Question

prove the identity. do not skip any steps. every step must flow directly from the step before.
sinx=cosx/cotx

Answer 1

cot(x) = cos(x)/sin(x)

Answer 2

that's all?

Answer 3

it's a hint to get you started

Answer 4

let me know how far you get

Answer 5

ok.. one sec

Answer 6

"prove the identity. do not skip any steps. every step must flow directly from the step before." <---- sounds like instructions to the paper holder

Answer 7

anyhow as jim_thompson5910 said, use that, simplify, be happy, eat ice-cream

Answer 8

next step is cotx=cosx^2 sinx?

Answer 9

you have this so far, right?

\[\Large \sin(x) = \frac{\cos(x)}{\cot(x)}\]
\[\Large \sin(x) = \frac{\cos(x)}{\cos(x)/\sin(x)}\]

Answer 10

no i didn't.. i am really struggling with this.. but i'm really trying to understand.. im thinking the next step is the cosx cancel out and leaves sinx=sinx

Answer 11

correct

Answer 12

hint: \(\bf \cfrac{\frac{a}{b}}{\frac{c}{{\color{blue}{ d}}}}\implies \cfrac{a}{b}\cdot \cfrac{{\color{blue}{ d}}}{c}\)

Answer 13

this is what the full step by step picture looks like

\[\Large \sin(x) = \frac{\cos(x)}{\cot(x)}\]
\[\Large \sin(x) = \frac{\cos(x)}{\cos(x)/\sin(x)}\]
\[\Large \sin(x) = \cos(x)*\frac{\sin(x)}{\cos(x)}\]
\[\Large \sin(x) = \cancel{\cos(x)}*\frac{\sin(x)}{\cancel{\cos(x)}}\]
\[\Large \sin(x) = \sin(x)\]

Answer 14

but sinx=sinx does not equal the original problem? I thought the question was wanting me to show how they are the same

Answer 15

i think i was wrong. it says probe and i guess thats what you did. is there more to the problem?

Answer 16

once you get to sin(x) = sin(x), you have shown that the two original sides are the same expression

Answer 17

alright thank you for your help. would you mind checking two problems that I already did?

Answer 18

that effectively proves/verifies the identity

Answer 19

sure

Answer 20

(sinx+cosx)^2=+sinx cosx

[sinx+cosx]^2
=sin^2+2sinx cosx+cos^2x
=[sin^2x+cos^2x]+2sinx cosx
sin^2x+cos^2x=1 by the pythagorean identity
[sin^2x+cos^2x]+2sinx cosx=1+2sinx cosx

Answer 21

@jim_thompson5910 are you stilll here?

Answer 22

is there a typo in your first line when you wrote "(sinx+cosx)^2=+sinx cosx" ?

Answer 23

oh yes i'm sorry.. my num lock was on.. it's (sinx+cosx)^2=1+2sinx cosx

Answer 24

it looks good, I would do it like this (see attached)

Answer 25

but you effectively have the right idea

Answer 26

thank you.. total i have 3 more problems.. I was able to do one, could you check that and help with the other two? ill attempt them myself again though

Answer 27

the one i did was 1+cos(8x)=2cos^2(4x)

cos(A+B)=cosAcosB-sinAsinB
A=B=4x
cos8x=cos^2 4x-sin^2 4x
=cos^2 4x-(-1-cos^2 4x)=2cos^2 4x-1
@jim_thompson5910

Answer 28

looks good, the +1 in the left side of the original equation will cancel with the -1 in 2cos^2 (4x)-1

Answer 29

so it would just be 2cos^2 4x

Answer 30

now the two problems i am having trouble with.. sinx(cosx cotx-sinx)=cos(2x)

Answer 31

recall that cot(x) = cos(x)/sin(x)

Answer 32

distribute the sine through. What do you get?

Answer 33

cotx(sinx)=cosx?

Answer 34

yes

Answer 35

the sines will cancel

Answer 36

then i have cotx=cosx

Answer 37

cotx=cosx is not an identity

Answer 38

cotx(sinx)=cosx is an identity

Answer 39

so that's the end of it? all i have is cotx=cosx/sinx
cotx(sinx)=cosx?

Answer 40

that's just part of it to help you do the whole thing

Answer 41

ok so after cotx(sinx)=cosx i will have
cotx=cosx

Answer 42

sinx(cosx cotx-sinx)=cos(2x)

sinx(cosx cotx) + sinx*(-sinx)=cos(2x)

cosx*cosx - sinx*sinx=cos(2x)

what's next?

Answer 43

wouldn't it be cosx^2 sinx^2=cos(2x)

Answer 44

i think you mean cos^2(x) - sin^2(x) = cos(2x)

Answer 45

im sorry. im really trying to understand this.. is that the end of that problem?

Answer 46

no, the ultimate end is where the two sides are the same expression

Answer 47

OH IT FINALLY CLICKED! i understan now,

Answer 48

here is a nontrig example of an identity (and proving/verifying it)

2x+3x = 5x

5x = 5x

Answer 49

soooo cos^2x-sin^2x=cos(2x)
then we need to get -sin^2x away... can i just add +sin^2x
to leav with cos^2x=cos(2x)

Answer 50

no, you cannot add things to both sides

you must only manipulate one side only

Answer 51

and cos^2x=cos(2x) is false

Answer 52

cos^2x-sin^2x=cos(2x)
cotx=cos^2x/-sin^2x=cos(2x)
which would then give me cos(2x) on both sides

Answer 53

i feel like im messing up more now @jim_thompson5910

Answer 54

@jim_thompson5910 please help me finish this problem

Answer 55

one sec and I'll type something up

Answer 56

See attached

Answer 57

my very last problem is different.. show all of the work to derive a real number value that will make this a true statement: if tanx=-5/12 and x is in quadrant 2, then sin 2x=______

Answer 58

i would plug -5/12 in for x.... to be.. sin 2(-5/12) then solve right

Answer 59

@jim_thompson5910 sin(-5/12)=-.404

Answer 60

no that's not correct

Answer 61

did i mess up from the beginning or just the -.404

Answer 62

if tan(x) = -5/12, then what is x?

Answer 63

isnt it -5/12

Answer 64

no