#precalculus help
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Sorry I edited it, it’s tan (t) = 1 and your finding tan (t + pie )
So one period
$$ \tan (x+\pi) = \frac{\tan x + \tan \pi}{1-\tan x\tan \pi} = \tan x $$
aL
@marsh harness
$$ \tan (-t + 2\pi) = \tan (-t) = \frac{\sin (-t)}{\cos (-t)} = -\frac{\sin t}{\cos t} = -\frac{\sqrt{1-\cos ^2t}}{\cos t} $$
aL
It’s by using the periodic properties and even/odd identities
what do you conclude about period of tan from these equalities?
That’s why I came here, lol
From seeing the questions I believe tan being 1, it equals to pie/4 + a period
But I’m uncertain if there is more to do or not
you dont have to know what t is
it is enough to know that tan (t) = 1 and tan is pi-periodic
conclusion?
But is the answer simply 1 then?
yes!
see this, as well
Alright I think I understand a bit
If the first answer is 1 what would be the others?
Yep
the question in the picture is different
$$ \cos (a+b) = \cos a\cos b - \sin a\sin b $$
aL
What is that equation?
relevant to 2.
But the question is simply asking of periodic properties and even/odd identities
what do you think the period of cosine is?
and if you're not certain, how do you prove it?
Seeing cos is an even property it would be 0.25
And seeing as sin is a negative property it would be 0.9 or -0.9
negative?
I’m uncertain, I have to revisit it again
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cos is 2pi-periodic and even, therefore cos (-t) = cos (-t +2pi) = 0.25
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sin is 2pi-periodic and odd, therefore sin (-t+2pi) = sin (-t) = -sin t = -0.9
you don't say "cos is an even property"
that is meaningless wordsoup
sure, if this one is solved, close it and open a new question 🍺
Alright