Question

Equate the imaginary parts of this equation:
cos^2θ+2isinθcosθ-sin^2θ=cos2θ+isinθ ,

Please show steps.
Thank you...

Answer 1

do you know which are the imaginary parts?

Answer 2

yes

The parts with the I - that is 2isinθ and isinθ

Answer 3

in the first one you are missing something out

Answer 4

is that the whole term - 2isinθcosθ

Answer 5

yep (but this "-" is minus or hyphen?)

Answer 6

sorry forget the minus

Answer 7

it is just 2isinθcosθ

Answer 8

Now you can equate them. What is next? You need to solve for "theta"?

Answer 9

Okay do I do transposition of formula?

Answer 10

My problem is that I am stuck at this particular point.

Answer 11

imaginary terms are \[2 \sin \theta \cos \theta \]
\[\sin \theta\]
now
\[2 \sin \theta \cos \theta=\sin \theta\]
so here
\[\sin 2 \theta= \sin \theta\]

Answer 12

theyatin thanks but my textbook gives a different answer, let me see what carlosgp says. :)

Answer 13

you have:\[2\sin(\theta)\cos(\theta)=\sin(\theta)\]
The trivial solution is theta=0 that will make sin(theta)=0
For another solution, suppose sin(theta)<>0, then cos(theta)=1/2, find the "theta" whose cosine is 0.5

Answer 14

okay, the answer from the textbook is sin2θ = 2sinθcosθ

Answer 15

now you will have to solve for value of theta and that is zero. . .
beacause \[2 \theta = \theta\]
its the only case if theta is zero

Answer 16

thanks carlos, theyatin

Answer 17

Any idea why the txt book gives this answer sin2θ = 2sinθcosθ

Answer 18

No clue

Answer 19

okay many thanks

Answer 20

u sure its not printing mistake??
we did everything right. . .

Answer 21

That's a typo, second part must be sin2θ

Answer 22

@theyatin , the question specifically says to equate the imaginary parts. It didn't require solving, just equating.

Kind regards....

Answer 23

@unseenoceans
oh goodness
then you have your answer already right?? i over explained. . .