Let $\alpha$, $\beta$, $\gamma$ be angles of acute triangle. How to prove that $(\tan(\frac{\alpha}{2}))^2 + (\tan(\frac{\beta}{2}))^2 + (\tan(\frac{\gamma}{2}))^2 \ge 1$? Does left side of equation have bigger than 1 bound?
121932
We have an acute triangle with angles $\alpha, \beta, \gamma$ and we need to tell whether the following inequality is true for all such triangles:  $$(\tan(\frac{\alpha}{2}))^2+(\tan(\frac{\beta}{2}))^2+(\tan(\frac{\gamma}{2}))^2\geq1$$  We are also asked if $1%$ is the greatest number, for which it is true.  I'm having huge problems when trying to solve such optimization calculus problems. I think we must somehow find triangle for which we can prove, that the upper sum is minimum, than evaluate it and prove that it is equal to $1$. As with most optimization problems, it turns out that if $\alpha=\beta=\gamma=\frac{\pi}{3}$, then the sum evaluates to $1$. But how to prove that this is infimum of this sum?
92868
Using double angle and compound angles formulae prove,  $$ \frac{1-\cos x}{\sin x}     = \tan\frac{x}{2} $$  Can someone please help me figure this question, I have no idea how to approach it?
1261
If $$\frac{\tan (\alpha + \beta - \gamma )}{\tan (\alpha - \beta + \gamma )}=\frac{\tan \gamma}{\tan \beta}$$ then prove  $$\sin (\beta - \gamma)=0$$ or $$\sin 2\alpha + \sin2\beta + \sin 2\gamma =0$$ here my try goes -  To prove $\sin (\beta - \gamma)=0$, sin x = 0, only when x = 0 or 360 degree, but for the sake of simplicity let's take x = 0, then we've to prove that $\beta - \gamma =0$ or $\beta = \gamma$.  Now,   If $\beta = \alpha$, then first equation will be $\frac{\tan \alpha}{\tan \alpha}= \frac{\tan\gamma}{\tan\beta}$  or $\frac{\tan\gamma}{\tan\beta}=1$  and   $$\frac{\tan x}{\tan y}=1$$, only if $$\tan x = \tan y$$  So, also in this case $\tan \gamma$ must be equal to $\tan \beta$ so, as to prove that $\sin (\beta - \gamma)=0$.  So, how to prove that $\beta = \gamma$?  and how to solve the second part of the question here?
3310
I want to prove  \begin{equation*} \tan A  + \tan B + \tan C = \tan A\tan B\tan C \quad\text{when } A+B+C = 180^\circ \end{equation*}  We know that   \begin{equation*} \tan(A+B) = \frac{\tan A+\tan B}{1-\tan A\tan B}~\text{and that}~A+B = 180^\circ-C. \end{equation*}  Therefore $\tan(A+B) = -\tan C.$ From here, I get stuck. Please help.
82310
To prove  $ \tan(A) + 2 \tan(2A) +4\tan4A + 8 \cot8A =\cot(A) $.  I tried to convert $\tan(4A)$ and $ \tan(8A)$ to $\tan(A$) and tried putting in L.H.S but it becomes a mess. Is there a easy and also intuitive  way to do this?  Thanks
92237
Hi all, I am interested to find elementary proof of tangent half angle formula.  My solutions are the following:  Triangle $AOB$ is such that $|AB|=1$ and $\angle AOB=\theta$. We then extend $OB$ to $P$ and $Q$ such that $|OP|=|OQ|=1$. Thus we will have two isosceles triangles: $AOP$ and $AOQ$.  From the picture, $\tan{\left(\frac{\theta}{2}\right)}=\frac{AB}{BP}=\frac{\sin{\left(\theta\right)}}{1+\cos{\left(\theta\right)}}\ \ \ =\frac{BQ}{AB}=\frac{1-\cos{\left(\theta\right)}}{\sin{\left(\theta\right)}}$  Could You guys please check my solution. I am also wondering if there are other elementary solutions, please share, thanks!
74372
Prove $\sin(x)\tan(x) &gt; x^2$ for $x \in ( \,0, \frac{\pi}{2}) \,$  So I did the following: Let $f(x) = \sin(x)\tan(x) - x^2$.  Then of course $f(0) = 0$ and I want to show that for $x \in ( \,0, \frac{\pi}{2}) \,$ $f'(x) &gt; 0$ but I don't really know where to go from here as I can't get anything reasonable done with the derivative.  I was thinking of approximating it using the following inequality $\sin(x) &lt; x &lt; \tan(x)$.   Is that a good direction?
