Proving identities

**Oasis1993** · December 2nd 2009, 06:28 AM

a) Prove the Identity cot 1/2 A - tan 1/2 A = 2cotA

b) By choosing a suitable numerical value for A, show that tan15 is a root of the quadratic equations t^2 + 2 sqrt.3t -1 = 0

thanks.

**qmech** · December 2nd 2009, 06:49 AM

Let B = A/2

$ cot B - tan B = \frac { cos B}{sin B } -\frac {sin B}{cos B} $

$ = \frac { cos^2 B - sin^2B}{sin B cos B} $

$ = \frac { cos2 B}{(sin 2B)/2} $

$ = 2 cot 2B $

**mrmohamed** · December 2nd 2009, 06:51 AM

Originally Posted by Oasis1993

a) Prove the Identity cot 1/2 A - tan 1/2 A = 2cotA

Originally Posted by Oasis1993

thanks.

Hi all

**mrmohamed** · December 2nd 2009, 06:54 AM

Originally Posted by qmech

Let B = A/2

$ cot B - tan B = \frac { cos B}{sin B } -\frac {sin B}{cos B} $

$ = \frac { cos^2 B - sin^2B}{sin B cos B} $

$ = \frac { cos2 B}{(sin 2B)/2} $

$ = 2 cot 2B $

sorry sir
i didnt see your solution
mrmohamed

**mathaddict** · December 2nd 2009, 07:08 AM

Originally Posted by Oasis1993

a) Prove the Identity cot 1/2 A - tan 1/2 A = 2cotA

b) By choosing a suitable numerical value for A, show that tan15 is a root of the quadratic equations t^2 + 2 sqrt.3t -1 = 0

thanks.

HI

$\cot \frac{1}{2}A-\tan \frac{1}{2}A=\frac{\cos \frac{1}{2}A}{\sin \frac{1}{2}A}-\frac{\sin \frac{1}{2}A}{\cos \frac{1}{2}A}$

$\frac{\cos^2 \frac{1}{2}A-\sin^2 \frac{1}{2}A}{\sin \frac{1}{2}A\cos \frac{1}{2}A}$

$\frac{\cos A}{\sin \frac{1}{2}A\cos \frac{1}{2}A}\cdot \frac{2}{2}$

$\frac{2\cos A}{\sin A}$

$2\cos A$ (proved)

Now , from this identity ..

$\tan \frac{1}{2}A=\cot \frac{1}{2}A-2\cot A$

$\tan \frac{1}{2}(30)=\cot \frac{1}{2}(30)-2\cot 30$

$\tan 15=\frac{1}{\tan 15}-\frac{2}{\tan 30}$

$\tan 15=\frac{1}{\tan 15}-2\sqrt{3}$

$\tan^2 15+2\sqrt{3}\tan 15-1=0$

let $\tan 15$ be t

$t^2+2\sqrt{3}t-1=0$

Then by using the quadratic formula

$t=\frac{-2\sqrt{3}\pm \sqrt{16}}{2}$

$t=-\sqrt{3}\pm 2$

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