Question

-cos135+isin135/sin105-icos105=?

Answer 1

\[-Cos(A)=Cos(-A)\]

Answer 2

sin(A+B)=sin A cos B + cos A sin B
sin(A-B)=sin A cos B - cos A sin B
cos(A+B)=cos A cos B - sin A sin B
cos(A-B)=cos A cos B + sin A sin B

try braking these cosines into exact numbers.

For example:
\[-Cos(135)=Cos(-135)=Cos(-45-90)\]

Answer 3

Do you know how to expand this?

Answer 4

Alright....

Answer 5

how to cis?

Answer 6

\[Cos(-45-90)=Cos(-45)~~Cos(90)~~-~~Sin(-45)Sin(90)\]
\[-Cos(45)~~Cos(90)~~+~~-Sin(45)Sin(90)\]\[-Cos(45)~~Cos(90)~~-~~Sin(45)Sin(90)\]

Answer 7

you can simplify this before anything

Cos45=Sin45 and
Cos(90)=0

Answer 8

And sin(90)=1

Answer 9

\[-Sin(45)~-Sin(45)\]\[-2~Sin(45)\]\[Sin(45)= \sqrt{2}/2\]So,\[-2 \times( \sqrt{2}/2)\]\[-\sqrt{2}\]So\[-Cos(135)= -\sqrt{2}\]

Answer 10

\[-\cos135+isin135/\sin105-icos105\]Now we have \[-\sqrt{2}+isin135/\sin105-icos105\]

Answer 11

Did you mean i sin(135) ?

Answer 12

yeap thanks so much