Question

how do you find tan x/2 = sinx when 0≤x<2pi in radians? need step by step help!!!

Answer 1

let z = x/2

So

2z = x

x = 2z

Answer 2

which means we get

tan(z) = sin(2z)

Answer 3

tan x/2 = sin2x?

Answer 4

Then we can use the identities

tan(A) = sin(A)/cos(A)

and

sin(2A) = 2*sin(A)cos(A)

--------------------------

tan(z) = sin(2z)

sin(z)/cos(z) = 2*sin(z)cos(z)

sin(z)/cos(z) - 2*sin(z)cos(z) = 0

sin(z)( 1/cos(z) - 2*cos(z)) = 0

sin(z) = 0 or 1/cos(z) - 2*cos(z) = 0

Answer 5

Solve those equations for z. Then use those solutions to find x.

Answer 6

Do you see how I'm getting all this?

Answer 7

z=x/2 right?

Answer 8

yes

Answer 9

how do you get sin(z) / cos(z) to sin(z)( 1/cos(z) ?

Answer 10

I factored out sin(z)

Answer 11

could you show me the ssteps:(??? i dont know how!!

Answer 12

wait i will draw it. could you please check!?

Answer 13

@jim_thompson5910

Answer 14

im so lost....

Answer 15

one sec

Answer 16

what you drew isn't valid (i don't think)

Answer 17

but you cant factor sin(z)( 1/cos(z) - 2*cos(z)) = 0 because then the denominator becomes sinzcosz....

Answer 18

so far we have this!

Answer 19

\[\Large \tan(z) = \sin(2z)\]

\[\Large \frac{\sin(z)}{\cos(z)} = 2\sin(z)\cos(z)\]

\[\Large \frac{\sin(z)}{\cos(z)} - 2\sin(z)\cos(z) = 0\]

\[\Large \sin(z)\left(\frac{1}{\cos(z)} - 2\cos(z)\right) = 0\]

With me so far ?

Answer 20

is it like this??

Answer 21

yes

Answer 22

We can go further...

\[\Large \tan(z) = \sin(2z)\]

\[\Large \frac{\sin(z)}{\cos(z)} = 2\sin(z)\cos(z)\]

\[\Large \frac{\sin(z)}{\cos(z)} - 2\sin(z)\cos(z) = 0\]

\[\Large \sin(z)\left(\frac{1}{\cos(z)} - 2\cos(z)\right) = 0\]

\[\Large \sin(z)\left(\frac{1}{\cos(z)} - \frac{2\cos^2(z)}{\cos(z)}\right) = 0\]

\[\Large \sin(z)\left( \frac{1-2\cos^2(z)}{\cos(z)}\right) = 0\]

\[\Large \sin(z)\left( \frac{-1(2\cos^2(z)-1)}{\cos(z)}\right) = 0\]

\[\Large \sin(z)\left( \frac{-cos(2z)}{\cos(z)}\right) = 0\]

\[\Large \sin(z)=0 \ \text{or} \ \frac{-cos(2z)}{\cos(z)} = 0\]

\[\Large \sin(z)=0 \ \text{or} \ -cos(2z) = 0\]

\[\Large \sin(z)=0 \ \text{or} \ cos(2z) = 0\]

Answer 23

i didnt get this step @jim_thompson5910 !!

Answer 24

it should be 2cos^2(z) - 1

that becomes cos(2z)

See this page for trig identities

http://www.sosmath.com/trig/Trig5/trig5/trig5.html

Answer 25

what....im still lost >< im so sorry!

Answer 26

one of the trig identities on that page is

\[\Large \cos(2u) = 2\cos^2(u)-1\]

do you see it

Answer 27

you have to scroll a bit

Answer 28

ahh okay!!

Answer 29

so do you see how i got all that?

Answer 30

im kind of lost on -cos2(z) / cos(z) = 0 how did it become just -cos2(z) = 0 ?

Answer 31

@jim_thompson5910

Answer 32

I multiplied both sides by the denominator cos(z)

Answer 33

Basically,

we have something like

A/B = 0

Multiply both sides by B to get

A = 0*B

A = 0

Answer 34

oh yesah ! okay!!

Answer 35

so your goal is to solve

sin(z) = 0

and

cos(2z) = 0

Answer 36

yes so since z= x/2 ill just plug in sinx/2 = 0

Answer 37

then itll be sin x = 2?

Answer 38

no

Answer 39

sin(x/2) = 0 does NOT become sin(x) = 2

Answer 40

sin(z) = 0

sin(x/2) = 0

x/2 = arcsin(0)

x/2 = 0 or x/2 = pi

x = 0 or x = 2pi

Answer 41

cos(2z) = 0

cos(2(x/2)) = 0

cos(x) = 0

x = arccos(0)

x = pi/2 or x = 3pi/2

Answer 42

does arcsin = \[\sin^-1\]

Answer 43

yes, arcsine and inverse sine are the same thing

Answer 44

now we have the following possible solutions

x = 0, x = pi/2, x = pi, x = 2pi

Answer 45

but when x = pi,

tan(x/2) = tan(pi/2) which is undefined

So x = pi is NOT a solution

Answer 46

also, x = 2pi isn't a solution either because it's not in the interval 0≤x<2pi

Answer 47

wait...x/2 = 0 or x/2 = pi i dont get this....

Answer 48

So only \[\Large x = \frac{\pi}{2}\] and \[\Large x =\frac{3\pi}{2}\] are solutions

Answer 49

oh i had

sin(x/2) = 0

and I took the arcsine of both sides to get

x/2 = arcsine(0)

then I evaluated the arcsine of 0 to get 0 or pi

Note: sin(0) = 0 and sin(pi) = 0

Answer 50

i got it!

Answer 51

ok great

Answer 52

@jim_thompson5910 im sorry again but could you please make it clear about the very beginning. you put z = x/2 so we plug in x/2 for all the (z) right ?

Answer 53

yes because we let z = x/2 at the top we replaced all copies of "z" with "x/2"

Answer 54

since z = x/2, we can multiply both sides by 2 to get x = 2z. So we can replace all copies of 'x' with '2z'

Answer 55

Then we simplify/solve to get

sin(z) = 0

or

cos(2z) = 0

Answer 56

from there, we can go ahead and replace 'z' with 'x/2' (since z = x/2) and 2z with x (since x = 2z) to get

sin(x/2) = 0

or

cos(x) = 0

Answer 57

Does that make sense?

Answer 58

yes thank you!!!

Answer 59

alright great