Question

Prove that:
(i) $cos{55}^{\circ }+cos{65}^{\circ }+cos{175}^{\circ }=0$
(ii) $sin{50}^{\circ }-sin{70}^{\circ }+sin{10}^{\circ }=0$
(iii) $cos{80}^{\circ }+cos{40}^{\circ }-cos{20}^{\circ }=0$
(iv) $cos{20}^{\circ }+cos{100}^{\circ }+cos{140}^{\circ }=0$
(v) $sin\frac{5\pi }{18}-cos\frac{4\pi }{9}=\sqrt{3}sin\frac{\pi }{9}$
(vi) $cos\frac{\pi }{12}-sin\frac{\pi }{12}=\frac{1}{\sqrt{2}}$
(vii) $sin{80}^{\circ }-cos{70}^{\circ }=cos{50}^{\circ }$
(viii) $sin{51}^{\circ }-cos{81}^{\circ }=cos{21}^{\circ }$

Answer 1

(i) $cos{55}^{\circ }+cos{65}^{\circ }+cos{175}^{\circ }=0cos{175}^{\circ }=-cos5$
substitute above value in the equation we get
$cos55+cos65=cos5$
applying rule $cosA+cosB=2cos$
$\left(\frac{A+B}{2}\right)cos\left(\frac{A-B}{2}\right)cos55+cos65=2cos\left(\frac{65+55}{2}\right)cos\left(\frac{65-55}{2}\right)=2cos60cos5=2\times \frac{1}{2}\times cos5=cos5$

(ii) $sin{50}^{\circ }-sin{70}^{\circ }+sin{10}^{\circ }=0(sin{50}^{\circ }-sin{70}^{\circ })+sin{10}^{\circ }\Rightarrow \left(2sin\left(\frac{{50}^{\circ }-{70}^{\circ }}{2}\right)\right)cos\left(2sin\left(\frac{{50}^{\circ }+{70}^{\circ }}{2}\right)\right)+sin{10}^{\circ }[\because sinc-sinD=2sin\left(\frac{C-D}{2}\right)cos\left(\frac{C+D}{2}\right)]=2isn(-{10}^{\circ })cos{60}^{\circ }+sin{10}^{\circ }=-2sin{10}^{\circ }\times \frac{1}{2}+sin{10}^{\circ }[\because cos{60}^{\circ }=\frac{1}{2}]=0=RHS$

(iii) $cos{80}^{\circ }+cos{40}^{\circ }-cos{20}^{\circ }=0(cos{80}^{\circ }+cos{40}^{\circ })-cos{20}^{\circ }=2cos\Rightarrow \left(\frac{{80}^{\circ }+{40}^{\circ }}{2}\right)cos\left(\frac{{80}^{\circ }-{40}^{\circ }}{2}\right)-cos{20}^{\circ }[\because cosC+cosD=2cos\left(\frac{C+D}{2}\right)cos\left(\frac{C-D}{2}\right)]=2cos{60}^{\circ }+cos{20}^{\circ }-cos{20}^{\circ }=2\times \frac{1}{2}cos{20}^{\circ }-cos{20}^{\circ }=cos{20}^{\circ }-cos{20}^{\circ }=0=RHS$

(iv) $cos{20}^{\circ }+cos{100}^{\circ }+cos{140}^{\circ }=0\Rightarrow (cos{20}^{\circ }+cos{100}^{\circ })+cos{140}^{\circ }=2cos\left(\frac{{20}^{\circ }+{100}^{\circ }}{2}\right)cos\left(\frac{{20}^{\circ }-{100}^{\circ }}{2}\right)+cos{140}^{\circ }[\because cosC+cosD=2cos\left(\frac{C+D}{2}\right)cos\left(\frac{C-D}{2}\right)]=2cos{60}^{\circ }+cos(-{40}^{\circ })+cos{140}^{\circ }=2\times \frac{1}{2}cos{40}^{\circ }+cos{140}^{\circ }[\because cos{60}^{\circ }=\frac{1}{2}]=cos{40}^{\circ }+cos({180}^{\circ }-{40}^{\circ })=cos{40}^{\circ }-cos{40}^{\circ }=0=RHS$

(v) $sin\frac{5\pi }{18}-cos\frac{4\pi }{9}=\sqrt{3}sin\frac{\pi }{9}LHS=sin\frac{5\pi }{18}-cos\frac{4\pi }{9}=sin{50}^{\circ }-cos{80}^{\circ }=sin{50}^{\circ }-sin{10}^{\circ }=2sin\left(\frac{{50}^{\circ }-{10}^{\circ }}{2}\right)cos\left(\frac{{50}^{\circ }-+{10}^{\circ }}{2}\right)=2sin{20}^{\circ }cos{30}^{\circ }=2sin{20}^{\circ }\times \frac{\sqrt{3}}{2}=\sqrt{3}sin\frac{\pi }{9}=RHS$.

(vi) $cos\frac{\pi }{12}-sin\frac{\pi }{12}=\frac{1}{\sqrt{2}}$
Multiplying and dividing by $\sqrt{2}$ on LHS
$=\sqrt{2}(\frac{1}{\sqrt{2}}cos\frac{\pi }{12}-\frac{1}{\sqrt{2}}sin\frac{\pi }{12})$
$=\sqrt{2}(sin\frac{\pi }{4}cos\frac{\pi }{12}-cos\frac{\pi }{4}sin\frac{\pi }{12})$
$[\because \frac{1}{\sqrt{2}}=cos\frac{\pi }{4}=sin\frac{\pi }{4}]$
$=\sqrt{2}\left(sin(\frac{\pi }{4}-\frac{\pi }{12})\right)$
$[\because sin(A-B)=sinAcosB-cosAsinB]$
$=\sqrt{2}\left(\frac{\pi }{6}\right)$

$=\sqrt{2}\times \frac{1}{2}=\frac{1}{\sqrt{2}}=RHS$

(vii) $sin{80}^{\circ }-cos{70}^{\circ }=cos{50}^{\circ }$
$LHS=sin{80}^{\circ }=cos{50}^{\circ }+cos{70}^{\circ }$
Now,
$cosC+cosD=2cos\frac{C+D}{2}cos\frac{C-D}{2}$

$RHS=cos{50}^{\circ }+cos{70}^{\circ }=2cos\left(\frac{{50}^{\circ }+{70}^{\circ }}{2}\right)cos\left(\frac{{50}^{\circ }-{70}^{\circ }}{2}\right)=2cos{60}^{\circ }+cos(-{10}^{\circ })=2\times \frac{1}{2}cos{10}^{\circ }\left[cos\right(-\theta )=cos\theta ]=cos{10}^{\circ }=sin{80}^{\circ }=LHS[\because cos-\theta =cos\theta ]=cos{10}^{\circ }=sin{80}^{\circ }=LHS[\because cos\theta =sin(90-\theta \left)\right]$

(viii) $sin{51}^{\circ }-cos{81}^{\circ }=cos{21}^{\circ }sin{51}^{\circ }=cos{21}^{\circ }-cos{81}^{\circ }RHS=cos{21}^{\circ }-cos{81}^{\circ }=-2sin\left({51}^{\circ }\right)sin(-{30}^{\circ })[\because cosC-cosD=-2sin\frac{C+D}{2}sin\frac{C-D}{2}]=+2sin\left({51}^{\circ }\right)sin(-{30}^{\circ })=2sin{51}^{\circ }\times \frac{1}{2}=sin{51}^{\circ }=LHS$