Question

Prove that:
(i) $cos3A+cosSA+cos7A+cos15A4cos4AcosSAcos6A$
(ii) $cosA+cos3A+cos5A+cos7A=4cosAcos2Acos4A$
(iii) $sinA+sin2A+sin4A+sin5A4cos\frac{A}{2}cos\frac{3}{A}sin3A$
(iv) $sin3A+sin2A+sinA-4sinAcos\frac{A}{2}cos\frac{3A}{2}$
(v) $cos{20}^{\circ }cos{100}^{\circ }+cos{100}^{\circ }cos{140}^{c}os200°=-\frac{3}{4}$

(vi) $sin\frac{\theta }{2}sin\frac{7\theta }{2}+sin\frac{3\theta }{2}sin\frac{11\theta }{2}=sin2\theta sin5\theta$
(vii) $sin\frac{\theta }{2}-cos3\theta cos=\frac{9\theta }{2}=sin7\theta sin8\theta$

Answer 1

(i) $cos3A+cosSA+cos7A+cos15A4cos4AcosSAcos6A$
We have,
$LHS=cos3A+cos5A+cos7A+cos15A=[cos5A+cos3A]+[cos15A+cos7A]=\left[2cos\frac{(5A+3A)}{2}cos\frac{(5A-3A)}{2}\right]+\left[2cos\frac{(15A+7A)}{2}cos\frac{(15A-7A)}{2}\right]=2cos4AcosA+2cos11Acos4A=2cos4A[cosA+cos11A]=2cos4A\left[2cos\frac{(11A+A)}{2}cos\frac{(11A-A)}{2}\right]=4cosA\left[cos6Acos5A\right]=4cos4Acos5Acos6A=RHS\therefore cos3A+cos5A+cos7A+cos15A=4cos4Acos5Acos6A$

(ii) $cosA+cos3A+cos5A+cos7A=4cosAcos2Acos4A$
$LHS=cosA+cos3A+cos5A+cos7A=(cos3A+cosA)-(cos7A+cos5A)\left[2cos\left(\frac{3A+A}{2}\right)cos\left(\frac{3A-A}{2}\right)\right]+\left[2cos\left(\frac{7A+5A}{2}\right)cos\left(\frac{7A-5A}{2}\right)\right]=2cos2AcosA+2cos6AcosA=2cosA[cos2A+cos6A]=2cosA[cos6A+cos2A]=2cosA\left[2cos\left(\frac{6A+2A}{2}\right)cos\left(\frac{6A-2A}{2}\right)\right]=4coaA\left[cis4Acos2A\right]\therefore cosA+cos3A+cos5A+cos7A=4cosAcos2‘Acos4A$.

(iii) $sinA+sin2A+sin4A+sin5A4cos\frac{A}{2}cos\frac{3}{A}sin3AWehave,LHS=sinA+sin2A+sin4A+sin5A=(sin2A+sinA)+(sin5A+sin4A)=\left[2cos\left(\frac{2A+A}{2}\right)cos\left(\frac{2A+A}{2}\right)\right]+\left[2sin\left(\frac{5A+4A}{2}\right)cos\left(\frac{5A-4A}{2}\right)\right]=2sin\frac{3A}{2}coss\frac{A}{2}+2sin\frac{9A}{2}cos\frac{A}{2}=2cos\frac{A}{2}[sin\frac{3A}{2}+sin\frac{9A}{2}]=2cos\frac{A}{2}[sin\frac{9A}{2}+2sin\frac{3A}{2}]=\left[2cos\frac{A}{2}2sin\left\{\frac{1}{2}(\frac{9A}{2}+\frac{3A}{2})\right\}cos\left\{\frac{1}{2}(\frac{9A}{2}-\frac{3A}{2})\right\}\right]=4cos\frac{A}{2}\left[sin\frac{12A}{4}cos\frac{6A}{4}\right]=4cos\frac{A}{2}sin3Acos\frac{3A}{2}=4cos\frac{A}{2}cos\frac{3A}{2}sin3A=RHS\therefore sinA+sin2A+sin4A+sin5A+=4cos\frac{A}{2}cos\frac{3A}{2}sin3A$

(iv) We have,
$sin3A+sin2A+sinA-4sinAcos\frac{A}{2}cos\frac{3A}{2}=sin3A+sin2A-sinA=sin3A-sin2A-sin2A=2sin\left(\frac{3A-A}{2}\right)cos\left(\frac{3A+A}{2}\right)+sin2A=2sinAcos2A+sin2A=2sinAcos2A+2sinAcosA=2sinA[cos2A+cosA]=2sinA\left[2cos\left(\frac{2A+A}{2}\right)cos\left(\frac{2A-A}{2}\right)\right]=4sinAcos\frac{3A}{2}cos\frac{A}{2}=4sinAcos\frac{A}{2}cos\frac{3A}{2}=RHS\therefore sin3A+sin2A-sinA=4sinAcos\frac{A}{2}cos\frac{3A}{2}$

