Question

Use a compound angle formula for cosine to show that cos2x=cos^2x-sin^2x

Answer 1

@kc_kennylau

Answer 2

\[\cos(2x)=\cos(x+x)\]

Answer 3

Then apply the compound angle formula :)

Answer 4

http://www.a-levelmathstutor.com/images/trigonometry/ca-identities01.jpg

Answer 5

can you help me get it with the compound angle formula??

Answer 6

It confuses me sooo much :(

Answer 7

@kc_kennylau

Answer 8

What is confusing you? :)

Answer 9

compound angle formula

Answer 10

But you just have to memorize the formulae by heart, I can't help you on this...

Answer 11

but how do I use it??

Answer 12

@kc_kennylau please heeeeeelp

Answer 13

I suck at math :(

Answer 14

well, you use it when the thing inside the trig funcs are two things added together...

Answer 15

or one thing subtracting one thing

Answer 16

can you do it for me.. so I can see hows it done??

Answer 17

pleaseee

Answer 18

I'll do \(\sin(2x)\) for you so that I don't break the rules and you can learn :)

\[\begin{array}{ll}
&\sin(2x)\\
=&\sin(x+x)\\
=&\sin x\cos x+\cos x\sin x\\
=&\sin x\cos x+\sin x\cos x\\
=&2\sin x\cos x
\end{array}\]

Answer 19

http://www.a-levelmathstutor.com/images/trigonometry/ca-identities01.jpg

Answer 20

Basically, this isn't technically cheating but you gotta see that when they have A and B, when you do cos(2x) its really just (x+x) and look at the addition formula, just replace all the As and Bs with xs

Answer 21

so now I have to do it for sin?

Answer 22

I mean cos! @kc_kennylau

Answer 23

yes :)

Answer 24

okay check this @kc_kennylau

Answer 25

cos(2x)
cos(x+x)
cosx sinx + cosx cosx
cosx sin x + cosx sinx
= 2cosx sinx

I KNOW ITS WRONG

Answer 26

http://www.a-levelmathstutor.com/images/trigonometry/ca-identities01.jpg

Answer 27

thats the same link

Answer 28

The identity is \(\cos(A+B)=\cos A\cos B-\sin A\sin B\)

Answer 29

yep, it's always the same link

Answer 30

The formula for sine is different from the formula of cosine :)

Answer 31

its wrong right?

Answer 32

can you start it off for me then? :)

Answer 33

\[\cos(2x)\]\[=\cos(x+x)\]\[=\fbox{Now apply the formula :D}\]

Answer 34

okay let me try it

Answer 35

will it be 2cosx2sinx ?

Answer 36

nope... what do you get after applying the formula without simplifying? :)

Answer 37

Just plug in A=x and B=x into the formula \(\cos(A+B)=\cos A\cos B-\sin A\sin B\)...

Answer 38

cosx cosx-sinx sinx

Answer 39

exactly :)

now simplify it :)

Answer 40

2sinxcosx

Answer 41

??

Answer 42

how the hell did you get \(2\sin x\cos x\) from \(\cos x\cos x-\sin x\sin x\) /(-_-)\

Answer 43

Idk

Answer 44

What do you get when you multiply a number by itself? :p

Answer 45

square

Answer 46

exactly :D

Answer 47

so what do you get when you multiply \(\cos x\) by itself? :)

Answer 48

cosx^2

Answer 49

and what do you get when you multiply \(\sin x\) by itself? :)

Answer 50

sinx^2

Answer 51

and what about \(\cos x\cos x-\sin x\sin x\)?

Answer 52

cosxsinx^2

Answer 53

just tell me-_-

Answer 54

Side-note here, notation problem, we say \(\large\sin^2x\) instead of \(\large\sin x^2\) to avoid confusion, but they're just the same thing :)

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

Answer 55

thank you :)

Answer 56

wait, typo, should be \(=\cos^2x-\sin^2x\)