જો $\cos \theta = \frac{1}{2}\left( {a + \frac{1}{a}} \right),$ તો $\cos 3\theta = . . .$
જો $\cos \theta = \frac{1}{2}\left( {a + \frac{1}{a}} \right),$ તો $\cos 3\theta = . . .$
- A
$\frac{1}{8}\left( {{a^3} + \frac{1}{{{a^3}}}} \right)$
- B
$\frac{3}{2}\left( {a + \frac{1}{a}} \right)$
- C
$\frac{1}{2}\left( {{a^3} + \frac{1}{{{a^3}}}} \right)$
- D
$\frac{1}{3}\left( {{a^3} + \frac{1}{{{a^3}}}} \right)$
Similar Questions
$1 + \cos \,{56^o} + \cos \,{58^o} - \cos {66^o} = $
$1 + \cos \,{56^o} + \cos \,{58^o} - \cos {66^o} = $
- [IIT 1964]
${\sin ^2}\frac{\pi }{8} + {\sin ^2}\frac{{3\pi }}{8} + {\sin ^2}\frac{{5\pi }}{8} + {\sin ^2}\frac{{7\pi }}{8}$ =
${\sin ^2}\frac{\pi }{8} + {\sin ^2}\frac{{3\pi }}{8} + {\sin ^2}\frac{{5\pi }}{8} + {\sin ^2}\frac{{7\pi }}{8}$ =
જો $\cos \left( {\alpha + \beta } \right) = \frac{4}{5}$ અને $\sin \left( {\alpha - \beta } \right) = \frac{5}{{13}}$,કે જ્યાં $0 \le \alpha ,\beta \le \frac{\pi }{4}$. તો $\tan 2\alpha $ મેળવો.
જો $\cos \left( {\alpha + \beta } \right) = \frac{4}{5}$ અને $\sin \left( {\alpha - \beta } \right) = \frac{5}{{13}}$,કે જ્યાં $0 \le \alpha ,\beta \le \frac{\pi }{4}$. તો $\tan 2\alpha $ મેળવો.
- [IIT 1979]
${\cos ^2}A{(3 - 4{\cos ^2}A)^2} + {\sin ^2}A{(3 - 4{\sin ^2}A)^2} = $
${\cos ^2}A{(3 - 4{\cos ^2}A)^2} + {\sin ^2}A{(3 - 4{\sin ^2}A)^2} = $
જો $\tan x = \frac{{2b}}{{a - c}}(a \ne c),$
$y = a\,{\cos ^2}x + 2b\,\sin x\cos x + c\,{\sin ^2}x$
અને $z = a{\sin ^2}x - 2b\sin x\cos x + c{\cos ^2}x,$ તો
જો $\tan x = \frac{{2b}}{{a - c}}(a \ne c),$
$y = a\,{\cos ^2}x + 2b\,\sin x\cos x + c\,{\sin ^2}x$
અને $z = a{\sin ^2}x - 2b\sin x\cos x + c{\cos ^2}x,$ તો