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Trig functions
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File: SineGoldSide.png (271 KB, 3728x1648) Image search: [iqdb] [SauceNao] [Google]
271 KB, 3728x1648
I'm having difficulty with this calculation.

Calculate the sinus of alpha:

sin alpha = 1/5
cos alpha = 2/7

Answers are 2/5 sqrt 6 and 3/7 sqrt 5

It should be really easy if I understood it but can't find anything on the net and the book really doesn't explain it well enough.
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Do you mean "calculate alpha"?
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On the second one, to get the sinus of alpha, you can use the fundamental theorem of trigonometry:

sin^2(alpha) + cos^2(alpha) = 1

If cos(alpha) = 2/7, its square is 4/49.

1 - 4/49 = sin^2(alpha), so sin(alpha) is the square root of 45/49.

45 can be written as 9*5 = (3^2)*5, so its square root is 3 sqrt 5.

sin(alpha) = (3 sqrt 5)/7, or 3/7 sqrt 5

Dunno why you gave sin(alpha) in the first one.
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>>32924
I guess? The book literally says ''calculate the sinus of alpha''
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>>32924
Yeah I meant alpha.
For sine in 1 and cosine in 2.
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>>32927
Yeah, you messed up your first line. You meant cos(alpha) = 1/5.

Applying what I wrote on >>32925, sin^2(alpha) = 24/25.
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>>32929
Alternatively you may use a triangle. Thatight help you visualize the problem. If you draw a rectangle triangle ABC, place alpha "on top" of side AB, measured 2, and hypotenuse measuring 7, you can use Pythagora's Theorem to find the other side's length and then the sinus.

I'd draw it for you, but I'm not near a computer right now, on the phone drawing sucks, but I could try.
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>>32931
Here, this is the best I could do. Hopefully it helps you better understand the problem.
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I understand it now.

I couldn't find that sin^2(alpha) + cos^2(alpha) = 1 line

It all makes sense.

Thank you
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Now I've stumbled upon a problem with arcsin/arccos

alpha = arcsin 1/3
calculate :

cos (alpha + 4/pi)
and
cos 1/2 (alpha)

I understand sin 2 alpha = 2 * sin alpha * cos alpha

But I don't understand those 2...

2/3 - 1/6 * sqrt 2)
1/6 sqrt(18+12*sqrt(2) )
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>>32978
arcsin/arccos or any inverse trig function works like a logarithm; you do the problem backwards.

where you have alpha = arcsin 1/3 you could say it as "some angle, alpha, who's sine is 1/3"

this is because you can perform the sine function to both sides and get "sin (alpha) = 1/3"

with this information you should be able to do the rest of the problem
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File: arcsin-graph.png (5 KB, 460x371) Image search: [iqdb] [SauceNao] [Google]
5 KB, 460x371
>>32978

arcsin and arccos are the inverse operations of sin and cos, respectively: You obtain one of the angles corresponding to a given sine / cosine value.

So alpha = arcsin 1/3 simply means that sin alpha = 1/3, and alpha ~ 0.34, but instead of putting this approximate value you need to use trigonometric identities to determine the exact value of the cosine.
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>>32978
also, what do you mean by cos 1/2 (alpha)?

do you mean cosine (alpha/2) or do you mean (1/2) * cos (alpha) ? or do you mean cos (sqrt(alpha))
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>>32978
also, dont forget that cosine is just sine with a phase of + pi/2 !
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>>32983

I understand it's the inverse. But applying to cos 1/2 alpha doesn't work out..

cos 1/2 (alpha)

My answer with the cosine half angle formula is sqrt (1 + 2 * (sqrt 2) / 3 ) /2)

Doesn't make any sense....

1/6 sqrt (18+12*sqrt(2) ) as mentioned before.
Pretty frustrating
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File: wiskunde.png (11 KB, 776x187) Image search: [iqdb] [SauceNao] [Google]
11 KB, 776x187
this notation
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>>33003
anon sqrt((1+2*(sqrt (2))/3)/2)

is equivalent to 1/6 sqrt(18+13*sqrt(2))
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>>33018
incoming horrible picture, but here you go >>33003
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>>33021
allright.. I think I get it now.
Thank you !