Trigonometry proof (help mehhh)

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

#Trigonometry

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
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Comments

We 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 - 6 years, 9 months ago

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

Aditya Raut - 6 years, 9 months ago

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 - 6 years, 9 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!

Sean Ty - 6 years, 9 months ago

Truly said, the formulas are really good looking so I actually remember them intuitively :P

Aditya Raut - 6 years, 9 months ago

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 - 6 years, 9 months ago

Thank you~ ^__^ That's too easy I didn't even think about it XD

Samuraiwarm Tsunayoshi - 6 years, 9 months ago

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 - 5 years, 7 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$