Question

Medal and Fan, if someone really good at pre calc could help me :)

Answer 1

derive this identity from the sum and difference formulas for cosine: sin a sin b= (1/2){cos(a-b)-cos(a+n)}

Answer 2

Alright here's the thing i got 69/100 on this practice test and my teacher is letting me redo for half a credit per correct answer. can someone help explain this one to me please :)

Answer 3

@iGreen could you please take a look at this

Answer 4

@Loser66 i can show you my work if that helps?

Answer 5

Go ahead

Answer 6

alright it might take a second there's a table and in the left hand side there's calculations, and on the right side is the reasoning

Answer 7

cos(a-b)-cos(a+b) over 2 reason: To manipulate the right side

Answer 8

yup

Answer 9

ok then -2 sin (-a+b)+a+b over 2) ^2 (a+a+b-b over 2)

Answer 10

nope

Answer 11

-cos s+ cost= dsin (stt over 2) sin (s-t over 2)

Answer 12

= sin a sin b ^2 reason : simplified

Answer 13

\(cos (a-b) = cos a cos b + sina sinb\\cos(a+b) = cosa cos b- sinasinb\\----------------------\\\dfrac{cos(a-b) - cos(a+b)}{2} =\dfrac{ cosacosb+sinasinb-cosacosb +sinasinb}{2}\\=\dfrac{\cancel {2} sinasinb}{\cancel 2}= sinasinb ~~=\text{left hand side}\)

Answer 14

So thats for the second part that i had gotten wrong right?

Answer 15

the first part was correct so i should keep that right? what about the reasonings?

Answer 16

I really appreciate your help by the way :)

Answer 17

\[cos (a-b) = cos a cos b + sina sinb ~~~~~(\text{identity})\\cos(a+b) = cosa cos b- sinasinb~~~(\text{identity})\\----------------------\\\dfrac{cos(a-b) - cos(a+b)}{2} \\=\dfrac{ cosacosb+sinasinb-cosacosb +sinasinb}{2}~~(\text{open parentheses}) \\=\dfrac{\cancel {2} sinasinb}{\cancel 2}= sinasinb ~~=\text{left hand side}\]

Answer 18

Thank you so much @Loser66, you're such a legend