16150
How prove  this inequality $$\tan{(\sin{x})}&gt;\sin{(\tan{x})},0&lt;x&lt;\dfrac{\pi}{2}$$  this  PDF give  a ugly methods :   I post this solution:let     $$f(x)=\tan{(\sin{x})}-\sin{(\tan{x})}$$   then we have   $$f'(x)=sec^2{(\sin{x})}\cos{x}-\cos{(\tan{x})}\sec^2{x}=\dfrac{\cos^3{x}- \cos{(\tan{x})}\cos^2{(\sin{x})}}{\cos^2{(\sin{x})}\cos^2{x}}$$   case 1:$0&lt;x&lt;\arctan{\dfrac{\pi}{2}}$, then we have    $$0&lt;\tan{x}&lt;\dfrac{\pi}{2},0&lt;\sin{x}&lt;\dfrac{\pi}{2}$$   so Use AM-GM inequality we have   $$\sqrt[3]{\cos{(\tan{x})}\cos^2{(\sin{x})}}\le\dfrac{1}{3}[\cos{(\tan{x})}+2\cos{(\sin{x})}]\le\cos{\dfrac{\tan{x}+2\sin{x}}{3}}$$   use   $$\tan{x}+2\sin{x}&gt;3x$$   so   $$f'(x)&gt;0\Longrightarrow f(x)&gt;0$$   case2: $\arctan{\dfrac{\pi}{2}}\le x\le\dfrac{\pi}{2}$, so   $$\sin{(\arctan{\dfrac{\pi}{2}})}&lt;\sin{x}&lt;1$$   since   $$\sin{(\arctan{\dfrac{\pi}{2}})}=\dfrac{\pi}{\sqrt{4+\pi^2}}&gt;\dfrac{\pi}{4}$$   $$\Longrightarrow \dfrac{\pi}{4}&lt;\sin{x}&lt;1$$,so   $$1&lt;\tan{(\sin{x})}&lt;\tan{1},$$,so   $$f(x)&gt;0$$   I think this inequality have other  simple methods.Thank you  It is said can use integral inequality,But I can't   In fact,we have $$\tan{(\sin{x})}-\sin{(\tan{x})}=\dfrac{1}{30}x^7+o(x^7),x\to 0$$ can see:
78848
If $A,B,C$ are the angle of a triangle, then show that $\sin A+\sin B-\cos C\le \dfrac32$ I tried substituting $C=180^\circ-(A+B)$ and got stuck. I also tried using the formula $\sin A+\sin B=2\sin \frac{A+B}{2}\cos \frac{A-B}{2}$ without any progress. Please help!
17339
If $A,B,C$ are angles of a triangle, find the extreme value of $\cos A\cos B\cos C$.   I have tried using $A+B+C=\pi$, and applying all and any trig formulas, also AM-GM, but nothing  helps.  On this topic we learned also about Cauchy inequality, but I have no experience with it.  The answer according to Mathematica is when $A=B=C=60$.  Any ideas?
48681
$$\frac {2\tan {x}}{1-(\tan {x})^2} = (\sin {2x})^2$$  I tried a lot but I get nowhere
53574
If angle $\alpha$ is reflex, and $\cos \alpha = -\frac{9}{41}$, without using a calculator, evaluate  (a) $\sin \alpha$ (b) $\tan \alpha$ (c) $\cos (\alpha - 180\deg)$     .
38407
$Q.14)$ If $\alpha,\beta$ are complex numbers where $\alpha\ne \beta$ and $|\alpha| = 1$. Prove that $\displaystyle\left|\frac{\alpha\overline\beta-1}{\alpha-\beta}\right|$. .  For $Q.14$ I tried separating the expression into real and imaginary components and then solving it but it does not lead to anywhere. What approach should I take to solve this question? Thanks.
107848
How to find this ?   I'm stuck at $1$ and how to $1$ to change the tangent associated with a,b or c.  I meant this ,everybody.  $\frac{tan(a)+tan(b)+tan(y)−tan(a)tan(b)tan(y)}{1−tan(a)tan(b)−tan(a)tan(y)−tan(b)tan(y)}$
58292
If $\alpha$ is a root of the equation $4x^2+2x-1=0$, then prove that $4\alpha^3-3\alpha$ is the other root. How do I proceed? The sum of the roots, the product of the roots lead me nowhere. Should I find the roots of the equation and substitute in the given two expressions of $\alpha$ and check whether manipulating the first root gives me the second root (which seems much complicated), or is there any other way?
807
Possible Duplicate:          Prove $$\cos(\alpha) + \cos(\alpha + \beta) + \cos(\alpha + 2\beta) + \dots + \cos[\alpha + (n-1)\beta] = \frac{\cos(\alpha + \frac{n-1}{2}\beta) \cdot \sin\frac{n\beta}{2}}{\sin\frac{\beta}{2}} $$
44697
if $\alpha$ , $\beta$  are the values of  $\theta$ satisfying  $$\displaystyle a\tan(\theta)+b\sec(\theta) = c$$     prove that $$\displaystyle\tan(\alpha+\beta) = \dfrac{2ac}{a^2-c^2}$$ How do i solve this, suggestions about how to go about doing this