How do you prove $\frac{1 - cos (x)}{sin (x)} = tan (\frac{x}{2})$ $[1 - cos(x)]/[sin(x)] = tan(x/2)$ ?

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How do you prove $\frac{1 - cos (x)}{sin (x)} = tan (\frac{x}{2})$ $[1 - cos(x)]/[sin(x)] = tan(x/2)$ ?

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Answer 1 · 2016-10-01T17:35:43

Please see below.

Explanation:

As $cos x = 1 - 2 {sin}^{2} (\frac{x}{2})$ $cosx=1-2sin^2(x/2)$ and $sin x = 2 sin (\frac{x}{2}) cos (\frac{x}{2})$ $sinx=2sin(x/2)cos(x/2)$

Hence $\frac{1 - cos x}{sin x}$ $(1-cosx)/sinx$

= $\frac{1 - (1 - 2 {sin}^{2} (\frac{x}{2}))}{2 sin (\frac{x}{2}) cos (\frac{x}{2})}$ $(1-(1-2sin^2(x/2)))/(2sin(x/2)cos(x/2))$

= $\frac{1 - 1 + 2 {sin}^{2} (\frac{x}{2})}{2 sin (\frac{x}{2}) cos (\frac{x}{2})}$ $(1-1+2sin^2(x/2))/(2sin(x/2)cos(x/2))$

= $\frac{2 {sin}^{2} (\frac{x}{2})}{2 sin (\frac{x}{2}) cos (\frac{x}{2})}$ $(2sin^2(x/2))/(2sin(x/2)cos(x/2))$

= $\frac{2 sin (\frac{x}{2}) sin (\frac{x}{2})}{2 sin (\frac{x}{2}) cos (\frac{x}{2})}$ $(2sin(x/2)sin(x/2))/(2sin(x/2)cos(x/2))$

= $\frac{sin (\frac{x}{2})}{cos (\frac{x}{2})}$ $sin(x/2)/cos(x/2)$

= $tan (\frac{x}{2})$ $tan(x/2)$

Answer 2 · 2016-10-01T22:58:29

Please follow the instructions below:

Explanation:

$sin (\frac{x}{2} + \frac{x}{2})$ $sin(x/2+x/2)$

$= sin (\frac{x}{2}) cos (\frac{x}{2}) + cos (\frac{x}{2}) sin (\frac{x}{2})$ $=sin(x/2)cos(x/2)+cos(x/2)sin(x/2)$

$= sin (\frac{x}{2}) {cos (\frac{x}{2}) + cos (\frac{x}{2})}$ $=sin(x/2){cos(x/2)+cos(x/2)}$

$= sin (\frac{x}{2}) \cdot 2 cos (\frac{x}{2})$ $=sin(x/2)*2cos(x/2)$

$= 2 sin (\frac{x}{2}) cos (\frac{x}{2})$ $=2sin(x/2)cos(x/2)$

$= sin (x)$ $=sin(x)$

This is because:

$sin (A + B) = sin (A) cos (B) + cos (A) sin (B)$ $sin(A+B)=sin(A)cos(B)+cos(A)sin(B)$

$cos (\frac{x}{2} + \frac{x}{2})$ $cos(x/2+x/2)$

$= cos (\frac{x}{2}) cos (\frac{x}{2}) - sin (\frac{x}{2}) sin (\frac{x}{2})$ $=cos(x/2)cos(x/2)-sin(x/2)sin(x/2)$

$= {cos}^{2} (\frac{x}{2}) - {sin}^{2} (\frac{x}{2})$ $=cos^2(x/2)-sin^2(x/2)$

$= 1 - {sin}^{2} (\frac{x}{2}) - {sin}^{2} (\frac{x}{2})$ $=1-sin^2(x/2)-sin^2(x/2)$

$= 1 - 2 {sin}^{2} (\frac{x}{2})$ $=1-2sin^2(x/2)$

$= cos (x)$ $=cos(x)$

This is because:

$cos (A + B) = cos (A) cos (B) - sin (A) sin (B)$ $cos(A+B)=cos(A)cos(B)-sin(A)sin(B)$

Which means that:

$L H S$ $LHS$

$= \frac{1 - cos (x)}{sin (x)}$ $=(1-cos(x))/sin(x)$

$= \frac{1 - {1 - 2 {sin}^{2} (\frac{x}{2})}}{2 sin (\frac{x}{2}) cos (\frac{x}{2})}$ $=(1-{1-2sin^2(x/2)})/(2sin(x/2)cos(x/2))$

$= \frac{1 - 1 + 2 {sin}^{2} (\frac{x}{2})}{2 sin (\frac{x}{2}) cos (\frac{x}{2})}$ $=(1-1+2sin^2(x/2))/(2sin(x/2)cos(x/2))$

$= \frac{2 {sin}^{2} (\frac{x}{2})}{2 sin (\frac{x}{2}) cos (\frac{x}{2})}$ $=(2sin^2(x/2))/(2sin(x/2)cos(x/2))$

$= \frac{2 sin (\frac{x}{2}) \cdot sin (\frac{x}{2})}{2 sin (\frac{x}{2}) cos (\frac{x}{2})}$ $=(2sin(x/2)*sin(x/2))/(2sin(x/2)cos(x/2))$

$= \frac{sin (\frac{x}{2})}{cos (\frac{x}{2})}$ $=(sin(x/2))/(cos(x/2))$

$= tan (\frac{x}{2})$ $=tan(x/2)$

$= R H S$ $=RHS$

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