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.
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.
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.
Here, this is the best I could do. Hopefully it helps you better understand the problem.
Now I've stumbled upon a problem with arcsin/arccos
alpha = arcsin 1/3
cos (alpha + 4/pi)
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) )
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
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.
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.