am i dumb, why cant i do this (1 Viewer)

yashbb · Oct 22, 2021

prove qs

011235 · Oct 22, 2021

c)

$LHS = \frac{\sin 2A}{1 - \cos 2A}$
$= \frac{2 \sin A \cos A}{1 - (\cos ^2 A - \sin ^2 A)}$
$= \frac{2 \sin A \cos A}{1 - ((1 - \sin ^2 A) - \sin ^2 A)}$
$= \frac{2 \sin A \cos A}{2 \sin ^2 A}$
$= \frac{\cos A}{\sin A}$
$= \cot A = RHS$

b)

$LHS = \cos 2x + \cos x$
$= \cos ^2 x - \sin ^2 x + \cos x$
$= \cos ^2 x - (1 - \cos ^2 x) + \cos x$
$= 2 \cos ^2 x + \cos x - 1$
$= 2 \cos ^2 x + 2 \cos x - \cos x - 1$
$= (\cos x + 1)(2 \cos x -1) = RHS$

Useful things to remember:

$\cos 2x = 2 \cos ^2 x - 1$
$\cos 2x = 1 - 2 \sin ^2 x$

CM_Tutor · Oct 22, 2021

In a case like (c), where it may not be obvious what to do, it sometimes helps to examine LHS - RHS and see if you can get 0, then note that if LHS - RHS = 0 then it necessarily follows that LHS - RHS. A good example like this is the proof that

$\cfrac{\sin\alpha}{1-\cos\alpha} = \cfrac{1+\cos\alpha}{\sin\alpha}$

CM_Tutor · Oct 22, 2021

Also, note that these identities can be linked to solving equations. For example, solving

$\cos{2x}=-\cos{x}$

is potentially easier by rewriting as

$\cos{2x}+\cos{x} = 0$

and substituting with the identity to form

$(\cos{x}+1)(2\cos{x}-1)=0$

which can be immediately simplified to

$\cos{x} = -1\qquad \qquad \text{OR} \qquad \qquad \cos{x}=\cfrac{1}{2}$

yashbb · Oct 22, 2021

CM_Tutor said:
Also, note that these identities can be linked to solving equations. For example, solving

$\cos{2x}=-\cos{x}$

is potentially easier by rewriting as

$\cos{2x}+\cos{x} = 0$

and substituting with the identity to form

$(\cos{x}+1)(2\cos{x}-1)=0$

which can be immediately simplified to

$\cos{x} = -1\qquad \qquad \text{OR} \qquad \qquad \cos{x}=\cfrac{1}{2}$

can i ask how to this one then?

CM_Tutor · Oct 22, 2021

Expand LHS using the double angle formula. To start with...

$\begin{align*} \tan{(\beta-\alpha)} &= \cfrac{\tan\beta - \tan\alpha}{1+\tan\beta\tan\alpha} \\ \text{So,} \qquad \tan{\left(\cfrac{\pi}{4}-\alpha\right)} &= \cfrac{\tan{\cfrac{\pi}{4}} - \tan\alpha}{1+\tan{\cfrac{\pi}{4}}\tan\alpha} \\ &= \cfrac{1 - \tan\alpha}{1+1 \times \tan\alpha} \\ &= \cfrac{1 - \tan\alpha}{1+\tan\alpha} \end{align*}$

Similarly, you should find that

$\tan{\left(\cfrac{\pi}{4}+\alpha\right)} = \cfrac{1 + \tan\alpha}{1- \tan\alpha}$

So:

$\begin{align*} \text{LHS}\ &= \tan{\left(\cfrac{\pi}{4}+\alpha\right)} - \tan{\left(\cfrac{\pi}{4}-\alpha\right)} \\ &= \cfrac{1 + \tan\alpha}{1 - \tan\alpha} - \cfrac{1 - \tan\alpha}{1 + \tan\alpha} \\ &= \cfrac{(1 + \tan\alpha)^2 - (1 - \tan\alpha)^2}{(1 - \tan\alpha)(1 + \tan\alpha)} \\ &= \cfrac{1 + 2\tan\alpha - \tan^2{\alpha} - (1 - 2\tan\alpha + \tan^2{\alpha})}{1 - \tan^2{\alpha}} \\ &= \cfrac{4\tan\alpha}{1 - \tan^2{\alpha}} \\ &= 2 \times \cfrac{2\tan\alpha}{1 - \tan^2{\alpha}} \\ &= 2\tan{2\alpha} \\ &=\ \text{RHS} \end{align*}$

am i dumb, why cant i do this (1 Viewer)

yashbb

Active Member

011235

Active Member

CM_Tutor

Moderator

CM_Tutor

Moderator

yashbb

Active Member

CM_Tutor

Moderator

Users Who Are Viewing This Thread (Users: 0, Guests: 1)