and you can prove that it is equal to 8.

Process: Calculate $\frac{b}{c}$ from the given equations. $$b = \frac{\sin (\theta + \phi)}{\cos \theta. \cos \phi}$$ $$c = \frac{\cos \theta + \cos \phi}{\cos \theta. \cos \phi}$$ $$\frac{b}{c} = \frac{2\sin(\frac{\theta + \phi}{2}).\cos(\frac{\theta + \phi}{2})}{2\cos(\frac{\theta + \phi}{2}).\cos(\frac{\theta - \phi}{2})}$$ $$\frac{b}{c} = \frac{\sin(\frac{\theta + \phi}{2})}{\cos(\frac{\theta - \phi}{2})}$$ You will end up with: $$\frac{b}{c}-\frac{c}{b}=-\frac{\cos \theta. \cos \phi}{\cos(\frac{\theta - \phi}{2}).sin(\frac{\theta + \phi}{2})}$$ $$bc(\frac{b}{c}-\frac{c}{b})^2 = \frac{\sin (\theta + \phi)(\cos \theta + \cos \phi)}{\cos^2 \theta. \cos^2 \phi}\frac{\cos^2 \theta. \cos^2 \phi}{\cos^2(\frac{\theta - \phi}{2}).\sin^2(\frac{\theta + \phi}{2})}$$ $$= \frac{2\sin(\frac{\theta + \phi}{2}).\cos(\frac{\theta + \phi}{2})2\cos(\frac{\theta + \phi}{2}).\cos(\frac{\theta - \phi}{2})}{\cos^2(\frac{\theta - \phi}{2}).\sin^2(\frac{\theta + \phi}{2})}$$ $$= \frac{4\cos^2 (\frac{\theta + \phi}{2})}{\cos(\frac{\theta - \phi}{2})\sin (\frac{\theta + \phi}{2})}$$ $$4\frac{b}{c} + bc(\frac{b}{c}-\frac{c}{b})^2 = \frac{4}{\cos (\frac{\theta - \phi}{2})}(\sin(\frac{\theta + \phi}{2})+\frac{\cos^2 (\frac{\theta + \phi}{2})}{\sin(\frac{\theta + \phi}{2})})$$ $$= \frac{4}{\cos (\frac{\theta - \phi}{2})\sin(\frac{\theta + \phi}{2})}$$ $$a[4\frac{b}{c} + bc(\frac{b}{c}-\frac{c}{b})^2] = 4\frac{\sin \theta + \sin \phi}{\cos (\frac{\theta - \phi}{2})\sin(\frac{\theta + \phi}{2})}$$ $$= 4\frac{2\cos (\frac{\theta - \phi}{2})\sin(\frac{\theta + \phi}{2})}{\cos (\frac{\theta - \phi}{2})\sin(\frac{\theta + \phi}{2})}$$ $$= 8$$

• I'm sorry but I couldn't see how you got the expression for $\frac{b}{c}$. – David Sebastian Dec 2 '14 at 13:19
• Ok. I'll post the steps for it. – lsp Dec 2 '14 at 13:20
• Try to get the result from where I left and let me know in case you need any help. – lsp Dec 2 '14 at 13:30
• @DavidSebastian Updated. You just need to know a few formulae like $\sin 2\theta = 2\sin \theta\cos \theta$ and $\cos A + \cos B = 2\cos (\frac{A+B}{2})\cos (\frac{A-B}{2})$ – lsp Dec 2 '14 at 15:47
• @DavidSebastian Provided with the entire answer. Hope you got it now. – lsp Dec 10 '14 at 9:12

