find cos(sin-1(3/5))
Draw a right angled triangle.
done
can't you just plug it in your calculator? find the inside function and then put that value into the cos equation?
no calculator allowed
Yes, you can put it in you calculator... and learn nothing.
OK so a triangle has an angle (x) - if sin of this angle is 3/5, then one side of the triangle is 3, and the other 5 (by definition). Use pythagoras to work out the other side, and then cos (by definition).
x = sin^-1(3/5), by the way
sin(a) = 3/5 sin^2 + cos^2 = 1 (3/5)^2 + cos^2 = 1 cos^2 = 1 - 9/25 cos^2 = 16/25 cos(a) = 4/5
cos-1(cos(a)) = 4/5
cos^-1(cos(a)) = a, actually ;)
hmm..... thats right. lol
there aint no simple angles from trig to give you a cosine of 4/5; or a sin of 3/5 :)
the answer is 4/5 right?
Yes (but my method was better :p)
what would cos(sin(3/5)) be?
That would be .. a lot harder (if neither are ^-1)
sorry cos-1(sin(3/5))
woops
\[\cos^{-1}\left[\sin\left(\frac{3}{5}\right)\right] = x \iff \sin\left(\frac{3}{5}\right) = \cos(x)\]
Are you working in radians or degrees?
radians i assume. the problem calls for the exact value
By definition: \[\sin(\alpha) = \cos\left(\frac{\pi}{2}-\alpha\right) \text{ and } \cos(\alpha) = \sin\left(\frac{\pi}{2}-\alpha\right) \]
So if sin(3/5) = cos(x), then x = ....
3/5 ?
im sorry those formulas are in my book and i can't understand them.
They aren't in your book, do you mean?
are. i have example problems i just wanted it broken down.
i have to go. thanks for the bit of help though! sorry i couldn't do more with it
Oh, OK. Well, they work because cos is just a translation of sin, by pi/2. The answer is x = pi/2 - 3/5, from the formulae
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