Question

cosx-sinx)^2+(cosx+sinx)^2=2
Please help
will give medal

Answer 1

heres what i have so far (cosx - sinx)^2 = cos^2x -2cosxsinx +sin^2x
(cosx+sinx)^2 = cos^2x+2cosxsinx+sin^2x

Answer 2

i need help proving the identity

Answer 3

good.
\[\large\rm \color{Red}{(cosx-sinx)^2}+\color{blue}{(cosx+sinx)^2}\]
\[\large\rm \color{Red}{ \cos^2x -2cosxsinx +\sin^2x} +\color{blue}{ \cos^2x+2cosxsinx+\sin^2x}\]
there is plus sign(addition ) so we don't need to put parentheses now just combine like terms

Answer 4

a*

Answer 5

cosxsinex

Answer 6

hmm what about them ???
what are like terms in that equation ?

Answer 7

the like terms would be cosine

Answer 8

it's not cos
cos^2x
and what else ?

Answer 9

sin^2x

Answer 10

alright what else ?

Answer 11

=1

Answer 12

hmm i'm saying what are like terms in this equation ? .
\[\large\rm \color{Red}{ \cos^2x -2cosxsinx +\sin^2x} +\color{blue}{ \cos^2x+2cosxsinx+\sin^2x}\]

Answer 13

do you know what are `like terms ` ?

Answer 14

like terms are equations that have the same varible

Answer 15

correct! combine like terms in that equation

Answer 16

1-2cosxsinx+cos^2x+sin2x

Answer 17

\[\large\rm \color{Red}{ \cos^2x} -2cosxsinx +\sin^2x +\color{red}{ \cos^2x}+2cosxsinx+\sin^2x\]
cos^2(x) and cos^2(x) are like terms
so combine them
cos^2(x)+cos^2(x)= ???

Answer 18

cos^4x

Answer 19

no
lets review `like terms`
how would you combine \[\large\rm x+x = ??\]

Answer 20

i would xx+xx

Answer 21

xx+xx

Answer 22

please explain what is that ?
how did you get 2 x on each side of plus sign

Answer 23

i went by your example

Answer 24

it would just be cos

Answer 25

when we combine `like terms` we should add/subtract their coefficients\[\large\rm 1x+1x= 2x\]
when we `multiply like terms` then we should add their exponents

Answer 26

2cos

Answer 27

right but we should keep the square \[\huge\rm \cos^2x+\cos^2x= 2\cos^2x\] \[\large\rm \color{Red}{ 2\cos^2x} -2cosxsinx +\sin^2x \color{red}{}+2cosxsinx+\sin^2x\]
now combine other like terms

Answer 28

cosx+2sinx

Answer 29

how did you get cos x ??

Answer 30

i added cosx-cosx sin+sinxcosxsinx+sinx

Answer 31

here are some example
\[\large\rm \rm 2x^2+4x^2\]
the variables of both terms are the same and same exponent
so i can combine their `coefficients`
\[(2+4)x^2=7x^2\]
i just added coefficients and keep the x^2 variable

Answer 32

let cos x = a
and sin x = b alright now
i'm gonna replace alll sin x with b
and cosx with a

Answer 33

\[\large\rm \color{red}{ a^2 -2ab+b^2} +\color{blue}{ a^2+2ab+b^2}\]
try to combine like terms now

Answer 34

cosx-cosinxsin+sinx+cosx+cosxsin+sinx

Answer 35

\[\large\rm \color{red}{ a^2 -2ab+b^2} +\color{blue}{ a^2+2ab+b^2}\]
what are like terms in this equation ?

Answer 36

cos sin

Answer 37

i replaced cosx with a
and sin x = b
to make it looks better and simple

Answer 38

\[\large\rm \color{red}{ a^2 -2ab+b^2} +\color{blue}{ a^2+2ab+b^2}\]
so what are like terms in this equation

Answer 39

remember
x and x^2 are NOT like terms

Answer 40

cos2 and sin2???

Answer 41

alright then keep sin and cos \[\large\rm \color{Red}{ \cos^2x} \color{pink}{-2cosxsinx }+\color{blue}{\sin^2x} +\color{red}{ \cos^2x}+\color{pink}{2cosxsinx}+\color{blue}{\sin^2x}\]
these are like terms

Answer 42

sin^2x+sin^2x= ??
cos^2x+cos^2= ??
2sinxcosx - 2sincosx= ??

Answer 43

2sin^2x
2cos^2x
sin2x−2sinxcosx

Answer 44

am i right

Answer 45

first two are right
2sinxcos-2sinxcos is same as
2ab -2ab = ??

Answer 46

sin2x−2cosxsinx

Answer 47

now what happen when we subtract like terms
like a-a=
x-x= ?
2-2= ?

Answer 48

we get 0

Answer 49

right so \[\rm 2sinxcosx - 2sinxcosx =?? \]
you're subtracting like terms

Answer 50

sin2x-2sinxcosx ?

Answer 51

how did you get that ?

Answer 52

i subtracted

Answer 53

we know cos^2+cos^2 = 2cos ^2
\[\large\rm \color{Red}{ 2\cos^2x} -2cosxsinx +\sin^2x \color{red}{}+2cosxsinx+\sin^2x\]
and sin^2x+sin^2x= 2sin^2x
\[\large\rm \color{Red}{ 2\cos^2x} -2cosxsinx +\color{blue}{2sin^2x}\color{red}{}+2cosxsinx\]
now we need to combine 2sinxcosx-2sinxcosx

Answer 54

Would it be 0

Answer 55

right so we left with \[\huge\rm 2\cos^2x+2\sin^2x\]
now take out the common factor

Answer 56

1

Answer 57

what about one ??

Answer 58

That's what it would equal

Answer 59

hmm now first we take out the common factor

Answer 60

no** not now

Answer 61

\[\huge\rm 2\cos^2x+2\sin^2x\]
2 is common so we should take it out

Answer 62

So I would get cosx+sinx

Answer 63

no

Answer 64

cos^2x and sin^2 not just cos sin
that's very big mistake

Answer 65

\[\huge\rm 2(\cos^2x+\sin^2x)\]and we should keep the common factor outside the parentheses

Answer 66

now use the special identity

Answer 67

I would get 2

Answer 68

right

Answer 69

So would my identity be false??

Answer 70

why ?

Answer 71

Or true

Answer 72

what do you think ?
we were working on left side
is that equal to right side ?

Answer 73

Yes

Answer 74

Wait no

Answer 75

no to what ?

Answer 76

It's not an identity

Answer 77

answer my question
is the left side equal to right side ?

Answer 78

Yes

Answer 79

so if both sides are equal then the equation is correct
true identity
if both sides aren't equal
like `4=6` then false

Answer 80

Ok I got thank you again

Answer 81

np :=))