Question

1) 4sinx * sin(pi/3+x)*sin(pi/3-x) = sin3x ?
2) Show sin 15 *cost 15=1/4 , cos 15-sin 15 = V2/2
V=Radical
without actually calculate sin 15 and cos 15.

Answer 1

for the first question do you want to show that
\[4\sin x\sin (\pi/3+x)\sin (\pi/3-x)=\sin 3x\]

Answer 2

yes

Answer 3

you might want to expand sin(pi/3+x) and sin(pi/3-x)
using the sum of two angles identity

Answer 4

\[\sin (a\pm b)=\sin a\cos b\pm \sin b\cos a\]

Answer 5

use this identity

Answer 6

and see where you are to get to right hand side

Answer 7

or use product to sum identity
sure you need to know your identities to do this

Answer 8

see the cheat sheet if you don't remember your identities
http://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf

Answer 9

\[4sinx* ( \sin \frac{ \pi }{ 3 }*cosx+sinx*\cos \frac{ \pi }{ 3 })(\sin \frac{ \pi }{ 3 }*cosx-sinx*\cos \frac{ \pi }{ 3 }) = how i continue?\]

Answer 10

evaluate sin pi/3 and cos pi/3
multiply out everything and goq

Answer 11

go*

Answer 12

how i evaluate? sin pi/3 = sin 60 , cos pi/3 = cos 60 ...

Answer 13

?

Answer 14

how i continue at \[sinx*3\cos ^{2}x-\sin ^{2}x\] ?

Answer 15

well hmm... we can also go the other way and see if we can get that
\[\sin(2x+x)=\sin(2x)\cos(x)+\sin(x)\cos(2x) \\ =2\sin(x)\cos(x)\cos(x)+\sin(x)(\cos^2(x)-\sin^2(x)) \]
maybe you can continue with this and show that this is also
\[3 \sin(x)\cos^2(x)-\sin^2(x)\]

Answer 16

but I think you are going to end up with
\[3\sin(x)\cos^2(x)-\sin^3(x)\]

Answer 17

so I think you might have made a tiny mistake in earlier work