Question

show that sin^2x-in^2y=sin(x+y)sin(x-y)

Answer 1

try expanding the rhs by \(\sin(\theta+\phi)=\sin\theta\cos\phi+\cos\theta\sin\phi\) ?

Answer 2

how would i use that

Answer 3

how come i wouldn't use sin(a-b)sin(a)cos(b)-cos(a)sin(b) instead

Answer 4

@Kainui
sorry I do not know how to solve this problem

Answer 5

its ok do you know how to solve this one csc-1 (2)

Answer 6

\[\begin{array}{rl}x&=&\csc^{-1}2\\\csc x&=&2\\\sin x&=&0.5\end{array}\]

Answer 7

gtg bye :)

Answer 8

let x=u+v, y=u-v
sin(u+v)^2-sin(u-v)^2=sin(2u)sin(2v)
(sin(u+v)+sin(u-v))(sin(u+v)-sin(u-v))=sin(2u)sin(2v)
(2sin(u)cos(v))(2cos(u)sin(v))=sin(2u)sin(2v)
(2sin(u)cos(u))(2sin(v)cos(v))=sin(2u)sin(2v)

Answer 9

2sin(u)cos(u)=sin(2u)
2sin(v)cos(v)=sin(2v)