Large Powers Just Vanished?

Geometry Level 1

Simplify
sin ⁡ 6 θ + cos ⁡ 6 θ + 3 sin ⁡ 2 θ cos ⁡ 2 θ {\sin}^6\theta + {\cos}^6\theta + 3{\sin}^2\theta{\cos}^2\theta

1 2 \frac12 1 1 1 1 2 1\frac12 2 2

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Ashish Menon by Ashish Menon , 20, India

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3 solutions

Ashish Menon
Jul 5, 2016

Relevant wiki: Proving Trigonometric Identities - Basic

sin ⁡ 6 θ + cos ⁡ 6 θ + 3 sin ⁡ 2 θ cos ⁡ 2 θ × 1 = ( sin ⁡ 2 θ ) 3 + ( cos ⁡ 2 θ ) 3 + 3 sin ⁡ 2 θ cos ⁡ 2 θ ( sin ⁡ 2 θ + cos ⁡ 2 θ ) = ( sin ⁡ 2 θ + cos ⁡ 2 θ ) 3 = 1 3 = 1 \begin{aligned} {\sin}^6\theta + {\cos}^6\theta + 3{\sin}^2\theta{\cos}^2\theta × 1 & = {\left({\sin}^2\theta\right)}^3 + {\left({\cos}^2\theta\right)}^3 + 3{\sin}^2\theta{\cos}^2\theta\left({\sin}^2\theta + {\cos}^2\theta\right)\\ \\ & = {\left({\sin}^2\theta + {\cos}^2\theta\right)}^3\\ \\ & = 1^3\\ \\ & = \color{#3D99F6}{\boxed{1}} \end{aligned}

32 Helpful 0 Interesting 0 Brilliant 0 Confused

A classic way to prove the equation equals 1.

but if you got to do it real quick(under 5 secs) then just put θ as 0° or 90°(even though if its not a foolproof method). and we are done.

Satyabrata Dash - 4 years, 11 months ago

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Thanks! Yep JEE style! But this question did not take even 5 seconds to come up with :P

Ashish Menon - 4 years, 11 months ago

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Ya the jee style. But still the % of solver has further decreased to 38%

Satyabrata Dash - 4 years, 11 months ago

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@Satyabrata Dash – Hehe lol! Maybe because it appears tough :P

Ashish Menon - 4 years, 11 months ago

Haha nice way to confuse people into thinking that it is a very HARD question......I got the trick when I saw it (At that instant :>) Nice question and solution!!!(+1)

abc xyz - 4 years, 11 months ago

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Thanks! :) :)

Ashish Menon - 4 years, 11 months ago

Right, so only 42% got it right :P

Ashish Menon - 4 years, 11 months ago
Hung Woei Neoh
Jul 6, 2016

sin ⁡ 6 θ + cos ⁡ 6 θ + 3 sin ⁡ 2 θ cos ⁡ 2 θ = sin ⁡ 6 θ + ( cos ⁡ 2 θ ) 3 + 3 sin ⁡ 2 θ ( 1 − sin ⁡ 2 θ ) = sin ⁡ 6 θ + ( 1 − sin ⁡ 2 θ ) 3 + 3 sin ⁡ 2 θ − 3 sin ⁡ 4 θ = sin ⁡ 6 θ + 1 − 3 sin ⁡ 2 θ + 3 sin ⁡ 4 θ − sin ⁡ 6 θ + 3 sin ⁡ 2 θ − 3 sin ⁡ 4 θ = 1 \sin^6 \theta+ \cos^6 \theta+3\sin^2\theta\cos^2\theta\\ =\sin^6 \theta + (\cos^2 \theta)^3 + 3\sin^2\theta(\color{magenta}{1-\sin^2\theta})\\ =\sin^6 \theta+ (\color{magenta}{1-\sin^2 \theta})^3 + 3\sin^2\theta - 3\sin^4\theta\\ =\color{#3D99F6}{\sin^6 \theta} + 1 \color{#EC7300}{-3\sin^2 \theta}\color{#D61F06}{+3\sin^4\theta}\color{#3D99F6}{-\sin^6\theta}\color{#EC7300}{+3\sin^2\theta}\color{#D61F06}{-3\sin^4\theta}\\ =\boxed{1}

Theorem used: Pythagorean Trigonometric Identity - sin ⁡ 2 θ + cos ⁡ 2 θ = 1 \color{magenta}{\sin^2\theta+\cos^2\theta=1}

9 Helpful 0 Interesting 0 Brilliant 0 Confused

Colors again!!

Ashish Menon - 4 years, 11 months ago

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Once you get used to it, it's pretty easy

Besides, a little color makes it nicer to read

Hung Woei Neoh - 4 years, 11 months ago

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Lol, yeah :]

Ashish Menon - 4 years, 11 months ago
Oli Hohman
Jul 6, 2016

Expand ( sin ⁡ 2 ( θ ) + cos ⁡ 2 ( θ ) ) 3 {\left(\sin^2(\theta) + \cos^2(\theta)\right)}^3 You know that this expression is equivalent to 1, but let's see what happens when we expand, anyway!

