Question

Prove this:

Answer 1

\[\large \sin(x+y)\sin(x-y)=\sin^2x-\sin^2y\]

Answer 2

ok, do u know the formula of 2 sinA sin B =...?

Answer 3

\[ LS=(sinxsosx+sinycosy)(sixcosx-sinycosy)\]

Answer 4

i think u need to revise your formulas....how'd u get that ?

Answer 5

sin(x+y) = sin x cosy + cos x sin y

Answer 6

sin (x+y) formula and then the sin (x-y)

Answer 7

sin(x-y) = sin x cosy - cos x sin y

Answer 8

but here its better to use 2sin A sin B =... ?
formula....

Answer 9

no wonder -.-

Answer 10

2sinxcosx

Answer 11

\(\large 2sinAsinB=cos(A-B)-cos(A+B)\)
use this.
A=x+y, B=x-y
and tell me what u get ?

Answer 12

ohh we didnt use that one

Answer 13

is this for LS ?

Answer 14

yes.

Answer 15

not sure but,
= (cosxcosy+sinxsiny) - ( cosxcosy-sinxsiny )

Answer 16

i think you are not comfortable with that formula, lets use the formula you know
sin(x+y) = sin x cosy + cos x sin y
sin(x-y) = sin x cosy - cos x sin y

so multiplying the above 2 equations , you get
LS = sin (x+y) sin (x-y)
=[ sin x cosy + cos x sin y ] [sin x cosy - cos x sin y]
=.....
can you figure out next step ??

Answer 17

its of the form [a+b][a-b] =....?

Answer 18

\[=\sin^2xcos^2y-sinxsinycosxcosy+cosxcosysinxsiny-\cos^2xsin^2y\]

Answer 19

yeah, that simplifies to ... ?

Answer 20

does any term gets cancelled ?

Answer 21

the middle two

Answer 22

correct!
now in RS you only want sine terms, right ?
so convert cos^2 y and cos^2 x into sine terms, do u know how ??

Answer 23

wait on the left side
= \[\sin^2xcos^2x-\cos^2xsin^2y\]

Answer 24

its
\(\sin^2xcos^2y-\cos^2xsin^2y\)

Answer 25

opps typo okay :)

Answer 26

so now convert all cos to sin...

Answer 27

\[\sin^2x(1-\sin^2y)-(1-\sin^2x)\sin^2y\]

Answer 28

absolutely correct
if you simplify , you directly get RS
try it.

Answer 29

Yup, i got it !
thankyou you are a great teacher ! :)

Answer 30

welcome ^_^