anonymous
  • anonymous
cosx-sinx)^2+(cosx+sinx)^2=2 Please help will give medal
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
heres what i have so far (cosx - sinx)^2 = cos^2x -2cosxsinx +sin^2x (cosx+sinx)^2 = cos^2x+2cosxsinx+sin^2x
anonymous
  • anonymous
i need help proving the identity
Nnesha
  • Nnesha
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

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Nnesha
  • Nnesha
a*
anonymous
  • anonymous
cosxsinex
Nnesha
  • Nnesha
hmm what about them ??? what are like terms in that equation ?
anonymous
  • anonymous
the like terms would be cosine
Nnesha
  • Nnesha
it's not cos cos^2x and what else ?
anonymous
  • anonymous
sin^2x
Nnesha
  • Nnesha
alright what else ?
anonymous
  • anonymous
=1
Nnesha
  • Nnesha
hmm i'm saying what are like terms in this equation ? . \[\large\rm \color{Red}{ \cos^2x -2cosxsinx +\sin^2x} +\color{blue}{ \cos^2x+2cosxsinx+\sin^2x}\]
Nnesha
  • Nnesha
do you know what are `like terms ` ?
anonymous
  • anonymous
like terms are equations that have the same varible
Nnesha
  • Nnesha
correct! combine like terms in that equation
anonymous
  • anonymous
1-2cosxsinx+cos^2x+sin2x
Nnesha
  • Nnesha
\[\large\rm \color{Red}{ \cos^2x} -2cosxsinx +\sin^2x +\color{red}{ \cos^2x}+2cosxsinx+\sin^2x\] cos^2(x) and cos^2(x) are like terms so combine them cos^2(x)+cos^2(x)= ???
anonymous
  • anonymous
cos^4x
Nnesha
  • Nnesha
no lets review `like terms` how would you combine \[\large\rm x+x = ??\]
anonymous
  • anonymous
i would xx+xx
anonymous
  • anonymous
xx+xx
Nnesha
  • Nnesha
please explain what is that ? how did you get 2 x on each side of plus sign
anonymous
  • anonymous
i went by your example
anonymous
  • anonymous
it would just be cos
Nnesha
  • Nnesha
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
anonymous
  • anonymous
2cos
Nnesha
  • Nnesha
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
anonymous
  • anonymous
cosx+2sinx
Nnesha
  • Nnesha
how did you get cos x ??
anonymous
  • anonymous
i added cosx-cosx sin+sinxcosxsinx+sinx
Nnesha
  • Nnesha
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
Nnesha
  • Nnesha
let cos x = a and sin x = b alright now i'm gonna replace alll sin x with b and cosx with a
Nnesha
  • Nnesha
\[\large\rm \color{red}{ a^2 -2ab+b^2} +\color{blue}{ a^2+2ab+b^2}\] try to combine like terms now
anonymous
  • anonymous
cosx-cosinxsin+sinx+cosx+cosxsin+sinx
Nnesha
  • Nnesha
\[\large\rm \color{red}{ a^2 -2ab+b^2} +\color{blue}{ a^2+2ab+b^2}\] what are like terms in this equation ?
anonymous
  • anonymous
cos sin
Nnesha
  • Nnesha
i replaced cosx with a and sin x = b to make it looks better and simple
Nnesha
  • Nnesha
\[\large\rm \color{red}{ a^2 -2ab+b^2} +\color{blue}{ a^2+2ab+b^2}\] so what are like terms in this equation
Nnesha
  • Nnesha
remember x and x^2 are NOT like terms
anonymous
  • anonymous
cos2 and sin2???
Nnesha
  • Nnesha
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
Nnesha
  • Nnesha
sin^2x+sin^2x= ?? cos^2x+cos^2= ?? 2sinxcosx - 2sincosx= ??
anonymous
  • anonymous
2sin^2x 2cos^2x sin2x−2sinxcosx
anonymous
  • anonymous
am i right
Nnesha
  • Nnesha
first two are right 2sinxcos-2sinxcos is same as 2ab -2ab = ??
anonymous
  • anonymous
sin2x−2cosxsinx
Nnesha
  • Nnesha
now what happen when we subtract like terms like a-a= x-x= ? 2-2= ?
anonymous
  • anonymous
we get 0
Nnesha
  • Nnesha
right so \[\rm 2sinxcosx - 2sinxcosx =?? \] you're subtracting like terms
anonymous
  • anonymous
sin2x-2sinxcosx ?
Nnesha
  • Nnesha
how did you get that ?
anonymous
  • anonymous
i subtracted
Nnesha
  • Nnesha
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
anonymous
  • anonymous
Would it be 0
Nnesha
  • Nnesha
right so we left with \[\huge\rm 2\cos^2x+2\sin^2x\] now take out the common factor
anonymous
  • anonymous
1
Nnesha
  • Nnesha
what about one ??
anonymous
  • anonymous
That's what it would equal
Nnesha
  • Nnesha
hmm now first we take out the common factor
Nnesha
  • Nnesha
no** not now
Nnesha
  • Nnesha
\[\huge\rm 2\cos^2x+2\sin^2x\] 2 is common so we should take it out
anonymous
  • anonymous
So I would get cosx+sinx
Nnesha
  • Nnesha
no
Nnesha
  • Nnesha
cos^2x and sin^2 not just cos sin that's very big mistake
Nnesha
  • Nnesha
\[\huge\rm 2(\cos^2x+\sin^2x)\]and we should keep the common factor outside the parentheses
Nnesha
  • Nnesha
now use the special identity
anonymous
  • anonymous
I would get 2
Nnesha
  • Nnesha
right
anonymous
  • anonymous
So would my identity be false??
Nnesha
  • Nnesha
why ?
anonymous
  • anonymous
Or true
Nnesha
  • Nnesha
what do you think ? we were working on left side is that equal to right side ?
anonymous
  • anonymous
Yes
anonymous
  • anonymous
Wait no
Nnesha
  • Nnesha
no to what ?
anonymous
  • anonymous
It's not an identity
Nnesha
  • Nnesha
answer my question is the left side equal to right side ?
anonymous
  • anonymous
Yes
Nnesha
  • Nnesha
so if both sides are equal then the equation is correct true identity if both sides aren't equal like `4=6` then false
anonymous
  • anonymous
Ok I got thank you again
Nnesha
  • Nnesha
np :=))

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