Question

Verify the identity.

4 csc 2x = 2 csc2x tan x

Answer 1

@satellite73

Answer 2

use the definitions of csc and tan.

Answer 3

cscx=1/sinx

Answer 4

\[\large \csc x=\frac{1}{\sin x}\qquad\text{and}\qquad \tan x=\frac{\sin x}{\cos x} \]

Answer 5

and i would begin from the right side.

Answer 6

\[\large 2\csc(2x)\tan x=2\cdot\frac{1}{\sin(2x)}\cdot\frac{\sin x}{\cos x}= \]

Answer 7

i am sure u can finish this

Answer 8

sorry its 2 csc^2x

Answer 9

so would that change it to 1/sin^2x

Answer 10

yes.

Answer 11

so then would it be 2cosx2sin^2x

Answer 12

@helder_edwin

Answer 13

the identity is
\[\large 4\csc(2x)=2\csc^2x\tan x \]
right?

Answer 14

well i meant for the right side but yes

Answer 15

ok. so the right side would be
\[\large 2\csc^2x\tan x=2\cdot\frac{1}{\sin^2x}\cdot\frac{\sin x}{\cos x} \]

Answer 16

do u see u can simplify one sin x?

Answer 17

no

Answer 18

on the right side there is one sin(x) in the numerator and sin^2(x) in the denominator. so u can simplify one sin(x)

Answer 19

so it becomes
\[\large 2\cdot\frac{1}{\sin x}\cdot\frac{1}{\cos x}=\frac{2}{\sin x\cdot\cos x} \]
agree?

Answer 20

ooooooh ok

Answer 21

now use this identity
\[\large \color{red}{\sin(2x)=2\sin x\cdot\cos x} \]

Answer 22

how?

Answer 23

this can also be written as
\[\large \color{red}{\frac{\sin(2x)}{2}=\sin x\cdot\cos x} \]
agree?

Answer 24

i guess

Answer 25

u guess???
the 2 on the right was multiplying, it passes to the left to divide. right?

Answer 26

anyway. if we substitute this in the identity we r trying to prove we get
\[\large =\frac{2}{\sin x\cos x}=\frac{2}{\frac{\sin(2x)}{2}}=\frac{4}{\sin(2x)}=
4\cdot\csc(2x) \]

Answer 27

ok yes yes

Answer 28

got it?

Answer 29

yeah i got it, can you help me with a few more?

Answer 30

sure

Answer 31

\[\frac{ 1-sinx }{ cosx }=\frac{ cosx }{ 1+sinx }\]

Answer 32

this one is easy. say we begin from the left. ok?

Answer 33

ok

Answer 34

\[\large \frac{1-\sin x}{\cos}=\frac{1-\sin x}{\cos x}\cdot\color{red}{\frac{1+\sin x}{1+\sin x}}= \]
can u finish?

Answer 35

\[\frac{ 1 }{ cosx+sinx}\]??

Answer 36

no. u have a difference of squares in the numerator, can u see it?

Answer 37

no

Answer 38

sorry this is like the only type of math i am bad at

Answer 39

no problem
\[\large (1-\sin x)(1+\sin x)=1^2-\sin^2x=1-\sin^2x \]
agree?

this is what u have in the numerator.

Answer 40

oh ok, i see

Answer 41

so we have
\[\large =\frac{1-\sin^2x}{\cos(1+\sin x)}=\frac{\cos^2x}{\cos x(1+\sin x)} \]
ok?

Answer 42

does -sinx=cosx?

Answer 43

no.

Answer 44

now we can simplify one cos(x) from the numerator and one cos(x) from the denominator and get
\[\large =\frac{\cos x}{1+\sin x} \]
ok?

Answer 45

ok

Answer 46

cool. i hope u r really getting this stuff.

Answer 47

sorta