$\color{red}{c^2=\sec^2\theta+\sec^2\phi+2\sec\theta\sec\phi\tag{1}}$

$\color{blue}{b^2=\tan^2\theta+\tan^2\phi+2\tan\theta\tan\phi\tag{2}}$

$(1)-(2)$ gives,

\begin{align}\left(c^2-b^2\right) & =\sec^2\theta+\sec^2\phi+2\sec\theta\sec\phi-\tan^2\theta-\tan^2\phi-2\tan\theta\tan\phi\\ &=2(1+\sec\theta\sec\phi-\tan\theta\tan\phi)\\&=2\left(1+\dfrac{1}{\cos\theta\cos\phi}-\dfrac{\sin\theta\sin\phi}{\cos\theta\cos\phi}\right)\\&=\dfrac{2}{\cos\theta\cos\phi}(1+\cos(\theta+\phi))\\&=\dfrac{4\cos^2\left(\dfrac{\theta+\phi}{2}\right)}{\cos\theta\cos\phi}\end{align}

$\color{darkgreen}{\therefore\left(c^2-b^2\right)^2=\dfrac{16\cos^4\left(\dfrac{\theta+\phi}{2}\right)}{\cos^2\theta\cos^2\phi}\tag{3}}$

$\color{brown}{4b^2=\dfrac{4\sin^2(\theta+\phi)}{\cos^2\theta\cos^2 \phi}=\dfrac{16\sin^2\left(\dfrac{\theta+\phi}{2}\right)\cos^2\left(\dfrac{\theta+\phi}{2}\right)}{\cos^2\theta\cos^2 \phi}\tag{4}}$

$$\boxed{4b^2+\left(b^2-c^2\right)^2=\dfrac{16\cos^2\left(\dfrac{\theta+\phi}{2}\right)}{\cos^2\theta\cos^2 \phi}}$$

$c=\dfrac{\cos\theta+\cos\phi}{\cos\theta\cos\phi}=\dfrac{2\cos\left(\dfrac{\theta+\phi}{2}\right)\cos\left(\dfrac{\theta-\phi}{2}\right)}{\cos\theta\cos\phi}$

$b=\dfrac{\sin(\theta+\phi)}{\cos\theta\cos\phi}=\dfrac{2\sin\left(\dfrac{\theta+\phi}{2}\right)\cos\left(\dfrac{\theta+\phi}{2}\right)}{\cos\theta\cos\phi}$

$a=2\sin\left(\dfrac{\theta+\phi}{2}\right)\cos\left(\dfrac{\theta-\phi}{2}\right)$

$$\boxed{\dfrac{8bc}{a}=\dfrac{32\sin\left(\dfrac{\theta+\phi}{2}\right)\cos^2\left(\dfrac{\theta+\phi}{2}\right)\cos\left(\dfrac{\theta-\phi}{2}\right)}{2\sin\left(\dfrac{\theta+\phi}{2}\right)\cos\left(\dfrac{\theta-\phi}{2}\right)\cos^2\theta\cos^2 \phi}=\dfrac{16\cos^2\left(\dfrac{\theta+\phi}{2}\right)}{\cos^2\theta\cos^2 \phi}}$$

It is much more convenient to simplify algebraic expressions using only two variables $t$ and $T$ together for tangent of each half angle than struggling with Trig functions:

$$t = \tan(\theta/2); T = \tan(\phi/2);$$

$$sth = 2 t/(1 + t^2) ; tth = 2 t/(1 - t^2);secth = (1 + t^2)/(1 - t^2);$$

$$sph = 2 T/(1 + T^2); tph = 2 T/(1 - T^2); secph = (1 + T^2)/(1 - T^2);$$

$$a = sth + sph; b = tth + tph; c = secth + secph;$$

Each of $8\,b\,c$ and $a[4b^2 + (b^2-c^2)^2]$ are brought to a common denominator simplifying them separately to :

