Question

Fill in the blanks.
(i) The value of the expression
$3\left[sin^{4}\left(\frac{3\pi}{2}-\alpha\right)+sin^{4}\left(3\pi+\alpha\right)\right]$
$-2\left[sin^{6}\left(\frac{\pi}{2}+\alpha\right)+sin^{6}\left(5\pi-\alpha\right)\right]$ is P .
(ii) The value of the expression
$cos^{4}\, \frac{\pi}{8}+cos^{4}\, \frac{3\pi}{8}+cos^{4}\, \frac{5\pi}{8}+cos^{4}\, \frac{7\pi}{8}$ is Q .
(iii) If $\theta$ lies in the first quadrant and $cos\, \theta=\frac{8}{17}$, then the value of
$cos\left(30^{\circ}+\theta\right)+cos\left(45^{\circ}-\theta\right)+cos\left(120^{\circ}-\theta\right)$ is R .

	$P\quad$	$Q\quad$	$R\quad$
(a)$\quad$	$2$	$\frac{1}{2}$	$\frac{13}{17}\left(\frac{\sqrt{3}-1}{2}+\frac{1}{\sqrt{3}}\right)$
(b)	$1$	$\frac{3}{2}$	$\frac{23}{17}\left(\frac{\sqrt{3}-1}{2}+\frac{1}{\sqrt{2}}\right)$
(c)	$3$	$\frac{3}{2}$	$\frac{23}{17}\left(\frac{\sqrt{3}-1}{2}+\frac{1}{\sqrt{3}}\right)$
(d)	$1$	$\frac{1}{2}$	$\frac{13}{17}\left(\frac{\sqrt{3}-1}{2}+\frac{1}{\sqrt{2}}\right)$

• A. (a)
• B. (b)
• C. (c)
• D. (d)

Answer 1

Answer: (B)

(i) We have,
$3\left[sin^{4}\left(\frac{3\pi}{2}-\alpha\right)+sin^{4}\left(3\pi+\alpha\right)\right]$
$-2\left[sin^{6}\left(\frac{\pi}{2}+\alpha\right)+sin^{6}\left(5\pi-\alpha\right)\right]$
$=3\left[cos^{4}\alpha+sin^{4}\alpha\right]-2\left[cos^{6}\alpha+sin^{6}\alpha\right]$
$=3\left[\left(cos^{2}\alpha+sin^{2}\alpha\right)^{2}-2sin^{2}\alpha \,cos^{2}\alpha \right]$
$- 2\left[\left(cos^{2}\alpha + sin^{2}\alpha\right)^{3} - 3\, cos^{2}\alpha\, sin^{2}\alpha\, \left(cos^{2}\alpha + sin^{2}\alpha\right)\right]$
$= 3\left[1-2 \,sin^{2}\alpha\, cos^{2}\alpha\right] - 2 \left[1 - 3 \,cos^{2}\,\alpha\, sin^{2}\,\alpha\right]$
$=3-2=1$

(ii) $cos^{4} \frac{\pi}{8}+cos^{4}\frac{3\pi }{8}+cos^{4}\frac{5\pi }{8}+cos^{4} \frac{7\pi}{8}$
$=cos^{4}\frac{\pi }{8}+cos^{4}\frac{3\pi }{8}+cos^{4}\left(\pi-\frac{3\pi}{8}\right)+cos^{4}\left(\pi-\frac{\pi}{8}\right)$
$=cos^{4}\frac{\pi }{8}+cos^{4}\frac{3\pi }{8}+cos^{4}\frac{3\pi }{8}+cos^{4}\frac{\pi }{8}$
$=2\left[cos^{4}\frac{\pi }{8}+cos^{4} \frac{3\pi}{8}\right]=2\left[cos^{4}\frac{\pi }{8}+cos^{4}\left(\frac{\pi}{2}-\frac{\pi}{8}\right)\right]$
$=2\left[cos^{4}\frac{\pi }{8}+sin^{4} \frac{\pi}{8}\right]$
$=2\left[\left(cos^{2}\frac{\pi }{8}sin^{4}\frac{\pi }{8}\right)^{2}-2\,cos^{2}\frac{\pi }{8}\cdot sin^{2}\frac{\pi }{8}\right]$
$=2\left[1-2\,cos^{2}\frac{\pi }{8}\cdot sin^{2} \frac{\pi}{8}\right]=2-\left(2 sin \frac{\pi}{8}\cdot cos \frac{\pi}{8}\right)^{2}$
$=2-\left(sin \frac{2\pi}{8}\right)^{2}=2-\left(\frac{1}{\sqrt{2}}\right)^{2}$
$=2-\frac{1}{2}=\frac{3}{2}$

(iii) Given, $cos\,\theta =\frac{8}{17}\quad \ldots \left(i\right)$
$\Rightarrow sin\,\theta=\sqrt{1-\frac{64}{289}}$
$\Rightarrow sin\,\theta=\frac{15}{17}$
[$\because \theta$ lies in first quadrant] $\quad \ldots(ii)$
Now, $cos\left(30^{\circ} + \theta\right) + cos\left(45^{\circ} - \theta\right) + cos\left(120^{\circ} - \theta\right)$
$= cos\left(30^{\circ} + \theta\right) + cos\left(45^{\circ} - \theta\right) + cos\left(90^{\circ} + \left(30^{\circ} - \theta\right)\right)$
$= cos30^{\circ}\, cos\theta - sin30^{\circ}\, sin\theta + cos45^{\circ}\, cos\theta$
$+ sin45^{\circ}\, sin\theta - sin30^{\circ} \,cos\theta + cos30^{\circ}\, sin\theta$
$\left(\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}-\frac{1}{2}\right) \frac{8}{17}+\left(\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}+\frac{1}{2}\right) \frac{15}{17}$
[Using $(i)$ and $(ii)$]
$=\frac{23}{17}\left(\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}-\frac{1}{2}\right)$
$=\frac{23}{17}\left(\frac{\sqrt{3}-1}{2}+\frac{1}{\sqrt{2}}\right)$