Prove that \(\cos{20^{\circ}}\cos{40^{\circ}}\cos{80^{\circ}} = \frac{1}{8}\)

Every solutions are all acceptable.

I feel bad for being suck at trigonometry and stuffs. T__T

Prove that \(\cos{20^{\circ}}\cos{40^{\circ}}\cos{80^{\circ}} = \frac{1}{8}\)

Every solutions are all acceptable.

I feel bad for being suck at trigonometry and stuffs. T__T

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TopNewestWe start with the trigonometric identity

\(Sin(2x)=2Sin(x)Cos(x)\)

which means

\(Cos(x)=\dfrac { Sin(2x) }{ 2Sin(x) } \)

So that

\(Cos(x)Cos(2x)Cos(4x)...Cos({ 2 }^{ n-1 }x)=\)

\(\dfrac { Sin(2x) }{ 2Sin(x) } \dfrac { Sin(4x) }{ 2Sin(2x) } \dfrac { Sin(8x) }{ 2Sin(4) } ...\dfrac { Sin({ 2 }^{ n }x) }{ 2Sin({ 2 }^{ n-1 }x) } =\)

\(\dfrac { Sin({ 2 }^{ n }x) }{ { 2 }^{ n }Sin(x) } \)

From this we can solve the posted problem, since \(Sin(160)=Sin(180-160)=Sin(20)\)

and so we're left with \( \dfrac { 1 }{ 8 } \) – Michael Mendrin · 2 years, 1 month ago

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– Aditya Raut · 2 years, 1 month ago

True, this is what Sean Ty used. But this is for the special case ! Luckily in this case we are getting "double angles"Log in to reply

There is a way that doesn't use multiplication too. There is a good direct formula, that

\[\displaystyle \sin\theta \cdot \sin (60^\circ- \theta) \cdot \sin(60^\circ + \theta) =\dfrac{ \sin (3\theta)}{4}\]

\[\displaystyle\cos\theta \cdot \cos (60^\circ- \theta) \cdot \cos(60^\circ + \theta) =\dfrac{ \cos (3\theta)}{4}\]

\[\displaystyle\tan\theta \cdot \tan(60^\circ- \theta) \cdot \tan(60^\circ + \theta) =\tan(3\theta)\]

They're very easy to prove (just expand) and very useful at any place.

For example, here it's okay that luckily your angles gave you double angles, as in @Sean Ty 's solution.

But this formula is useful even if you were asked to find \(\displaystyle\sin 5^\circ \cdot \sin 55^\circ sin 65^\circ\).

That would simply be \(\displaystyle\dfrac{\sin 15^\circ}{4} =\dfrac{\sqrt{3}-1}{2\sqrt{2}}\cdot \dfrac{1}{4} = \dfrac{\sqrt{3}-1}{8\sqrt{2}}\)

\(\Biggl(\sin15^\circ = \sin (\frac{30^\circ}{2}) = \sqrt{\dfrac{1-\cos 30^\circ}{2}} = \sqrt{\dfrac{2-\sqrt{3}}{4}} = \dfrac{\sqrt{4-2\sqrt{3}}}{2\sqrt{2}}= \dfrac{\sqrt{3}-1}{2\sqrt{2}} \Biggr)\)

@Samuraiwarm Tsunayoshi , was this helpful ? – Aditya Raut · 2 years, 2 months ago

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– Sean Ty · 2 years, 2 months ago

Cool! That'll save me time from deriving formulas (I do that if I forget them) in a contest. We all learn from each other!Log in to reply

– Aditya Raut · 2 years, 1 month ago

Truly said, the formulas are really good looking so I actually remember them intuitively :PLog in to reply

Multiply and divide by 2sin20 and write the result in the numerator as sin40cos40cos80. Now multiply and divide by 2, followed by the same step as before. I suggest you also read a liitle on Morrie's Law from Wikipedia. – Rahul Sethi · 11 months, 2 weeks ago

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Alright, so I'm using mobile and there's this bug where if I finish typing and press preview or post, it will open something that's 'behind' it. If you can't understand what I mean, you can ask me to picture what's happening (screenshot) and post it as a note. If possible. Well, going on to the problem. (I'm gonna give hints because when I attempted to the solution thrice, everything got wiped out, thrice.)

Hints:

Step 1: Set the expression equal to some variable let's say... \(a\).

Step 2: Multiply both sides by \(\sin 20^\circ\).

Step 3: Double-Angle Formula.

And I think you can go from there :) – Sean Ty · 2 years, 2 months ago

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– Samuraiwarm Tsunayoshi · 2 years, 2 months ago

Thank you~ ^__^ That's too easy I didn't even think about it XDLog in to reply