Question

prove the following identities:
@zepdrix

Answer 1

Given that\[\cos~2x-\sin~2x=\sqrt{2}\sin~2x\]Prove that\[\cos~2x+\sin~2x=\sqrt{2}\cos~2x\]

Answer 2

@skullpatrol

Answer 3

\[\cos~2x=\sin~2x+\sqrt{2}\sin~2x\]\[\cos~2x=\sin~2x(1+\sqrt{2})\]\[\frac{ 1 }{ (1+\sqrt{2}) }\cos~2x=\sin~2x\]

Answer 4

i'm stuck at here

Answer 5

sorry,i don't get u..can u pls show me the working? :)

Answer 6

Ok, just take your expression for sin2x and substitute it into the other equation.

Answer 7

ok

Answer 8

\[\cos~2x+\frac{ 1 }{ (1+\sqrt{2}) }\cos~2x=\sqrt{2}\cos~2x\]

Answer 9

so,what should i do next?

Answer 10

So all you need to do now is to show that this equation is true, right?

Answer 11

right

Answer 12

Do you see any common factor on both sides?

Answer 13

Must we factorise cos 2x on the LHS?

Answer 14

Try it and see what you get :)

Answer 15

okay,hold on.

Answer 16

\[\cos~2x(1+\frac{ 1 }{ 1+\sqrt{2} })=\sqrt{2}\cos~2x\]like this ?

Answer 17

Yep, but it doesn't look too promising.

Answer 18

okay,what should i do next?

Answer 19

The other option is to work in terms of sin2x, right?

Answer 20

yep

Answer 21

Try it and see what you get :)

Answer 22

\[\cos~2x=\sin~2x(\sqrt{2}-1)\]\[\cos~2x(1-\sqrt{2})+\sin~2x=0\]\[\sin~2x(\sqrt{2}+1)(1-\sqrt{2})+\sin~2x=0\]like this?

Answer 23

Actually, if you use a calculator on your first attempt you can see that the two coefficients are equal.

Answer 24

oh i c.. i hv done the working for this question but not sure whether the working is acceptable..can u take a look?

Answer 25

Sure.

Answer 26

ok,hold on..

Answer 27

\[x=\sqrt{2-(\cos2x-\sin2x)^2}\]\[=\sqrt{2-(\cos^{2}2x+\sin^{2}2x-2\sin2xcos2x)}\]

Answer 28

\[=\sqrt{2-(1-2\sin2xcos2x)}\]\[=\sqrt{1+2\sin2xcos2x}\]

Answer 29

\[=\sqrt{\cos^{2}2x+\sin^{2}2x+2\sin2xcos2x}\]\[=\sqrt{(\cos2x+\sin2x)^2}\]\[=\cos2x+\sin2x\]

Answer 30

\[\cos2x=\frac{ \cos2x+\sin2x }{ \sqrt{2} }\]\[\cos2x+\sin2x=\sqrt{2}\cos2x\]

Answer 31

This is my working but i'm not sure whether the working is acceptable

Answer 32

Sorry, I don't know.

Answer 33

it's okay.

Answer 34

hmm,there is some tricky stuff i hvn't learn yet
\[\cos2x=\sin2x+\sqrt{2}\sin2x\]\[\cos2x=\sin2x(1+\sqrt{2})\]\[\frac{ 1 }{ (1+\sqrt{2}) }=\sin2x\](this is my teacher's working but not completed yet)

Answer 35

edit\[\frac{ 1 }{ (1+\sqrt{2}) }\cos2x=\sin2x\]

Answer 36

wait a minute..i think i got it..

Answer 37

\[\frac{ 1 }{ (1+\sqrt{2}) }\cos2x=\sin2x\]\[\frac{ 1-\sqrt{2} }{ 1-\sqrt{2} }\times \frac{ 1 }{ (1+\sqrt{2}) }\cos2x=\sin2x\]

Answer 38

\[\frac{ 1-\sqrt{2} }{ -1 }\cos2x=\sin2x\]\[(1-\sqrt{2})\cos2x=-\sin2x\]

Answer 39

\[\cos2x-\sqrt{2}\cos2x=-\sin2x\]\[\cos2x+\sin2x=\sqrt{2}\cos2x\]

Answer 40

Thanks @skullpatrol