$$\dfrac{32\, (t + T)\, ( t\, T-1)^2 ( t\, T+1)} {(t^2 -1)^2 (T^2 -1)^2}$$

• I must confess that I'm a bit irritated to see that your answer, based on the same core idea as mine, has gained more upvotes in less time. Not that I begrudge you the reputation, but I'd really like to understand what people like about your answer as opposed to mine. Is it the shortness due to the fact that you omit expanded forms for $a,b,c$ and have the final value in unexpanded factors? Or is it the fact that I had omitted the initial $t=\tan(\theta/2)$ in my answer until just now? In any case, it seems you contributed something people like, and I'm happy for you about that. – MvG Dec 11 '14 at 21:35
• @MvG: I cannot say what appealed to others.. I can only mention what I did; the $8 b c$ part I did by hand, where (t+T), ( t T + 1) etc. are preserved throughout as separate bracketable quantities till the end. Second part,left it to CAS. By alloying a heavy expression one loses sight of individual elements also elegance of algebraic handling. You mentioned factoring though at the end of your post, and it factors out to same result ( u = T ).The half angle suggests itself from the post as fairly straight elementary Trig. Anyways for your last comment, schoenen Dank! – Narasimham Dec 12 '14 at 21:33
• @MvG: In actual working (that seems to be long struggle) using trigonometric quantities we are still doing algebraic manipulations. In this case better use algebra from the beginning itself. – Narasimham Dec 12 '14 at 21:41

I'd use the tangent half-angle substitution:

\begin{align*} t&=\tan\frac\theta2 & \sin\theta&=\frac{2t}{1+t^2} & \tan\theta&=\frac{2t}{1-t^2} & \sec\theta&=\frac{1+t^2}{1-t^2} \\ u&=\tan\frac\phi2 & \sin\phi&=\frac{2u}{1+u^2} & \tan\phi&=\frac{2u}{1-u^2} & \sec\phi&=\frac{1+u^2}{1-u^2} \end{align*}

Then you have

\begin{align*} a &= \frac{2 t^{2} u + 2 t u^{2} + 2 t + 2 u}{t^{2} u^{2} + t^{2} + u^{2} + 1} \\ b &= \frac{-2 t^{2} u - 2 t u^{2} + 2 t + 2 u}{t^{2} u^{2} - t^{2} - u^{2} + 1} \\ c &= \frac{-2 t^{2} u^{2} + 2}{t^{2} u^{2} - t^{2} - u^{2} + 1} \end{align*}

\begin{multline*} 8bc=a[4b^2 + (b^2-c^2)^2] =\\ 32\frac{t^{4} u^{3} + t^{3} u^{4} - t^{3} u^{2} - t^{2} u^{3} - t^{2} u - t u^{2} + t + u}{t^{4} u^{4} - 2 t^{4} u^{2} - 2 t^{2} u^{4} + t^{4} + 4 t^{2} u^{2} + u^{4} - 2 t^{2} - 2 u^{2} + 1} \end{multline*}

The great benefit here is that after that first step, choosing to use that formulation, you almost don't have to think at all any more. From there on it's a straight-forward computation in rational functions. I left that to my computer algebra system, but if you are careful you can certainly do it by hand as well. The fact that my CAS expanded all those polynomials automatically makes the formulas above perhaps look more complicated than they really are, haven't tried.

• Since I received a downvote for this, I'd be happy for an explanation why you consider this answer not useful. I don't mind the loss in reputation, but I'm genuinely curious about the rationale behind that vote. I hope the voter is still around and will drop me a line. – MvG Dec 12 '14 at 16:53

\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*} $cos^2(\theta)\cos^2(\phi)(b^2 - c^2)^2 = 16\cos^4((\theta + \phi)/2$ $cos^2(\theta)\cos^2(\phi)(4b^2 + (b^2 - c^2)^2) = 16\cos^2((\theta + \phi)/2$ \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*}

The desired result follows.

First, observe that adding or subtracting a tangent with a secant of like variables simplifies much more readily than adding tangents of different variables or secants of different variables (just look at the nasty formulas you obtained for $b$ and $c$!). We find.

$$\tan{x}+\sec{x}=\frac{\sin{x}}{\cos{x}}+\frac{1}{\cos{x}}=\frac{\sin{x}+1}{\cos{x}},$$

and

$$\tan{x}-\sec{x}=\frac{\sin{x}}{\cos{x}}-\frac{1}{\cos{x}}=\frac{\sin{x}-1}{\cos{x}}.$$

We can take advantage of this and avoid most of the masochistic clutter produced by trying to use the angle addition formulas first.

