Question

evaluate: sin[arcsin(1/4)+arctan(-4)]

Answer 1

so sin=1/4, tan=-4 i guess i would make a triangle?

Answer 2

or two triangles because i have to use sin(a+b)?

Answer 3

\[\sin(\alpha +\beta)=\sin(\alpha)\cos(\beta)+\sin(\beta)\cos(\alpha)\] you are given \(\sin(\alpha)=\frac{1}{4}\) and so \(\cos(\alpha)=\frac{\sqrt{15}}{4}\)

Answer 4

thanks for help so far! got up to finding cos(a) got the same answer as you=p

Answer 5

but tangent is different triangle because it cannot be -4 if sin is 1/4 and cos is sqrt15/4 no?

Answer 6

also \(\tan(\beta)=-4\) and so \(\sin(\beta)=-\frac{4}{\sqrt{17}}\)

Answer 7

yeah it is a different triangle, i was wrong

Answer 8

i am often wrong, but it looks like you know what you are doing

Answer 9

o its ok! i really appreciate it, and i guess i figured it out but you helped out alot!!

Answer 10

yw
all that is left is \(\cos(\beta)\) and plugging in the numbers

Answer 11

sooo i cam out with sqrt17 - 4sqrt85/4

Answer 12

came* seems to be correct but going to do it over since theres alot of radicals in it