Question

Simplify: (sin Θ − cos Θ)^2 + (sin Θ + cos Θ)^2

Answer 1

(sin Θ − cos Θ)^2 + (sin Θ + cos Θ)^2

(sin^2-cos^2) + (sin^2+cos^2)

(y^2-x^2) + (y^2+x^2) eliminate x^2

(y^2+y^2)= 2sin^2

Answer 2

please check my answer

Answer 3

@Luigi0210 @KyanTheDoodle @mathmath333 @sleepyjess

Answer 4

\[\huge\rm (x+y)^2 \cancel= x^2 + y^2\]

Answer 5

nope
let sin = x
cos = y
so \[(x -y)^2=(x-y)(x-y)\]
now foil

Answer 6

My options:

A. −sin2 Θ

B. −cos2 Θ

C. 0

D. 2

Answer 7

alright so u didn't foil it right \[(x-y)^2 \cancel= x^2 -y^2\]
(x-y)^2 is same as (x-y)(x-y )
can you foil these ?

Answer 8

one moment so i can foil

Answer 9

yes right

Answer 10

Open the square
Then put sin^2x+cos^x=1
You get
1-2sinxcosx +1+2sinxcosx
=1+1=2
(I'm putting theta as x)

Answer 11

is it x^2+2xy+y^2

Answer 12

Yes

Answer 13

yes right (x+y) = x^2 +2xy +y^2 :-)

Answer 14

what about (x-y)^2 ??

Answer 15

the same thing

Answer 16

(x-y)^2 is what i solved for

Answer 17

expand you have to foil it
(X-y)^2 is same as (x-y)(x-y)

Answer 18

No 2xy is replaced by-2xy

Answer 19

right

Answer 20

just like (x+y)(x+y)

Answer 21

\[\huge\rm x^2 -2xy +y^2 + x^2 +2xy +y^2\]\]
combine like terms

Answer 22

2x^2 + 2y^2

Answer 23

I hope you can add the two square carefully.

Answer 24

yep now replace x and y by sin and cos \[\huge\rm 2 \sin^2 + 2\cos^2\]
take out common factor apply this identity \[\large\rm sin^2 \theta + \cos^2 \theta = 1 \]

Answer 25

=2

Answer 26

yes right \[\huge\rm \color{reD}{ 2} \sin^2 +\color{reD}{ 2}\cos^2\]
\[\huge\rm \color{reD}{ 2} (\sin^2 \theta +cos^2 \theta)\]
\[\huge\rm \color{reD}{ 2} (1) =2\]

Answer 27

Thank you so much for your help @Nnesha i really appreciate it.

Answer 28

my pleasure :-)

Answer 29

<3