Question

\[−(√3/2)^2+1=\cos2x\] cos x = 1/2

I'd like to know how the negative sign distributes to make it 1/2 and not -1/2. Can someone please explain?

Answer 1

hey @amistre :)

Answer 2

oh no it should = cos x not cos 2x

Answer 3

\[-(a)^2=-(a*a) = -a*a\]hi

Answer 4

1-3/4 = 1/4

cos(2x) = 1/4

Answer 5

i typed it wrong

Answer 6

−(√3/2)2+1=cos^2x
cos x = 1/2

Sorry man!! that's how it should look

Answer 7

−(√3/2)^2+1=cos^2x cos x = 1/2 <-- OMG i mean this one!!

Answer 8

\[cos^2(x)=1/4\]
sqrt each side

cos(x) = 1/2

Answer 9

ahh that makes sense, but what happens to the negative sign?

Answer 10

its used

Answer 11

it stays

Answer 12

wont it as it is outside

Answer 13

1 - (sqrt(3)/2)^2 = 1 - 3/4 = 1/4

Answer 14

but it isnt 3/2 square
it is root3/2 square

Answer 15

see d question again

Answer 16

if it's -(sqrt3/2)^2 + 1 = cos^2 x ----> how does it go to 1 - (sqrt(3)/2)^2 = 1 - 3/4 = 1/4 ? (sorry!)

Answer 17

look up: commutative property of addition

Answer 18

OK, so basically -(sqrt3/2)^2 + 1 = 1 - (sqrt3/2)^2 ?

Answer 19

ya josh d ans comes out to be 1/4 for cos^2x but wen v root both sides the ans comes out to be 1/2

Answer 20

\[a+b=b+a\]
\[(-a)+b=b+(-a)=b-a\]

Answer 21

d ans simple
it comes out easily

Answer 22

Thank you so much! I understand now