Question

Prove tan(θ / 2) = sin θ / (1 + cos θ) for θ in quadrant 1

Answer 1

I substituted the tan(theta/2) with a half angle identity but I don't know where to go from there.

Answer 2

\[\frac{ \sin \theta }{ 1+\cos \theta }\]
\[\frac{ \ 2sin ( \theta /2 )* cos(\theta/2) }{ 2cos^2(\theta /2) }\]
\[\frac{ \ 2sin ( \theta /2 )* cos(\theta/2) }{ 2cos(\theta /2)*cos(\theta /2) }\]
\[\frac{ \sin \theta/2 }{ \cos \theta/2 }\]
\[ tan (\theta/2) \]

Answer 3

Q.E.D!

Answer 4

how did you get from sinθ/1+cosθ to [2sin(θ/2)∗cos(θ/2)]/[2cos2(θ/2)]

Answer 5

do you know the formulae for sin( A+B) and Cos(A+B) ??

Answer 6

yeah:
sin (a+b)=sinacosb+cosasinb
and cos(a+b)= cosacosb-sinasinb

Answer 7

But how does that relate? @SandeepReddy

Answer 8

we have formulae:
sin(x+y)=sinx.cosy+cosx.siny
sin(x−y)=sinx.cosy−cosx.siny
cos(x+y)=cosx.cosy−sinx.siny
cos(x−y)=cosx.cosy+sinx.siny
tan(x+y)=tanx+tany1−tanx.tany
tan(x−y)=tanx−tany1+tanx.tany
remember these
so sin(x+x) = sin(2x) = 2sinx.cosx and
cos(x+x) = cos^2x - sin^2x = 2cos^2x - 1

Answer 9

i used the last two formulae

Answer 10

\[ sin \theta = 2 sin\theta/2 * cos \theta/2 \]

Answer 11

as we have taken 2x = theta/ then x becomes theta/2 right?
same applies for cos2x

Answer 12

do you know the half angle formulas for:
\[\sin \frac{ \theta }{ 2 }\] and \[\cos \frac{ \theta }{ 2 }\] ?
if yes, then you can prove it by going back to the formula of tangent.
\[\tan \theta =\frac{ \sin \theta }{ \cos \theta }\]
\[\tan \frac{ \theta }{2 }=\frac{\sin \frac{ \theta }{2 } }{ \cos \frac{ \theta }{ 2 } }\]

Answer 13

substitute the half-angle formula of sine and cosine.
\[\tan \frac{ \theta }{ 2 }=\frac{ \sqrt{\frac{ 1-\cos \theta }{ 2 }} }{\sqrt{\frac{ \cos \theta+1 }{ 2 }} }\] cancel out two.
\[\tan \frac{ \theta }{ 2 }=\frac{ \sqrt{1-\cos \theta} }{ \sqrt{1+\cos \theta} }\] rationalize the denominator.

Answer 14

\[\tan \frac{ \theta }{ 2 }=\frac{ \sqrt{1-\cos ^{2}\theta} }{ 1+cos }\]
\[\tan \frac{ \theta }{ 2 }=\frac{ \sqrt{\sin ^{2}\theta} }{ 1+\cos \theta }\]
\[\tan \frac{ \theta }{ 2 }=\frac{ \sin \theta }{ 1+\cos \theta }\]
PROVED. :)

Answer 15

@ECEstudent9405 how did you get to tanθ/2=√1−cos^2θ/1+cosθ?

Answer 16

oh ok-ok. sorry.
let's just focus on the numerator ok?
when you rationalize it, it will be:
\[(\sqrt{1-\cos \theta})(\sqrt{1+\cos \theta})\]

Answer 17

right?

Answer 18

ok

Answer 19

o i get it! Thanks :)

Answer 20

wow. You're good. :)
no problem.