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If $\displaystyle A=3\sin^{-1}\left(\frac{6}{11}\right)$ and $\displaystyle B = 3\cos^{-1}\left(\frac{4}{9}\right),$ then proving $\cos (A+B)>0$

Attempt: $$ 3\sin^{-1}\left(\frac{1}{2}\right)<3\sin^{-1}\left(\frac{6}{11}\right)<3\sin^{-1}\left(\frac{\sqrt{3}}{2}\right)\Rightarrow \frac{\pi}{2}<A<\pi$$

same way $$\displaystyle 3\cos^{-1}\left(0\right)<3\cos^{-1}\left(\frac{4}{9}\right)<3\cos^{-1}\left(\frac{1}{2}\right)\Rightarrow \frac{\pi}{2}<B<\frac{3\pi}{2}$$

so $$\pi<A+B<\frac{5\pi}{2}$$

could some help me how to prove $\cos (A+B)>0,$ thanks

DXT
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  • Try using identity for $\cos (A+B)$ –  Mar 29 '17 at 02:55
  • It is not necessary to actually find the value of $\cos(A+B)$ since $A+B$ can be shown to lie in either quadrant IV or quadrant I by the approach which you have taken. You just need tighter bounds on angles $A$ and $B$ than those you give. See my answer below. – John Wayland Bales Mar 30 '17 at 16:38

3 Answers3

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Hint:

$\cos(3(A/3+B/3)) = 4 \cos^3 (A/3+B/3) - 3 \cos (A/3+B/3) $

Now expand this and you will gets terms only in $\cos (A/3), \cos (B/3),\sin (A/3), \sin (B/3)$ which we can get from given hypothesis.

User8976
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2

$A = 3\sin^{-1}(\frac{6}{11}), B = 3\cos^{-1}(\frac{4}{9})$

Let $C = \cos(\frac{A+B}{3})$

Note that $\cos (3x) = 4\cos^3x - 3\cos x$

Let $x=\frac{A+B}{3} \implies \cos x = \cos \frac{A}{3} \cos\frac{B}{3} -\sin\frac{A}{3}\sin\frac{B}{3}$

$\cos x = \cos (\sin^{-1}(\frac{6}{11}))\cos(\cos^{-1}(\frac{4}{9})) - \sin(\sin^{-1}(\frac{6}{11}))\sin(\cos^{-1}(\frac{4}{9}))$

Consider $y = \sin^{-1}(\frac{6}{11}) \implies \cos y = \frac{\sqrt{85}}{11}$

Consider $y = \cos^{-1}(\frac{4}{9}) \implies \sin y = \frac{\sqrt{65}}{9}$

Then $\cos x = \frac{\sqrt{85}}{11}\cdot \frac{4}{9} -\frac{6}{11}\cdot \frac{\sqrt{65}}{9} = \frac{4\sqrt{85}-6\sqrt{65}}{99}$

So then $\cos (3x) = 4\cdot (\frac{4\sqrt{85}-6\sqrt{65}}{99})^3-3\cdot (\frac{4\sqrt{85}-6\sqrt{65}}{99}) = 4\cdot (\frac{4\sqrt{85}-6\sqrt{65}}{99})^3+3\cdot (\frac{-4\sqrt{85}+6\sqrt{65}}{99})$

$\cos (3x) = \frac{134080\sqrt{85}-154080\sqrt{65})}{99^3} +\frac{-12\sqrt{85}+18\sqrt{65}}{99}$

$\cos (3x) = \cos(A+B) = \frac{(134080-117612)\sqrt{85} +(176418-154080)\sqrt{65}}{99^3} > 0$

and we're done

mrnovice
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Let $\displaystyle \alpha=\sin^{-1}\left(\frac{6}{11}\right)=\frac{A}{3}$ and let $\displaystyle\beta=\cos^{-1}\left(\frac{4}{9}\right)=\frac{B}{3}$

Then since $\frac{1}{2}<\frac{6}{11}<\frac{\sqrt{2}}{2}$ we have $\frac{\pi}{6}<\alpha<\frac{\pi}{4}$.

And since $0<\frac{4}{9}<\frac{1}{2}$ we have $\frac{\pi}{3}<\beta<\frac{\pi}{2}$.

Addition of the two inequalities gives $\frac{\pi}{2}<\alpha+\beta<\frac{3\pi}{4}$.

Multiplication by $3$ gives $\frac{3\pi}{2}<3\alpha+3\beta<\frac{9\pi}{4}$.

Thus $\frac{3\pi}{2}<A+B<\frac{9\pi}{4}$ meaning that $A+B$ lies in either quadrant IV or quadrant I.

Thus $\cos(A+B)>0$.

What has been shown here is that, referring to the following diagram, for any standard angle $A$ such that the terminal side of $\dfrac{A}{3}$ lies in the blue region and for any standard angle $B$ such that the terminal side of $\dfrac{B}{3}$ lies in the green region it will be the case that $\cos(A+B)>0$. The "red" lines are the terminal sides of the particular $\dfrac{A}{3}$ and $\dfrac{B}{3}$ given in the question.

Image of angles A/3 and B/3

John Wayland Bales
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