Sine, Cosine, and Tangent of 45 degrees?

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Sine, Cosine, and Tangent of 45 degrees?

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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|dw:1439360650195:dw|
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we need to set the lengths of the sides of the triangle for simplicity we can set the adjacent side to be 1

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Other answers:

what will the other sides of the triangle be if the adjacent side is 1?
Wait what do you mean?
This is actually a timed test and I'm super tired
|dw:1439361178381:dw|
(notice it is an isosceles triangle )
I forgot how to do this and I don't have a calculator with me. Is there another way?
this way doesn't need a calculator
[isosceles triangle have a pair of sides of equal length]
[the sides of equal length, will be opposite to the angles of equal measure]
Ohhhhhhh so 1
yes
|dw:1439361833342:dw|
now \[\tan\theta = \frac{\text{opp}}{\text{adj}}\]
so tan 45° = / =
What do you get?
so confused but 1/ over something, I'm not sure what the hypotenuse is
we don't need the hypotenuse for tan (we will need it for sine and cosine)
oh opposite adjacent
sorry
1/1
Great! which simplifies to 1 So we have tan 45° = 1
now we need to work out the length of that hypotenuse. we have a right angled triangle, so we can use Pythagorus a^2 + b^2 = c^2 where a and b are the adj and opp, and c is the hypo
so hypo = √( adj^2 + opp^2 )
plug in adj = 1, opp = 1
that's one of the standard triangles. |dw:1439362733572:dw|
why remember, when you can work it out
\[\tan 45 = 1 , \sin 45 = \frac{1}{\sqrt{2}}, \cos 45 = \frac{1}{\sqrt{2}} \]
why bother to learn pythag. and triangle geometry ?
Thank you both :)

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