Question

use sum-to-product to find EXACT VALUE of sin150degrees + sin30degrees

Answer 1

why can't you just add them?

Answer 2

\[\sin(150)=\frac{1}{2}\]
\[\sin(30)=\frac{1}{2}\]
\[\frac{1}{2}+\frac{1}{2}=1\] finis

Answer 3

It seemed too easy lol. and I kept getting either 1 or 0 .

Answer 4

it is pretty darned easy considering the other ones

Answer 5

Well it says I have to use the sum- to- product formula

Answer 6

then pretend to use it and write 1 as the answer

Answer 7

xD kay . I got a verifying identity i can't solve >.<

Answer 8

ok ok
so we have to have
\[\sin(\alpha+\beta)+\sin(\alpha-\beta)=2\sin(\alpha)\cos(\alpha)\] right?

Answer 9

mhmm

Answer 10

so lets see
\[\alpha +\beta=150\]
\[\alpha -\beta=30\] solve and get
\[2\alpha =180\implies \alpha =90\] and so
\[\beta =60\]

Answer 11

and in it goes:
\[2\sin(90)\cos(60)=2\times 1\times \frac{1}{2}=1\]

Answer 12

lotta work for that simple answer i would say

Answer 13

xD I appreciate it though