If $m\tan (\theta-30°)=n\tan (\theta+120°)$ then prove that : $$\cos 2\theta=\frac{m+n}{2(m-n)}$$
My attempt\ Here, $$m\tan (\theta-30°)=n\tan (\theta+120)$$ $$\frac{\tan (\theta-30°)}{\tan (\theta+120°)}=\frac{n}{m}$$.
Now, what should I do next?
Asked
Active
Viewed 1,893 times
2
-
Tan 120 = -tan 60 – Saikat Feb 07 '16 at 13:03
-
Are supplying all the homework, here? Btw $$\dfrac{m-n}{m+n}$$ reminds me of https://brilliant.org/wiki/componendo-and-dividendo/ – lab bhattacharjee Feb 07 '16 at 13:03
-
Expand both the numerator and the denominator in terms of tan(a-B) – Saikat Feb 07 '16 at 13:04
-
And remember cos 2A = 1-tan^2(A) / 1+ tan^2(A). – Saikat Feb 07 '16 at 13:05
-
See also: http://math.stackexchange.com/questions/1829636/if-m-tan-theta-pi-6-n-tan-theta-2-pi-3-then-find-cos-2-theta – Martin Sleziak Jun 17 '16 at 14:18
3 Answers
1
$\tan(\theta +120)$ = -$\cot(\theta +30)$ = - $1\over {\tan(\theta +30)}$
From this we get
$$\tan(30-\theta) * \tan(30+\theta) $$ = $n\over m$
Expand and proceed.
Win Vineeth
- 3,504
0
$$m\tan (\theta-30^\circ)=n\tan (\theta+120^\circ)$$ $$\frac{\tan (\theta+120^\circ)}{\tan (\theta-30^\circ)}=\frac{m}{n}$$ Applying Componendo-Dividendo, we get $$\frac{\tan (\theta+120^\circ)+\tan (\theta-30^\circ)}{\tan (\theta+120^\circ)-\tan (\theta-30^\circ)}=\frac{m+n}{m-n}$$
Now expand the L.H.S., keeping in mind that $\tan 30^\circ=\frac{1}{\sqrt3}$ and $\tan 120^\circ=-\sqrt3$
As Lab suggested, you can go as follows:
$$\frac{\frac{\sin (\theta+120^\circ)}{\cos (\theta+120^\circ)}+\frac{\sin (\theta-30^\circ)}{\cos (\theta-30^\circ)}}{\frac{\sin (\theta+120^\circ)}{\cos (\theta+120^\circ)}-\frac{\sin (\theta-30^\circ)}{\cos (\theta-30^\circ)}}=\frac{m+n}{m-n}$$
Then you have, $$\frac{\frac{\sin (\theta+120^\circ)\cos (\theta-30^\circ)+\cos (\theta+120^\circ)\sin (\theta-30^\circ)}{\cos (\theta+120^\circ)\cos (\theta-30^\circ)}}{\frac{\sin (\theta+120^\circ)\cos (\theta-30^\circ)-\cos (\theta+120^\circ)\sin (\theta-30^\circ)}{\cos (\theta+120^\circ)\cos (\theta-30^\circ)}}=\frac{m+n}{m-n}$$
The denominators cancel out to again give you $$\frac{\sin (\theta+120^\circ)\cos (\theta-30^\circ)+\cos (\theta+120^\circ)\sin (\theta-30^\circ)}{\sin (\theta+120^\circ)\cos (\theta-30^\circ)-\cos (\theta+120^\circ)\sin (\theta-30^\circ)}=\frac{m+n}{m-n}$$
$$\frac{\sin (2\theta + 90^\circ)}{\sin 150^\circ}=\frac{m+n}{m-n}$$
Can you complete the rest now?
Hope this helps.
SchrodingersCat
- 24,472
-
Expressing $\tan$ as $\dfrac\sin\cos$ will be more easier, right? – lab bhattacharjee Feb 07 '16 at 13:10
-
@labbhattacharjee Do you mean from the step where I have left off? Yes surely. That's the best trick to do it. – SchrodingersCat Feb 07 '16 at 13:12
-
@SchrodingersCat, Using $Tan120=-√3$ and $Tan30=\frac{1}{√3}$ and then using the compound angle formula for $Tan$, it becomes very complex.. – Iaamuser user Feb 07 '16 at 13:16
-
@Iaamuseruser Don't use compound angle formula for $\tan$. Do what lab bhattacharjee has commented. From the step where I have left, break $\tan$ into $\sin$ and $\cos$. And then simplify. I am updating my answer. – SchrodingersCat Feb 07 '16 at 13:18
-
@Iaamuseruser Please check my edit. I have updated it properly. – SchrodingersCat Feb 07 '16 at 13:29
-
-
@SchrodingersCat, The division of $\tan$ as $$\dfrac\sin\cos$$ should ideally be done even before applying Componendo-Dividendo, – lab bhattacharjee Feb 07 '16 at 14:43
0
Hint: Use $$\tan(a\pm b)=\frac{\tan(a)\pm \tan(b)}{1 \mp \tan(a)\tan(b)}$$ ie if $a+b$ then denominator has $-$ve sign and vice-versa, then $\cos 2x=\frac{1-\tan^2x}{1+\tan^2x}$ to get the result.
Shailesh
- 3,789
Archis Welankar
- 15,910