الجمعة، 3 يونيو 2016

Problem Study (Trigonometric Identity)

يونيو 03, 2016 TargetMathematics factor formulae, grade11 math, trigonometric identity, trigonometry, ဆယ္တန္းသခၤ်ာ, ဆယ္တန္းသခၤ်ာေမးခြန္း No comments Edit

Given that $\alpha +\beta +\gamma =180^\circ$ , Prove that $\cos\alpha +\cos\beta +\cos\gamma =1+4\sin\frac{\alpha }{2}\sin\frac{\beta}{2}\sin\frac{\gamma}{2}$ .

Solution

    $\alpha +\beta +\gamma =180^\circ$
   $\alpha +\beta =180^\circ -\gamma$
   $\frac{\alpha +\beta}{2} =90^\circ -\frac{\gamma}{2}$
   $\frac{\gamma}{2} =90^\circ -\frac{\alpha +\beta}{2}$
     $\cos\alpha +\cos\beta +\cos\gamma$
$=(\cos\alpha +\cos\beta) +\cos2\left ( \frac{\gamma }{2} \right )$
$=2\cos\frac{\alpha +\beta }{2}\cos\frac{\alpha -\beta }{2}+1-2\sin^2\frac{\gamma }{2}$
$=2\cos(90^\circ-\frac{\gamma }{2} )\cos\frac{\alpha -\beta }{2}+1-2\sin^2(\frac{\gamma }{2})$
$=2\sin(\frac{\gamma }{2})\cos\frac{\alpha -\beta }{2}+1-2\sin^2(\frac{\gamma }{2})$
$=1+2\sin(\frac{\gamma }{2})\cos\frac{\alpha -\beta }{2}-2\sin^2(\frac{\gamma }{2})$
$=1 + 2\sin \frac{\gamma }{2}\left( {\cos \frac{{\alpha - \beta }}{2} - \sin \frac{\gamma }{2}} \right)$
$=1 + 2\sin \frac{\gamma }{2}\left( {\cos \frac{{\alpha - \beta }}{2} - \sin \left( {{{90}^ \circ } - \frac{{\alpha + \beta }}{2}} \right)} \right)$
$=1 + 2\sin \frac{\gamma }{2}\left( {\cos \frac{{\alpha - \beta }}{2} - \cos \frac{{\alpha + \beta }}{2}} \right)$
$=1 + 2\sin \frac{\gamma }{2}\left( { - 2\sin \frac{{\frac{{\alpha - \beta }}{2} + \frac{{\alpha + \beta }}{2}}}{2}\sin \frac{{\frac{{\alpha - \beta }}{2} - \frac{{\alpha + \beta }}{2}}}{2}} \right)$
= $= 1 - 4\sin \frac{\gamma }{2}\sin \frac{\alpha }{2}\sin \left( { - \frac{\beta }{2}} \right$
= $= 1 + 4\sin \frac{\gamma }{2}\sin \frac{\alpha }{2}\sin \frac{\beta }{2}$ $(\because \sin\left ( -\frac{\beta }{2} \right )=-\sin\frac{\beta }{2})$

الخميس، 2 يونيو 2016

Problem Study (Analytic Geometry)

يونيو 02, 2016 TargetMathematics analytic geometry, calculus, challenge problem, grade-11 math, ဆယ္တန္းသခၤ်ာ, ဆယ္တန္းသခၤ်ာေမးခြန္း No comments Edit

Find the equation of the line passing through the vertices of this curve $f(x)=\frac{-2x-9}{x+5}$ .

Solution
$f(x)=\frac{-2x-9}{x+5}$

Let the vertices be (a,f(a))

and

where $a\neq b$ .

∴ $f (a) = \frac{-2a-9}{a+5}$ and $f (b) = \frac{-2b-9}{b+5}$
The gradient of tangent to the curve is $f ' (x) =\frac{1}{{{(x+5)}^{2}}}.$

At vertices, the tangents are parallel.
∴ f ' (a) = f ' (b).

$-\frac{1}{{{(a+5)}^{2}}}=-\frac{1}{{{(b+5)}^{2}}}$
(a + 5)^2 = (b + 5)^2

Since $a\neq b, a-b\neq 0$ .
$\therefore a + b + 10 = 0$ and b = -a-10

.
$\therefore f (b) = \frac{-2b-9}{b+5}= \frac{2a+11}{-(a+5)}.$
Since the line passing through vertices is perpendicular to the respective tangents, its gradient is (a+5)^2

.
Hence $\frac{f(a)-f(b)}{a-b}= (a + 5)^2$
   $\frac{\frac{-2a-9}{a+5}+\frac{2a+11}{a+5}}{2a+10}= (a + 5)^2$
   $\frac{1}{{{(a+5)}^{2}}}= (a + 5)^2$
   (a + 5)^4 = 1