How do I prove this seemingly simple trigonometric identity

Question

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

Answer 1

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: $$\frac{b}{c}-\frac{c}{b}=-\frac{\cos \theta. \cos \phi}{\cos(\frac{\theta - \phi}{2}).sin(\frac{\theta + \phi}{2})}$$ $$bc(\frac{b}{c}-\frac{c}{b})^2 = \frac{\sin (\theta + \phi)(\cos \theta + \cos \phi)}{\cos^2 \theta. \cos^2 \phi}\frac{\cos^2 \theta. \cos^2 \phi}{\cos^2(\frac{\theta - \phi}{2}).\sin^2(\frac{\theta + \phi}{2})}$$ $$ = \frac{2\sin(\frac{\theta + \phi}{2}).\cos(\frac{\theta + \phi}{2})2\cos(\frac{\theta + \phi}{2}).\cos(\frac{\theta - \phi}{2})}{\cos^2(\frac{\theta - \phi}{2}).\sin^2(\frac{\theta + \phi}{2})}$$ $$ = \frac{4\cos^2 (\frac{\theta + \phi}{2})}{\cos(\frac{\theta - \phi}{2})\sin (\frac{\theta + \phi}{2})}$$ $$4\frac{b}{c} + bc(\frac{b}{c}-\frac{c}{b})^2 = \frac{4}{\cos (\frac{\theta - \phi}{2})}(\sin(\frac{\theta + \phi}{2})+\frac{\cos^2 (\frac{\theta + \phi}{2})}{\sin(\frac{\theta + \phi}{2})})$$ $$= \frac{4}{\cos (\frac{\theta - \phi}{2})\sin(\frac{\theta + \phi}{2})}$$ $$a[4\frac{b}{c} + bc(\frac{b}{c}-\frac{c}{b})^2] = 4\frac{\sin \theta + \sin \phi}{\cos (\frac{\theta - \phi}{2})\sin(\frac{\theta + \phi}{2})}$$ $$ = 4\frac{2\cos (\frac{\theta - \phi}{2})\sin(\frac{\theta + \phi}{2})}{\cos (\frac{\theta - \phi}{2})\sin(\frac{\theta + \phi}{2})}$$ $$= 8$$

Answer 2

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

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

$(1)-(2)$ gives,

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

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

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

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

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

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

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

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

Answer 3

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 :

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

Answer 4

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.

Answer 5

\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.

Answer 6

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.

Answer 7

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

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

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 $\cos(\phi-\theta)+\cos(\phi+\theta)=2\cos\phi\cos\theta$