(v) $cos{20}^{\circ }cos{100}^{\circ }+cos{100}^{\circ }cos{140}^{c}os200°=-\frac{3}{4}LHS=cos{20}^{\circ }cos{100}^{\circ }+cos{100}^{\circ }cos{140}^{c}os200°=\frac{1}{2}[2cos{100}^{\circ }cos{20}^{\circ }+2cos{140}^{\circ }cos{100}^{\circ }-2cos{200}^{\circ }cos{140}^{\circ }]=\frac{1}{2}\left[cos\right({100}^{\circ }+{20}^{\circ })+cos({100}^{\circ }-{20}^{\circ })+cos({140}^{\circ }+{100}^{\circ })+cos({140}^{\circ }-{100}^{\circ })-\left\{cos\right({200}^{\circ }+{140}^{\circ })+cos({200}^{\circ }-{140}^{\circ }\}]=\frac{1}{2}[cos{120}^{\circ }+cos{80}^{\circ }+cos{240}^{\circ }+cos{40}^{\circ }-cos{340}^{\circ }-cos{60}^{\circ }]=\frac{1}{2}\left[cos\right({90}^{\circ }+{30}^{\circ })+cos{80}^{\circ }+cos{40}^{\circ }-cos({80}^{\circ }+{60}^{\circ })-cos({360}^{\circ }-{20}^{\circ })-\frac{1}{2}]=\frac{1}{2}[-sin{30}^{\circ }+2cos\left(\frac{{80}^{\circ }+{40}^{\circ }}{2}\right)cos\left(\frac{{80}^{\circ }-{40}^{\circ }}{2}\right)-cos{60}^{\circ }-cos{20}^{\circ }-\frac{1}{2}]=\frac{1}{2}[-\frac{1}{2}+2cos{60}^{\circ }cos{20}^{\circ }-\frac{1}{2}-cos{20}^{\circ }-\frac{1}{2}]=\frac{1}{2}[-\frac{3}{2}+2\times \frac{1}{2}cos{20}^{\circ }-cos{20}^{\circ }]=\frac{1}{2}[-\frac{3}{2}+cos{20}^{\circ }-cos{20}^{\circ }]=\frac{1}{2}[-\frac{3}{2}+0]=-\frac{3}{4}=RHS\therefore cos{20}^{\circ }cos{100}^{\circ }+cos{100}^{\circ }cos{140}^{\circ }-cos{140}^{\circ }cos{200}^{\circ }=-\frac{3}{4}$

(vi) $sin\frac{\theta }{2}sin\frac{7\theta }{2}+sin\frac{3\theta }{2}sin\frac{11\theta }{2}=sin2\theta sin5\theta$
We have,
$LHS=sin\frac{\theta }{2}sin\frac{7\theta }{2}+sin\frac{3\theta }{2}sin\frac{11\theta }{2}=\frac{1}{2}[2sin\frac{7\theta }{2}sin\frac{\theta }{2}+2sin\frac{11\theta }{2}sin\frac{3\theta }{2}]=\frac{1}{2}[cos(\frac{7\theta }{2}-\frac{\theta }{2})-cos(\frac{7\theta }{2}-\frac{\theta }{2})+cos(\frac{11\theta }{2}-\frac{3}{\theta }2)-cos(\frac{11\theta }{2}-\frac{3\theta }{2})]=\frac{1}{2}[cos\frac{6\theta }{2}-cos\frac{8\theta }{2}+cos\frac{8\theta }{2}-cos\frac{14\theta }{2}]=\frac{1}{2}[cos3\theta -cos4\theta +cos4\theta -cos7\theta ]=\frac{1}{2}[cos3\theta -cos7\theta ]=\frac{-1}{2}[cos7\theta -cos3\theta ]=\frac{-1}{2}[-2sin\left(\frac{7\theta +3\theta }{2}\right)sin\left(\frac{7\theta -3\theta }{2}\right)]=sin\frac{10\theta }{2}sin\frac{4\theta }{2}=sin5\theta sin2\theta =sin2\theta sin5\theta =RHS\therefore sin\frac{\theta }{2}sin\frac{7\theta }{2}+sin\frac{3\theta }{2}sin\frac{11\theta }{2}=sin2\theta sin5\theta$

(vii) $cos\theta cos\frac{\theta }{2}-cos3\theta cos=\frac{9\theta }{2}=sin7\theta sin8\theta LHS=cos\theta cos\frac{\theta }{2}-cos3\theta cos\frac{9\theta }{2}=\frac{1}{2}[2cos\theta .cos\frac{\theta }{2}-2cos3\theta .cos\frac{9\theta }{2}]=\frac{1}{2}[cos(\theta +\frac{\theta }{2})+cos(\theta -\frac{\theta }{2})-cos(3\theta +\frac{9\theta }{2})-cos(3\theta -\frac{9\theta }{2})]=\frac{1}{2}(cos\frac{3\theta }{2}+cos\frac{\theta }{2}-cos\frac{15\theta }{2}-cos\frac{3\theta }{2})=\frac{1}{2}[cos\frac{\theta }{2}-cos\frac{15\theta }{2}]=-\frac{1}{2}\left[2sin\left(\frac{\theta +15\theta }{2}\right)sin\left(\frac{\theta -15\theta }{2}\right)\right]-[\because cosx-cosy=-2sin\frac{x+y}{2}.sin\frac{x-y}{2}]=+(sin8\theta .sin7\theta )=RHS\therefore LHS=RHS$