${(3{\cos ^2}x + 2\sin x + 1)_{\max }} = ?$
The given expression = $- 3{\sin ^2}x + 2\sin x + 4$
= – $\left[ {{{(\sqrt 3 \sin x)}^2} – 2.\frac{1}{{\sqrt 3 }}.\sqrt 3 \sin x + \frac{1}{3} – \frac{1}{3}} \right]$ + 4
$ = – \left[ {{{\left( {\sqrt 3 \sin x – \frac{1}{{\sqrt 3 }}} \right)}^2} – \frac{1}{3}} \right] + 4$ Continue reading $3{\cos ^2}x + 2\sin x + 1$
(A) Triangle is isosceles
(B) Triangle is equilateral
(C) Triangle is right angled
(D) Triangle is scalene
We have, $\left| {\begin{array}{*{20}{c}}{4{{\cos }^3}A – 3\cos A}&{4{{\cos }^3}B – 3\cos B}&{4{{\cos }^3}C – 3\cos C}\\{\cos A}&{\cos B}&{\cos C}\\{\sin A}&{\sin B}&{\sin C}\end{array}} \right| = 0$
$ \Rightarrow \left| {\begin{array}{*{20}{c}}{4{{\cos }^3}A}&{4{{\cos }^3}B}&{4{{\cos }^3}C}\\{\cos A}&{\cos B}&{\cos C}\\{\sin A}&{\sin B}&{\sin C}\end{array}} \right| – \left| {\begin{array}{*{20}{c}}{3\cos A}&{3\cos B}&{3\cos C}\\{\cos A}&{\cos B}&{\cos C}\\{\sin A}&{\sin B}&{\sin C}\end{array}} \right| = 0$
$ \Rightarrow \left| {\begin{array}{*{20}{c}}{{{\cos }^3}A}&{{{\cos }^3}B}&{{{\cos }^3}C}\\{\cos A}&{\cos B}&{\cos C}\\{\sin A}&{\sin B}&{\sin C}\end{array}} \right| = 0$ as the 2nd determinant is 0.
$ \Rightarrow \cos A.\cos B.\cos C\left| {\begin{array}{*{20}{c}}{{{\cos }^2}A}&{{{\cos }^2}B}&{{{\cos }^2}C}\\1&1&1\\{\tan A}&{\tan B}&{\tan C}\end{array}} \right| = 0$ Continue reading In $\Delta ABC$,
$\left| {\begin{array}{*{20}{c}}{\cos 3A}&{\cos 3B}&{\cos 3C}\\{\cos A}&{\cos B}&{\cos C}\\{\sin A}&{\sin B}&{\sin C}\end{array}} \right| = 0$
Select the right option
Using $A.M. \ge G.M.$,
$\frac{{{e^x} + {e^{ – x}}}}{2} \ge \sqrt {{e^x}.{e^{ – x}}} $
$ \Rightarrow {e^x} + {e^{ – x}} \ge 2$
$\therefore \sin ({\pi ^x}) \ge 2$ which is not possible for any real x.
So, no solution.
L.H.S.=$\cos \frac{\pi }{7} – \cos \frac{{2\pi }}{7} + \cos \frac{{3\pi }}{7}$
= $\cos \frac{\pi }{7} + \cos \frac{{3\pi }}{7} – \cos \left( {\pi – \frac{{5\pi }}{7}} \right)$
= $\cos \frac{\pi }{7} + \cos \frac{{3\pi }}{7} + \cos \frac{{5\pi }}{7}$
= $\cos \left( {\frac{\pi }{7} + \frac{{3 – 1}}{2}.\frac{{2\pi }}{7}} \right).\frac{{\sin \left( {3.\frac{{2\pi /7}}{2}} \right)}}{{\sin \left( {\frac{{2\pi /7}}{2}} \right)}}$ Continue reading Prove That,
$\cos \frac{\pi }{7}-\cos \frac{{2\pi }}{7}+\cos \frac{{3\pi}}{7}=\frac{1}{2}$
A ladder of length $l=2\sqrt 6$ unit is resting against a wall just touching the cube of edge 1 unit as shown in the figure. Find h above the cube.
Solution
We have, $\tan \theta = h$ and $\sin \theta = \frac{{h + 1}}{l}$
So, $2\sqrt 6 \sin \theta = 1 + \tan \theta $ Continue reading Ladder, Cube & Wall Problem
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 $
Solve ${\cos ^2}x + {\cos ^2}2x + {\cos ^2}3x = 1$ and select correct option(s). x is odd multiple of …
(A) $\frac {\pi}{6}$
(B) $\frac {\pi}{4}$
(C) $\frac {\pi}{3}$
(D) $\frac {\pi}{2}$ Continue reading ${\cos ^2}x + {\cos ^2}2x + {\cos ^2}3x = 1$