The difference of squares $b^2-c^2$ then simplifies as follows:

\begin{align} b^2-c^2 &=\left(b+c\right)\cdot\left(b-c\right)\\ &=\left(\frac{\sin{\theta}+1}{\cos{\theta}}+\frac{\sin{\varphi}+1}{\cos{\varphi}}\right)\cdot\left(\frac{\sin{\theta}-1}{\cos{\theta}}+\frac{\sin{\varphi}-1}{\cos{\varphi}}\right)\\ &=\frac{\sin^2{\theta}-1}{\cos^2{\theta}}+\frac{\sin{\theta}+1}{\cos{\theta}}\cdot\frac{\sin{\varphi}-1}{\cos{\varphi}}+\frac{\sin{\varphi}+1}{\cos{\varphi}}\cdot\frac{\sin{\theta}-1}{\cos{\theta}}+\frac{\sin^2{\varphi}-1}{\cos^2{\varphi}}\\ &=-1+\frac{\left(\sin{\theta}+1\right)\left(\sin{\varphi}-1\right)+\left(\sin{\theta}-1\right)\left(\sin{\varphi}+1\right)}{\cos{\theta}\cos{\varphi}}-1\\ &=\frac{2\sin{\theta}\sin{\varphi}-2}{\cos{\theta}\cos{\varphi}}-2\\ &=2\left[\tan{\theta}\tan{\varphi}-\sec{\theta}\sec{\varphi}-1\right]. \end{align}

As you can see, this does a much better job of streamlining the algebraic manipulations of the problem.

• excellent answer ! – lsp Dec 3 '14 at 8:05
• Thank you very much for your answer but I have a problem. You've got an expression for $b^2-c^2$. I'm sure there's good reason but I don't see any. So how could I get to showing $8bc=a[4b^2+(b^2-c^2)^2]$ from there ? – David Sebastian Dec 4 '14 at 2:46

$$(c-\sec\theta)^2-(b-\tan\theta)^2=1$$

$$\implies b^2-c^2=2b\tan\theta-2c\sec\theta$$

Put $\tan2A=\dfrac{2\tan A}{1-\tan^2A},\sec2A=1/\cos2A=\dfrac{1+\tan^2A}{1-\tan^2A}$ to form Quadratic Equation in $\tan\dfrac\theta2$

We shall reach at the same equation if we start with $$(c-\sec\phi)^2-(b-\tan\phi)^2=1$$

So the roots of the Quadratic Equation will be $\tan\dfrac\theta2,\tan\dfrac\phi2$

Utilize Vieta's formula

in $\tan\dfrac{\theta+\phi}2=\dfrac{\tan\dfrac\theta2+\tan\dfrac\phi2}{1-\tan\dfrac\theta2\tan\dfrac\phi2}\cdots$

So, we can find $\sin(\theta+\phi)$ using $\sin2B=\dfrac{2\tan B}{1+\tan^2B}$

From $b=\tan\theta+\tan\phi=\dfrac{\sin(\theta+\phi)}{\cos\phi\cos\theta},$ we can find $\cos\phi\cos\theta$

From $c=\sec\theta+\sec\phi=\dfrac{\cos\theta+\cos\phi}{\cos\phi\cos\theta},$ we can find $\cos\phi+\cos\theta$

Now, $(\cos\phi+\cos\theta)^2+(\sin\phi+\sin\theta)^2=2+2\cos(\phi-\theta),$ which will give us $\cos(\phi-\theta)$

Finally use Werner's formula $\cos(\phi-\theta)+\cos(\phi+\theta)=2\cos\phi\cos\theta$