Question

prove: \[\cos ^{2}x-\frac{ 1 }{ 2 }tanxsin(2x)=\cos(2x)\]

Answer 1

If \(\color{#000000 }{ \displaystyle \cos^2x-\frac{1}{2}\tan x \sin(2x)=\cos(2x) }\)

then, \(\color{#0000f0 }{ \displaystyle \frac{1}{2}\tan x\sin(2x) =\sin^2x}\)

since \(\color{#000000 }{ \displaystyle \cos(2x)=\cos^2x-\sin^2x }\)

Answer 2

So, if you prove

\(\color{#0000f0 }{ \displaystyle \frac{1}{2}\tan x\sin(2x) =\sin^2x}\)

then your hypothesis is true

Answer 3

Note that \(\color{#000000 }{ \displaystyle \sin(2x) =2\sin x \cos x}\)

Answer 4

From there should be really easy...

Answer 5

yes i got it now and i have 1 more proof its : \[\frac{ cosxtanx+sinx }{ cosx }=2tanx\]

Answer 6

@SolomonZelman

Answer 7

\(\color{#000000 }{ \displaystyle \frac{ax+b }{a}=\frac{ax }{a} +\frac{b}{a}=x+\frac{b}{a} }\)

Answer 8

you would agree with this kind of statement, right?

Answer 9

So, just separate the fraction (as I did with a,b,x), and simplify.

Answer 10

Note that you know sin(x)/cos(x)=tan(x)
And regardless of what Z is, Z+Z=2Z.

Answer 11

yes i aree with that

Answer 12

ok, then you should be able to simplify that, apply basic trig and you are good to go:)

Answer 13

but when you said separate do you mean like this \[\frac{ cosx }{ cosx}+\frac{ tanx }{ cosx }+\frac{ sinx }{ cosx }\]

Answer 14

absolutely not

Answer 15

well with the cos and tan together sorry

Answer 16

yes, they are together, so you can just replace the first + you put, with ×

Answer 17

yea and tangent is just sinx/cosx

Answer 18

yes, but I would ay it the other way aroundfor the purpose of your identity

Answer 19

but on one side its \[\frac{ cosxtanx }{ cosx}+\frac{ sinx }{ cosx }\]

Answer 20

ok, so the first time you cancel the cos(x) on top and bottom, and you get?

sin(x)/cos(x) is same thing as?

then add tangents.

look at the right side

Answer 21

ohhhhhhhh!! once it cancels you just get 2 ttangents basicaly

Answer 22

yes, that is exactly right!

Answer 23

thanks so much

Answer 24

yw

Answer 25

hey the first proof we did i dont really understand

Answer 26

\[\cos ^{2}x-\frac{ 1 }{ 2 }tanxsin(2x)=\cos(2x)\]

Answer 27

Ok, there is a rule:

\(\color{#000000 }{ \displaystyle\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b) }\)

Answer 28

We can prove it geometrically if you want, or if you want to take that as given, then that is fine too

Answer 29

okay

Answer 30

And also,
\(\color{#000000 }{ \displaystyle\cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b) }\)

Answer 31

ill take the second option lol

Answer 32

no no, the first one is the one we apply, I am just letting you know of it

Answer 33

\(\color{#000000 }{ \displaystyle\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b) }\)

and what happens when a and b are the same number, let say that both are x.

Answer 34

2x

Answer 35

\(\color{#000000 }{ \displaystyle\cos(2x)=\cos(x+x)=\cos(x)\cos(x)-\sin(x)\sin(x) }\)

Answer 36

and that is same as

\(\color{#000000 }{ \displaystyle\cos(2x)=\cos^2(x)-\sin^2(x) }\)

Answer 37

yes so can you say sin(2x)=2sinxcosx

Answer 38

So,

\(\color{#000000 }{ \displaystyle \cos^2(x)-\frac{1}{2}\sin(2x)\tan(x)=\cos(2x)}\)

\(\color{#000000 }{ \displaystyle \cos^2(x)-\frac{1}{2}\sin(2x)\tan(x)=\cos^2(x)-\sin^2(x)}\)

\(\color{#000000 }{ \displaystyle \frac{1}{2}\sin(2x)\tan(x)=\sin^2(x)}\)

Answer 39

yes, and then you apply

\(\sin(2x)=2\sin(x)\cos(x)\)

Answer 40

And that rule comes from

\(\color{#000000 }{ \displaystyle \sin(a+b)=\sin(a)\cos(b)+\sin(b)\cos(a)}\)

(after setting a and b =x)

Answer 41

yes but how to conclude that that equals cos(2x)

Answer 42

\(\color{#000000 }{ \displaystyle \cos^2(x)-\frac{1}{2}\sin(2x)\tan(x)=\cos^2(x)-\sin^2(x)}\)

\(\color{#000000 }{ \displaystyle \cos^2(x)-\frac{1}{2}\cdot (2\sin(x)\cos(x))\cdot \tan(x)=\cos^2(x)-\sin^2(x)}\)

\(\color{#000000 }{ \displaystyle \cos^2(x)-\frac{1}{2}\cdot (2\sin(x)\cos(x))\cdot \frac{\sin(x)}{\cos(x)}=\cos^2(x)-\sin^2(x)}\)

Answer 43

simplify that and boom boom :)

Answer 44

i got it now

Answer 45

cool