This will leave you with

sin ⁡ 6 ( θ ) + 3 sin ⁡ 4 ( θ ) cos ⁡ 2 ( θ ) + 3 cos ⁡ 4 ( θ ) sin ⁡ 2 ( θ ) + cos ⁡ 6 ( θ ) \sin^6(\theta) + 3\sin^4(\theta)\cos^2(\theta) + 3\cos^4(\theta)\sin^2(\theta)+\cos^6(\theta)

The original expression contains sin ⁡ 6 ( θ ) + cos ⁡ 6 ( θ ) \sin^6(\theta) + \cos^6(\theta) , so let's factor out the greatest common factor (GCF) out of the two middle terms and see what we're left with:

3 sin ⁡ 2 ( θ ) cos ⁡ 2 ( θ ) ( sin ⁡ 2 ( θ ) + cos ⁡ 2 ( θ ) ) = 3 sin ⁡ 2 ( θ ) cos ⁡ 2 ( θ ) × 1 3\sin^2(\theta)\cos^2(\theta)(\sin^2(\theta) + \cos^2(\theta)) = 3\sin^2(\theta)\cos^2(\theta) \times 1

So the middle term matches the original expression, too, so therefore,

( sin ⁡ 2 ( θ ) + cos ⁡ 2 ( θ ) ) 3 = cos ⁡ 6 ( θ ) + 3 sin ⁡ 2 ( θ ) × cos ⁡ 2 ( θ ) + sin ⁡ 6 ( θ ) = 1 {\left(\sin^2(\theta)+\cos^2(\theta)\right)}^3 = \cos^6(\theta) + 3\sin^2(\theta) \times \cos^2(\theta) + \sin^6(\theta) = \boxed{1} .

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Usage of latex is preferable.

Satyabrata Dash - 4 years, 11 months ago

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Additionally, any criticism/opinion of the solution itself, rather than the formatting?

Oli Hohman - 4 years, 11 months ago

I apologize I am not familiar with Latex yet. Do you know of a good link to help learn it?

Oli Hohman - 4 years, 11 months ago

No the solution is fine. and you can see the formatting guide and the solution writing guide(thats actually a note by mention[11:Calvin Lin]) just below the solution writing space its helpful. Latex is quite easy you will learn it eventually.

@Calvin Lin you can edit this.

Cheers!!

Satyabrata Dash - 4 years, 11 months ago

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@Oli Hohman, read the instructions to upload in your solution at the bottom of this comment.

Expand ( sin ⁡ 2 ( θ ) + cos ⁡ 2 ( θ ) ) 3 {\left(\sin^2(\theta) + \cos^2(\theta)\right)}^3 You know that this expression is equivalent to 1, but let's see what happens when we expand, anyway!

This will leave you with

sin ⁡ 6 ( θ ) + 3 sin ⁡ 4 ( θ ) cos ⁡ 2 ( θ ) + 3 cos ⁡ 4 ( θ ) sin ⁡ 2 ( θ ) + cos ⁡ 6 ( θ ) \sin^6(\theta) + 3\sin^4(\theta)\cos^2(\theta) + 3\cos^4(\theta)\sin^2(\theta)+\cos^6(\theta)

The original expression contains sin ⁡ 6 ( θ ) + cos ⁡ 6 ( θ ) \sin^6(\theta) + \cos^6(\theta) , so let's factor out the greatest common factor (GCF) out of the two middle terms and see what we're left with:

3 sin ⁡ 2 ( θ ) cos ⁡ 2 ( θ ) ( sin ⁡ 2 ( θ ) + cos ⁡ 2 ( θ ) ) = 3 sin ⁡ 2 ( θ ) cos ⁡ 2 ( θ ) × 1 3\sin^2(\theta)\cos^2(\theta)(\sin^2(\theta) + \cos^2(\theta)) = 3\sin^2(\theta)\cos^2(\theta) \times 1

So the middle term matches the original expression, too, so therefore,

( sin ⁡ 2 ( θ ) + cos ⁡ 2 ( θ ) ) 3 = cos ⁡ 6 ( θ ) + 3 sin ⁡ 2 ( θ ) × cos ⁡ 2 ( θ ) + sin ⁡ 6 ( θ ) = 1 {\left(\sin^2(\theta)+\cos^2(\theta)\right)}^3 = \cos^6(\theta) + 3\sin^2(\theta) \times \cos^2(\theta) + \sin^6(\theta) = \boxed{1} .

Instructions:-

  • Click your icon at the top right corner your screen and click "Toggle Latex" button in the dropdown menu. - Scroll to this thread and you can see the Latex.
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Cheers!

Ashish Menon - 4 years, 11 months ago

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Latex is cool. Well are you a moderator or not??

Satyabrata Dash - 4 years, 11 months ago

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@Satyabrata Dash – Nope, I am not a moderator.

Ashish Menon - 4 years, 11 months ago

@Ashish Siva Wow, thank you so much! That was a great help! I edited the solution!

Oli Hohman - 4 years, 11 months ago

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My pleasure :)

Ashish Menon - 4 years, 11 months ago

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