Question

Fill in the blanks.
(i) The value of the expression
$3\left[si{n}^4\left(\frac{3\pi }{2}-\alpha \right)+si{n}^4\left(3\pi +\alpha \right)\right]$
$-2\left[si{n}^6\left(\frac{\pi }{2}+\alpha \right)+si{n}^6\left(5\pi -\alpha \right)\right]$ is P .
(ii) The value of the expression
$co{s}^4\frac{\pi }{8}+co{s}^4\frac{3\pi }{8}+co{s}^4\frac{5\pi }{8}+co{s}^4\frac{7\pi }{8}$ 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 .

	$P$	$Q$	$R$
(a)	$2$	$\frac{1}{2}$	$\frac{13}{17}\left(\frac{\sqrt{3}-1}{2}+\frac{1}{\sqrt{3}}\right)$
(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,
$3\left[si{n}^4\left(\frac{3\pi }{2}-\alpha \right)+si{n}^4\left(3\pi +\alpha \right)\right]$
$-2\left[si{n}^6\left(\frac{\pi }{2}+\alpha \right)+si{n}^6\left(5\pi -\alpha \right)\right]$
$=3\left[co{s}^4\alpha +si{n}^4\alpha \right]-2\left[co{s}^6\alpha +si{n}^6\alpha \right]$
$=3\left[{\left(co{s}^2\alpha +si{n}^2\alpha \right)}^2-2si{n}^2\alpha co{s}^2\alpha \right]$
$-2\left[{\left(co{s}^2\alpha +si{n}^2\alpha \right)}^3-3co{s}^2\alpha si{n}^2\alpha \left(co{s}^2\alpha +si{n}^2\alpha \right)\right]$
$=3\left[1-2si{n}^2\alpha co{s}^2\alpha \right]-2\left[1-3co{s}^2\alpha si{n}^2\alpha \right]$
$=3-2=1$

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

(iii) Given, $cos\theta =\frac{8}{17}\dots \left(i\right)$
$\Rightarrow sin\theta =\sqrt{1-\frac{64}{289}}$
$\Rightarrow sin\theta =\frac{15}{17}$
[$\because \theta$ lies in first quadrant] $\dots (ii)$
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 $(ii)$]
$=\frac{23}{17}\left(\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}-\frac{1}{2}\right)$
$=\frac{23}{17}\left(\frac{\sqrt{3}-1}{2}+\frac{1}{\sqrt{2}}\right)$