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.
Do you mean "calculate alpha"?
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.
>>32924
I guess? The book literally says ''calculate the sinus of alpha''
>>32924
Yeah I meant alpha.
For sine in 1 and cosine in 2.
>>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.
>>32931
Here, this is the best I could do. Hopefully it helps you better understand the problem.
I understand it now.
I couldn't find that sin^2(alpha) + cos^2(alpha) = 1 line
It all makes sense.
Thank you
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...
Answers to them are respectively
2/3 - 1/6 * sqrt 2)
1/6 sqrt(18+12*sqrt(2) )
>>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
>>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.
>>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))
>>32978
also, dont forget that cosine is just sine with a phase of + pi/2 !
>>32983
I understand it's the inverse. But applying to cos 1/2 alpha doesn't work out..
I read this: http://www.intmath.com/analytic-trigonometry/4-half-angle-formulas.php
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....
Answer should be
1/6 sqrt (18+12*sqrt(2) ) as mentioned before.
Pretty frustrating
this notation
>>33003
anon sqrt((1+2*(sqrt (2))/3)/2)
is equivalent to 1/6 sqrt(18+13*sqrt(2))
>>33018
incoming horrible picture, but here you go >>33003
>>33021
allright.. I think I get it now.
Thank you !
