Proving that $\cos^{-1}\frac{4}{5}+\cos^{-1}\frac{12}{13}=\cos^{-1}\frac{33}{65}$

Question

Prove

$$ \cos^{-1}\frac{4}{5}+\cos^{-1}\frac{12}{13}=\cos^{-1}\frac{33}{65} $$

My Attempt:

Let $\alpha=\cos^{-1}\frac{4}{5}\implies0<\alpha\leq\pi$ and $\beta=\cos^{-1}\frac{12}{13}\implies0<\beta<\pi$,

thus $0\leq\alpha+\beta\leq2\pi$ $$ \cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta=\frac{4}{5}.\frac{12}{13}-\frac{3}{5}.\frac{5}{13}=\frac{33}{65}=\cos\Big[\cos^{-1}\frac{33}{65}\Big]\\ \implies \cos^{-1}\frac{33}{65}=2n\pi\pm(\alpha+\beta) $$ case 1: If $0\leq\alpha+\beta\leq\pi$, $$ \cos^{-1}\frac{33}{65}=\alpha+\beta=\cos^{-1}\frac{4}{5}+\cos^{-1}\frac{12}{13} $$ case 2: If $\pi<\alpha+\beta\leq 2\pi$, $$ \cos^{-1}\frac{33}{65}=2\pi-(\alpha+\beta) $$

Is there any thing wrong with my approach and how do I eliminate case 2 in similar problems ?

Aren't both $\alpha$ and $\beta$ between 0 and $\pi/2$ since they're arccos of positive numbers? — minimalrho, Jan 30 '18 at 17:18
@minimalrho could u pls confirm my case 2 is valid right ?.i mean whn solving such problems i must consider and eliminate that case right ? — Sooraj S, Jan 30 '18 at 17:48

score 5 · Accepted Answer · answered Jan 30 '18 at 17:17

5

Because $$\cos\left(\cos^{-1}\frac{4}{5}+\cos^{-1}\frac{12}{13}\right)=\frac{4}{5}\cdot\frac{12}{13}-\frac{3}{5}\cdot\frac{5}{13}=\frac{33}{65}$$ and $$0^{\circ}<\cos^{-1}\frac{4}{5}+\cos^{-1}\frac{12}{13}<90^{\circ}.$$

answered Jan 30 '18 at 17:17

Michael Rozenberg

194,933

2

how do you know $0<\cos^{-1}\frac{4}{5}+\cos^{-1}\frac{12}{13}<\frac{\pi}{2}$ ? – Sooraj S Jan 30 '18 at 17:32
@ss1729 since both arguments of $\arccos$ are greater than $\arccos (1/ \sqrt 2) $, so both terms are less than $\pi/4$ – King Tut Jan 30 '18 at 17:36
Or $\arccos\frac{4}{5}+\arccos\frac{12}{13}<\arccos\frac{1}{2}+\arccos\frac{\sqrt3}{2}=90^{\circ}.$ – Michael Rozenberg Jan 30 '18 at 17:39
thanx. is it the right approach whn dealing with such problems?. i mean i need to consider and eliminate all cases right ? – Sooraj S Jan 30 '18 at 17:56
Yes, in this problem it's so. But there are problems where we need something more. – Michael Rozenberg Jan 30 '18 at 18:15
you mean in other problems i may not be able to eliminate other cases right as in this problem? – Sooraj S Jan 30 '18 at 18:19
1

Yes. For example, we can get that the sum, which depends on a parameter, is moved between $0^{\circ}$ and $180^{\circ}.$ If so we need cases. – Michael Rozenberg Jan 30 '18 at 18:24

Guy Fsone · Answer 2 · 2018-01-30T17:48:37.270

Since $$\frac{\sqrt2}{2}<\frac{4}{5}<1\implies \cos^{-1}\frac{\sqrt2}{2} >\cos^{-1}\frac{4}{5}> \cos^{-1} 1 $$

that is $$\frac{\pi}{4}>\cos^{-1}\frac{4}{5}> 0 $$ similarly $$\frac{\pi}{4}>\cos^{-1}\frac{12}{13}> 0 $$

Hence, $$0<\cos^{-1}\frac{4}{5}+\cos^{-1}\frac{12}{13}<\frac{\pi}{2}$$

Using $$\cos(\cos^{-1}x) = x~~~and~~~~\sin(\cos^{-1}x) =\sqrt{1-\cos^2(\cos^{-1}x)} =\color{red}{\sqrt{1-x^2}}$$

you easily arrive at $$\cos\left(\cos^{-1}\frac{4}{5}+\cos^{-1}\frac{12}{13}\right)\\=\cos\left(\cos^{-1}\frac{12}{13}\right)\cos\left(\cos^{-1}\frac{4}{5}\right)-\sin\left(\cos^{-1}\frac{12}{13}\right)\sin\left(\cos^{-1}\frac{4}{5}\right)\\=\frac{4}{5}\cdot\frac{12}{13}-\frac{3}{5}\cdot\frac{5}{13}=\frac{33}{65}$$ Done

pls check my post i have already written what u have mentioned. thats not my doubt. — Sooraj S, Jan 30 '18 at 17:36

lab bhattacharjee · Answer 3 · 2018-02-01T06:04:16.347

As $\cos(A+B)=?$

using Principal values,

$$0\le\arccos x,\arccos y,\arccos(xy-\sqrt{(1-x^2)(1-y^2)})\le\pi$$ $$\implies0\le\arccos x+\arccos y\le2\pi$$

$$\implies\arccos x+\arccos y$$

$$=\begin{cases}\arccos(xy-\sqrt{(1-x^2)(1-y^2)}) &\mbox{if } 0\le\arccos x+\arccos y\le\pi \\ 2\pi-\arccos(xy-\sqrt{(1-x^2)(1-y^2)}) & \mbox{ otherwise } \end{cases}$$

Now $0\le\arccos x+\arccos y\le\pi$

As $\arccos(u)$ lies in $\in[0,\pi],$

$\arccos x+\arccos y\ge0\forall x,y\in[0,1]$

So, we only need $$\arccos x+\arccos y\le\pi\iff\arccos x\le\pi-\arccos y=\arccos(-y)$$

As $\arccos$ is decreasing function $\in[0,1],$ we need $$x\le-y$$

See also: Proof for the formula of sum of arcsine functions $ \arcsin x + \arcsin y $

score 0 · Answer 4 · answered Jan 31 '18 at 00:09

When we multiply complex numbers, their angles are added. $(4+3i)$ and $(12+5i)$ have angles $\cos^{-1}(\frac{4}{5})$ and $\cos^{-1}(\frac{12}{13})$ respectively. $$(4+3i)(12+5i) = 33+56i$$ The argument of $33+56i$ is $\cos^{-1}(\frac{33}{65})$.

Proving that $\cos^{-1}\frac{4}{5}+\cos^{-1}\frac{12}{13}=\cos^{-1}\frac{33}{65}$

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