We have, ${\sin ^3}18^\circ + {\sin ^2}18^\circ = {\sin ^2}18^\circ (\sin 18^\circ + 1)$
$ = {\left( {\frac{{\sqrt 5 – 1}}{4}} \right)^2}\left( {\frac{{\sqrt 5 – 1}}{4} + 1} \right)$
$ = \frac{{6 – 2\sqrt 5 }}{{16}} \times \frac{{\sqrt 5 + 3}}{4}$
$ = \frac{{3 – \sqrt 5 }}{8} \times \frac{{3 + \sqrt 5 }}{4}$
$ = \frac{{9 – 5}}{{8 \times 4}} = \frac{1}{8} = {\sin ^3}30^\circ $
$ \Rightarrow {\sin ^2}18^\circ = {\sin ^3}30^\circ – {\sin ^3}18^\circ $
The required area = shaded area in the figure = $\frac{1}{2}({5^2} – {3^2})\left( {\frac{\pi }{4} – \frac{\pi }{8}} \right)$
$ = \frac{1}{2} \times 16 \times \frac{\pi }{8} = \pi $ square unit.
$x=-i$ satisfies the given equation,
$\therefore x^2(x+i)-xi(x+i)+2(x+i)=0$
$\Rightarrow (x+i)(x^2-xi+2)=0$
Now consider, $x^2-xi+2=0$
$x=-i$ again satisfies the above equation,
$\therefore x(x+i)-2i(x+i)=0$
$\Rightarrow (x+i)(x-2i)=0$
$x=2i$ is another solution,
So, $x=-i$, $x=-i$, $x=2i$ are the three solutions.
Let $a+b+c=k$. So,
$k(a+b)=15$
$k(b+c)=6$
$k(c+a)=-3$
Adding all of the above, $k.2k=18$
$\Rightarrow k=\pm 3$ Continue reading $(a+b)(a+b+c)=15$$(b+c)(a+b+c)=6$$(c+a)(a+b+c)=-3$
$\{ a,b,c\} \equiv ?$
$z = \frac{{1 – i\sin \theta }}{{1 + i\cos \theta }} \times \frac{{1 – i\cos \theta }}{{1 – i\cos \theta }}$
$ = \frac{{1 – i(\cos \theta + \sin \theta ) + {i^2}\sin \theta \cos \theta }}{{1 – {i^2}{{\cos }^2}\theta }}$
$ = \frac{{1 – \sin \theta \cos \theta – i(\cos \theta + \sin \theta )}}{{1 + {{\cos }^2}\theta }}$
$\because {\mathop{\rm Im}\nolimits} (z) = 0,\cos \theta + \sin \theta = 0$
$\Rightarrow \tan \theta = -1 $
$\Rightarrow \theta = n\pi – \frac {\pi}{4}$,$n\in I$
We have, $x^4y^4(x+y)=810$ and
$x^3y^3(x^3+y^3)=945$ or $x^3y^3 (x+y)(x^2-xy+y^2)=945$
On division, $\frac {x^2-xy+y^2}{xy}=\frac {945}{810}=\frac {7}{6}$
$\Rightarrow \frac {x}{y}-1+\frac {y}{x}=\frac {7}{6}$ Continue reading $x^4y^5+x^5y^4=810$
$x^3y^6+x^6y^3=945$
$\{x,y\}=?$
Let $f: (0, 2) \to R $ be defined as $f(x) = {\log _2}\left( {1 + \tan \left( {\frac{{\pi x}}{4}} \right)} \right)$. Then $\mathop {\lim }\limits_{n \to \infty } \frac{2}{n}\left( {f\left( {\frac{1}{n}} \right) + f\left( {\frac{2}{n}} \right) + ………… + f(1)} \right)$ is equal to _ _ _ _ . Continue reading $f(x) = {\log _2}\left( {1 + \tan \left( {\frac{{\pi x}}{4}} \right)} \right)$
$\mathop {\lim }\limits_{n \to \infty } \frac{2}{n}\left( {f\left( {\frac{1}{n}} \right) + f\left( {\frac{2}{n}} \right) + ………… + f(1)} \right)$=?
