Question

sin2x-cos2x=(1+√3)/2

Answer 1

do you mean

\[\sin^2(x)-\cos^2(x)\]

or

\[\sin(2x)-\cos(2x)\]

Answer 2

sin(2x)-cos(2x)

Answer 3

ok one sec

Answer 4

thx :)

Answer 5

\[\sin(2x)-\cos(2x)=\frac{(1+\sqrt3)}{2}\]Square both sides, then we have
\[[\sin(2x)-\cos(2x)]^2=(\frac{1+\sqrt3}{2})^2\]Expand the left side, we have
\[\sin^2(2x)-2\sin(2x)\cos(2x)+\cos^2(2x)=(\frac{1+\sqrt3}{2})^2\]Rearrange the terms, we have
\[(\sin^2(2x)+\cos^2(2x))-2\sin(2x)\cos(2x)=(\frac{1+\sqrt3}{2})^2\]Use \(\sin^2a+\cos^2a=1\) and \(\sin(2a) = 2\sin a\cos a\) to simplify the left side,
we have
\[1-\sin(4x)=(\frac{1+\sqrt3}{2})^2\]Rearrange the terms again, we have
\[\sin(4x)=1-(\frac{1+\sqrt3}{2})^2\]
I guess you can do the rest :)

Answer 6

Awesome!! thx a lot.. :)))

Answer 7

wait a sec,what im doing now?i really hate this subject ;( can you tell me please?

Answer 8

Which part do you not understand?

Answer 9

sin(4x)=-0.866? if yes what now?