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## #1 2005-10-30 07:10:43

Lloyd W
Guest

### Trigonometry help

Hi there, I am studying Trig at the moment and am a bit confused by this question i have been asked:

Using the definitions of Sine and Cosine in a righthanded triangle calculate the value of sin² +cos² . Use pythagoras theorem to help you.

ok, so adj² +opp² = hyp²
and sin+tanxcos and cos=sin/tan
i guess im supposed to use that. but what exactly is this VALUE im supposed to be finding.

- Lloyd

## #2 2005-10-30 07:42:01

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

### Re: Trigonometry help

You're on the right lines. cos = adj/hyp and sin = opp/hyp, so cos² = adj²/hyp² and sin² = opp²/hyp². So cos² + sin ² = (adj² + opp²)/hyp².

But Pythagoras's Theorem says that a² + b² = c², where c is the hypotenuse, so that means that adj² + opp² = hyp². Therefore, cos² + sin² = hyp²/hyp² = 1.

That's an identity that you'll be using a lot while you're studying trig, so remember it well.

Why did the vector cross the road?
It wanted to be normal.

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## #3 2005-10-30 07:55:57

Lloyd W
Guest

### Re: Trigonometry help

wicked, i should have figured that out. cheers