Question

What identity is used in this function, and how do I solve it: sin^2[1/2cos^-1(4/5)]

My book has no examples of this and I am lost

Answer 1

let u=arccos(4/5)
then cos(u)=4/5

find the opp
also
but use the triangle here to find
sin^2(1/2*u)

Answer 2

the opp is: 3

Answer 3

recall
\[\sin^2(\frac{u}{2})=\frac{1}{2}(1-\cos(u))\]

Answer 4

and right so that makes cos(u)=3/5
just plug in

Answer 5

and evaluate

Answer 6

no I'm dumb

Answer 7

cos(u)=4/5

Answer 8

you didn't have to find the opp side
don't know what I was thinking there

Answer 9

so it involves the half angle formula. Therefore i would take the angle of cos(u)=4/5 and evaluate it at \[(1/2)(1-\cos(4/5)\]

Answer 10

forgot an extra ) on my equation

Answer 11

well you mean cos(u) is 4/5
not cos(u) is cos(4/5)

Answer 12

\[\frac{1}{2}(1-\cos(u)) \\ \text{ and we have} \cos(u)=\frac{4}{5} \\ \text{ so we have } \\ \frac{1}{2}(1-\frac{4}{5})\]

Answer 13

oh yes. so: \[(1/2)(1-4/5)\]

Answer 14

little arithmetic and you are done

Answer 15

That was much more simple than I believed it would be. Thank you so much!

Answer 16

np