Question

csc(6x) + cot(6x)) / (csc(6x) - cot(6x)) = ________.


1) cot^2 (6x)

2) cot(3x)

3) cot^2(3x)

4) tan(6x)

Answer 1

rafaeli ,

nominator = csc(6x) + cot(6x))

denominator = (csc(6x) - cot(6x))

correct?

Answer 2

I see.

Answer 3

yes that's correct!

Answer 4

\(\Large\color{blue}{ \sf \frac{Csc(x) +Cot(6x)}{Csc(6x) - Cot(6x)} }\)

you can see that top and bottom are CONJUGATES ?

Answer 5

yes!

Answer 6

so where do I go from that? :)

Answer 7

Actually, I would simplifty top and bottom first.


\(\Large\color{blue}{ \sf \frac{Csc(6x) +Cot(6x)}{Csc(6x) - Cot(6x)} }\)


\(\Large\color{red}{ \sf Csc(6x)=\frac{1}{Sin(6x)} }\) \(\Large\color{red}{ \sf Cot(6x)=\frac{Cos(6x)}{Sin(6x)} }\)

substitute.....

\(\Huge\color{blue}{ \sf \frac{\frac{1}{Sin(6x)} +\frac{Cos(6x)}{Sin(6x)}}{\frac{1}{Sin(6x)} - \frac{Cos(6x)}{Sin(6x)}} }\)


\(\Huge\color{blue}{ \sf \frac{\frac{1+Cos(6x)}{Sin(6x)} }{\frac{1-Cos(6x)}{Sin(6x)} } }\)

Answer 8

Oooh I understood that. :) so how do i solve the rest now?

Answer 9

do I flip the denominator fraction and multiply??

Answer 10

YES !!!

Answer 11

so i got 1+cos(6x) / 1-cos(6x)

Answer 12

\(\Huge\color{blue}{ \sf \frac{ \frac{ 1+Cos(x) }{ Sin(x) } }{ \frac{ 1-Cos(x) }{ Sin(x) } }}\)



\(\Huge\color{blue}{ \sf \frac{ 1+Cos(x) }{ Sin(x) } \div \frac{ 1-Cos(x) }{ Sin(x) } =}\)


\(\Huge\color{blue}{ \sf \frac{ 1+Cos(x) }{ Sin(x) } \times \frac{ Sin(x) }{ 1-Cos(x) } }\)

Answer 13

@suhoping you had one mistake, you flip the fraction ....

Answer 14

i thought i flipped it haha sorry

Answer 15

it's alright :)

Answer 16

the way you wrote it, you can just simplify the 6 out?

Answer 17

Well ... it's really sin(6x).

Last year, when I was doing calculations I was just writing Sin/Cos

without saying Sin OF X and Cos OF X..

Answer 18

Anyway, can you finish the problem off?

Answer 19

oooh okay so I got this: 1+cos(6x) * 1-cos(6x)

Answer 20

\(\Huge\color{blue}{ \sf \frac{1+Cos(6x)}{Sin(6x)} \times \frac{Sin(6x)}{1-Cos(6x)} }\)


\(\Huge\color{blue}{ \sf \frac{1+Cos(6x)}{\cancel{ Sin(6x)} } \times \frac{\cancel{ Sin(6x)} }{1-Cos(6x)} }\)

\(\Huge\color{blue}{ \sf \frac{1+Cos(6x)}{1-Cos(6x)} }\)

CORRECT!

Answer 21

so wouldn't the answer really just be 1?

Answer 22

Now

\(\Huge\color{blue}{ \sf \frac{(1+Cos~6x~~)\color{red} { (1+Cos~6x~~) } }{(1-Cos~6x~~)\color{red} { (1+Cos~6x~~) }} }\)

Answer 23

No, -cos on the bottom and +cos on the top, they are not even

Answer 24

I am multiplying tp and bottom times

[1+ cos 6x ]

Answer 25

so the top would be 1+2cos(6x) ?

Answer 26

don't forget about the last part of

(a+b)(a+b)=a^2+2ab+b^2

Answer 27

1+2cos(6x)+ cos ^2 (6x)

Answer 28

@suhoping what would you get on the bottom?

Answer 29

okay i see what you did. give me a second to figure out the answer

Answer 30

can you tell me what is the bottom please

Answer 31

1 + cos^2 (6x) ?

Answer 32

Minus, not plus
\[(a-b)(a+b)=a^2-b^2\]

Answer 33

1^2-cos^2 (6x) = 1 -cos^2x = Sin^2 (6x)

Answer 34

so 1-cos^2 (6x) ?

Answer 35

yes, and that's equal to sin^2 (6x)

becuase

cos^2 (6x) + sin^2 (6x) =1
- cos^2(6x) -cos^2 (6x)

sin^2 (6x)=1-cos^2(6x)

Answer 36

oh okay i got that

Answer 37

so what do i do with the sin^2 (6x) and the 1+2cos(6x)+cos^2(6x) ?

Answer 38

i divide them right?

Answer 39

I am thinking to divide top and bottom by sin(6x)

Answer 40

I mean ^2(6x)

Answer 41

cos divided by sin is cot right?

Answer 42

(yes, you are right)