Question

I need help, i have to graph this:
cos^2x-2sinx*cosx-sin^2x=0
But i dont know how I would graph this.
do I graph the factored form of this equation?

Answer 1

use trig identities

Answer 2

yea i got the factored form, i just want to know if i have to graph the factored form, cuz that way seems more easier

Answer 3

what does it factor to?

Answer 4

cuz i got (sinx+1)(2sinx-1)=0

Answer 5

would i graph it like y=sinx +1 and then another graph y=2sinx-1?

Answer 6

Hello, Quebec friend

Answer 7

lol hey ^_^

Answer 8

im from Ontario :P

Answer 9

i think you need to find a more simplified form.

wolfram says this is equal to

\[\sqrt{2} \sin \left( \frac{ \pi }{ 4 } - 2x\right)\]

Answer 10

habs > your hockey teams ^_^ lol jk

Answer 11

lol

Answer 12

is that wat you got when u factored?

Answer 13

i didn't really touch the problem. i typed the original equation in wolfram (a website) and got different forms of it.

using trig id's you an apparently reduce it to that

i can try figuring it out

Answer 14

can*

Answer 15

ok

Answer 16

ill show u what i did

Answer 17

\[\cos^{2}x-2sinx*cosx-\sin^{2}x=0\]\[1-\sin^{2}x-\sin2x-\sin ^{2}x=0\]\[-2\sin ^{2}x-\sin2x+1=0\]\[-1(\sin^{2}x-\sin2x+1)=0(-1)\]\[2\sin ^{2}x+\sin2x-1=0\]\[2\sin ^{2}x+2sinx-sinx-1=0\]\[(\sin^{2}x+2sinx)-(sinx-1)=0\]\[2sinx(sinx+1)-1(sinx+1)=0\]
therefore \[(sinx+1)(2sinx-1)=0\]
\[sinx+1=0\]\[sinx=-1\]\[x=\sin^{-1} (-1)\]\[x=\frac{ -\pi }{ 2 }\]
or

\[2sinx-1=0\]\[sinx=\frac{ 1 }{ 2 }\]\[x=\sin^{-1}( \frac{ 1 }{ 2 })\]\[x=\frac{ \pi }{ 6 }\]

Answer 18

woo that took a while

Answer 19

first i had to solve for x, and now i have to graph the equation

Answer 20

i'm not sure that that is right.

check this out: http://wolfr.am/1kEvDGL

if you click "more" in the alternative forms place you'll see them

i can't really help with this problem sorry

Answer 21

okay thanks

Answer 22

what do u think loser, do i graph the factored form? or the what euler said?

Answer 23

I don't know why you take a biiiiiiiiig U turn to back to original one like this, to me,
cos^2 - 2 sinx cos x - sin^2 =0
rearrange cos^2 - sin^2 - 2 sinx cosx =0
and first 2 terms is cos (2x) so, it is cos (2x) - sin (2x) =0
and then solve for x if you want to get the roots, by the step cos (2x) = sin(2x) iff 2x = pi/4 + kpi
If you want to graph, then, you don't have equal sign, I mean you have function f(x) = cos (2x) - sin (2x)
from this, to graph, just give values to x and find value of f . for example
x = 0, f (x) = 1
x = pi/2, f(x) = 0
x =pi , f(x) =....
to graph

Answer 24

how did u get rid of 2sinx that was being multiplied by cosx?

Answer 25

2sinx cosx = sin 2x

Answer 26

Yes, you have f(x), clumsy as it is. Choose some x values close together over the radian range 0 to 2 pi, evaluate f(x) there and plot them up, then join with smooth curve.

Answer 27

yes and that wat i did

Answer 28

but your process is not correct at line 6, that step is NOT right

Answer 29

to me that looks like a quadratic trigonometric equation

Answer 30

@douglaswinslowcooper hehehe, I am not useless as I am in physics, right?

Answer 31

@M0j0jojo not that, because sin 2x and 2 sinx is not the same

Answer 32

ooh okay i see that now

Answer 33

for example, if x = pi/3 , then sin (2x) = sin (2pi/3) = sqrt(3)/2 and 2 sin x = 2* sqrt(3)/2 = sqrt(3)

Answer 34

oh ok

Answer 35

good,

Answer 36

i get confused between the two

Answer 37

not get, got

Answer 38

no more now

Answer 39

yea lol thanks

Answer 40

can i use the product and sum method on
2sin^2x+sin2x-1?

Answer 41

i\[2\sin ^{2}x+\sin2x-1=0\] kind of lost my path of thought, so i how would go about solving for x from this point on

Answer 42

u still there?

Answer 43

this is a decent resource for trig id's: http://www.sosmath.com/trig/Trig5/trig5/trig5.html
you could use the id: cos2x = 1 - 2sin^2(x) making what you have:

sin2x - cos2x = 0

now solve for x