Question

sine of x divided by one minus cosine of x + sine of x divided by one minus cosine of x = 2 csc x

Answer 1

Verify each trigonometric equation by substituting identities to match the right hand side of the equation to the left hand side of the equation.

Answer 2

\[\huge\rm \frac{ \sin(x) }{ 1-cosine(x) } +\frac{ \sin(x) }{ 1 -cosine(x) }\]
what is the common denominator ?

Answer 3

1-cosine(x)?

Answer 4

but the second one is sinx/1+cosx

Answer 5

well the common denominator is 1-cosine(x) obviously
both denominator are same u will get same numerator (sinx)\[\huge\rm \frac{ \sin(x) }{ 1-cosine(x) } \]

Answer 6

huh uh-oh are you sure check your question again

Answer 7

yea i checked it 1+cosx

Answer 8

http://prntscr.com/7vfzly

Answer 9

\[\huge\rm \frac{ \sin(x) }{ 1-cosine(x) } +\frac{ \sin(x) }{ 1 +cosine(x) }\]
so common denominator is (1-cosinex)(1+cosinex)

Answer 10

but the picture of the equation in the book says 1+cosx

Answer 11

alright you typed it wrong

Answer 12

it did that itself it copy pasted that way lol

Answer 13

really ? :o you didin't type it ?

Answer 14

shocking how that can translate :o :/

Answer 15

lol no that was weird tho right?

Answer 16

\[\huge\rm \frac{ sinx(1+cosx) +sinx(1-cosx)}{ (1-cosx)(1+cosx)}\]
common denominator is (1-cosx)(1+cosx)
and multiply numerator of first fraction by denominator of 2nd fraction
*and multiply numerator of 2nd fraction by denominator of first fraction

Answer 17

ye ....

Answer 18

and then you cancel out?

Answer 19

no no NO!
no then u will get the original question

Answer 20

distribute!

Answer 21

ohhh

Answer 22

wait what

Answer 23

yes familiar withe the foil method ? apply that at the denominator

Answer 24

oh okay so i leave the numerator alone righ tnow

Answer 25

1-cos2x?

Answer 26

yep right now distribute both parentheses by sinx 't the numerator

Answer 27

what? that last sentence you said threw me off i guess the wording did

Answer 28

\[\huge\rm \frac{ \color{reD}{sinx}(1+cosx) +\color{red}{sinx}(1-cosx)}{ (1-cosx)^2}\]

distribute both parentheses by red sin x

Answer 29

1sinx+sinxcosx+1sinx-sinxcosx

Answer 30

2sinx/1-cos2x

Answer 31

and cos^2x equal to what ?
do u remember sin/cos identities ?

Answer 32

csc?

Answer 33

nope

Answer 34

special identity \[\huge\rm sin^2x + \cos^2 x = 1\] solve this for cos^2x

Answer 35

some codes came out for when you said special identity nothing else?

Answer 36

sin^2x + cos^2x = 1 solve for cos^2 x

Answer 37

cos^2x=1-sin^2x

Answer 38

yes right so replace cos^2x by 1-sin^2x

Answer 39

okay

Answer 40

\[\huge\rm \frac{ \color{reD}{2sinx}}{ 1-(1-sin^2x)}\]

distribute parentheses by negative one

Answer 41

its coding

Answer 42

ahh
2sinx over 1-(1-sin^2x)

Answer 43

sorry lol im being a pain

Answer 44

2sinx/sin^2x

Answer 45

yes sin^2x is same as sinx times sin x
so 2sin over sinx sin x
simplfy

Answer 46

1/sin2x

Answer 47

is it sin^2 x ?

Answer 48

yea but with one over it

Answer 49

no what about 2 ?
it's 2sinx over sinx times sin x