Question

Write the expression as a product and simplify sin2alpha - sin8 alpha.
Will give medal!

Answer 1

@optiquest

Answer 2

Sum to Product Formulas

http://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf

Answer 3

Which one is it?

Answer 4

2nd one yeah?

Answer 5

Sum to Product Formula

Answer 6

yea sin - sin

Answer 7

em..in this case alpha = 2 beta = 8?

Answer 8

or does
alpha = 2alpha
beta = 8 alpha
??

Answer 9

everything inside the first sin in alpha

Answer 10

is

Answer 11

2cos[(alpha+beta)/2]sin[(alpha-beta)/2]

would be

2cos[(2alpha + 8alpha)/2]sin[(2alpha - 8alpha)/2]
??

Answer 12

yea

Answer 13

um so in the parenthesis when its ((2alpha + 8 beta)/2) that would become 5alpha?

Answer 14

and sin((2alpha-8alpha)/2 = -3alpha ?

Answer 15

yes

Answer 16

So we have

2cos(5alpha)sin(-3alpha)
now what?

Answer 17

should be it

Answer 18

what? really? o.0

Answer 19

it can't be simplified further?

Answer 20

dont think so

Answer 21

what about double angle formulas? those don't work here?

Answer 22

try it and see

Answer 23

I don't know, that's why i'm asking :P
double angle formula = sin(2theta) = 2sin (theta) cos(theta)
what we have is 2cos(5alpha)sin(-3alpha)

fit or no?

Answer 24

think it needs 2 theta like the name

Answer 25

i dont remember the derivation but i dont think it applies here

Answer 26

okies. Thank you!

Answer 27

@zepdrix dude..I know you're probably tired of me now. but can you just check reaaal quick if this is the most it can be simplified?

Answer 28

I will love you forever! and (maybe..not) stop bugging you!

Answer 29

I guess you could apply one more tiny tiny step if you wanted :)
Sine is an `odd function` therefore \(\Large\rm \sin(-x)=-\sin(x)\)
Do you see how we can use that in our problem here?\[\Large\rm 2\sin(-3\alpha)\cos(5\alpha)\]

Answer 30

-2sin(3alpha)cos(5alpha)

Answer 31

Good \c:/
That's probably as far as you need to simplify it.
If we had even multiples of alpha, then yes we could go further and apply our sine and cosine double angle formulas.
But I don't think they want us to do that here. That's a good place to stop!

Answer 32

yay! thanks :)