Question

Verify each trigonometric equation by substituting identities to match the right hand side of the equation to the left hand side of the equation.
sin x/ 1 - cos x + sin x/ 1 + cos x = 2 csc x

Answer 1

sin(x)/(1-cos(x))+sin(x)/(1+cos(x)) for the left hand side ?
If so I think it should first be thought to combine the fractions since our goal is to right the left hand side as one term (we do after all have one term on the right)

Answer 2

to write the left*

Answer 3

What would I do to combine?

Answer 4

find a common denominator

Answer 5

or force once

Answer 6

\[\frac{a}{b}+\frac{c}{d}=\frac{ad+cb}{bd} \text{ \to force a common denominator }\]

Answer 7

ahh I see! Give me one moment to do that

Answer 8

\[sinx(1 + cosx) + sinx(1 - cosx)/ (1 - cosx)(1 + cosx)\]

Answer 9

multiply the top out and the bottom out
something should cancel on top
and you should be able to use a really cool identity on bottom after the multiplying part

Answer 10

What would I do from here?

Answer 11

sorry I lost connection...

Answer 12

Did you try to multiply the top and bottom out?

Answer 13

For the botton I got 2-cos^2x

Answer 14

I'm not sure what to do with the top

Answer 15

\[\frac{\sin(x)+\sin(x) \cos(x)+\sin(x)-\sin(x)\cos(x)}{1-\cos^2(x)}\]
the bottom should be 1-cos^2(x)

Answer 16

oh ok

Answer 17

\[(1-\cos(x))(1+\cos(x)) \\ =1(1+\cos(x))-\cos(x)(1+\cos(x)) \\ =1+\cos(x)-\cos(x)-\cos^2(x) \\ =1+0-\cos^2(x) \\=1-\cos^2(x) \]
so that is how I got the bottom

Answer 18

for the top I just multiplied as well

Answer 19

sin(x)(1+cos(x))
is sin(x)+sin(x)cos(x)
and
sin(x)(1-cos(x))
is sin(x)-sin(x)cos(x)

Answer 20

you should see something cancel on top

Answer 21

and you should also be able to combine any like terms on top

Answer 22

and 1-cos^2(x) should look familiar (hint pythagorean identity)

Answer 23

hold on I finally understand it

Answer 24

1-cos^2x = sin^2x

Answer 25

yep

Answer 26

so I should now get \[2\sin^2x \div \sin x^2\]

Answer 27

correct?

Answer 28

or is it just 2sinx?

Answer 29

do you know that x+x=2x
and not 2x^2?

Answer 30

like 2+2=4
not 2(2)^2

Answer 31

ok I thought I had made that mistake

Answer 32

if you have 1 smiley and you add another smiley you say you have 2 smiley
not 2 smiley square

Answer 33

\[\frac{2 \sin(x)}{\sin^2(x)}\]

Answer 34

do you know where to go from here?

Answer 35

what would we do now?

Answer 36

hmmm

Answer 37

do you not see a common factor on top and bottom

Answer 38

yes, so simplify?

Answer 39

yep

Answer 40

2\[\frac{ 2 }{ sinx}\]

Answer 41

yeah

Answer 42

2csc(x)

Answer 43

and since sin x = 1/csc x you end up getting 2csc(x) correct?

Answer 44

yea

Answer 45

Thank you so much! I have been really struggling with this unit.