# Lovely, simple, trigonometric puzzle

Sometimes a puzzle can look complicated,  but be rather simple (see this geometry puzzle). I love puzzles like this and I particularly like to test them out on classes to try and build their problem solving ability.

Just now, I saw the following trig puzzle from brilliant.org and I love it! It’s amazing! Have you done it yet?

How long did it take you to spot it?

My initial thought was, it’s got three terms,  it’s bound to be a disguised quadratic that will factorise. A few seconds later I realised that it wasn’t. I saw the – sin^4 and suspected a difference of two squares but then a few seconds later it became clear.

If you haven’t spotted it yet, have a look at the expression rearranged:

Sin^6 + sin^4 cos^2 – sin^4

See it now? What if I rewrite it as:

Sin^4 sin^2 + sin^4 cos^2 – sin^4

I’m sure you have seen it now, but to be complete,  take the common factor of the first two terms:

Sin^4 (sin^2 + cos^2) – sin^4

Obviously sin^2 + cos^2 = 1, so we’re left with:

Sin^4 – sin^4 = 0

A lovely, satisfying, simple answer to a little brain teaser. Hope you liked it as much as I did.

Cross-posted to Cavmaths here.

## 5 thoughts on “Lovely, simple, trigonometric puzzle”

1. howardat58

Try this:
sin^6 – sin^4 +sin^4.cos^2
sin^6 – sin^4.(1 – cos^2)
sin^6 – sin^4.(sin^2)
sin^6 – sin^6 = 0
Three steps

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• srcav

Nice. A more concise solution!

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• howardat58

The notation can really get in the way with these things. I wanted to write it as s^6 – s^4 +s^4.c^4 with the exponents in the proper places. Makes life much easier.

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• howardat58

Especially when written without the typos !

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2. gasstationwithoutpumps

My first thought was to factor out the monomial:
sin^4 (sin^2 + cos^2 -1)
then the 0 jumped out at me.

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