Question

Prove that :
(i) $sin{38}^{\circ }+sin{22}^{\circ }=sin{82}^{\circ }$
(ii) $cos{100}^{\circ }+cos{20}^{\circ }=cos{40}^{\circ }$
(iii) $sin{50}^{\circ }+sin{10}^{\circ }=cos{20}^{\circ }$
(iv) $sin{23}^{\circ }+sin{37}^{\circ }=cos{7}^{\circ }$
(v) $sin{105}^{\circ }+cos{105}^{\circ }=cos{45}^{\circ }$
(vi) $sin{40}^{\circ }+sin{20}^{\circ }=cos{10}^{\circ }$

Answer 1

(i) $sin{38}^{\circ }+sin{22}^{\circ }=sin{82}^{\circ }\because sinC+sinD=2sin\frac{C+D}{2}cos\frac{C-D}{2}\Rightarrow sin{38}^{\circ }+sin{22}^{\circ }=2sin\frac{{60}^{\circ }}{2}cos\frac{{16}^{\circ }}{2}=2sin{30}^{\circ }cos{8}^{\circ }=2\times \frac{1}{2}cos{8}^{\circ }=cos(90-82{)}^{\circ }=sin{82}^{\circ }=RHS[\because cos\theta =sin(90-\theta )]$

(ii) $cos{100}^{\circ }+cos{20}^{\circ }=cos{40}^{\circ }[\because cosC+cosD=2cos\frac{C+D}{2}cos\frac{C-D}{2}]\Rightarrow 2cos\frac{({100}^{\circ }+{20}^{\circ })}{2}cos\frac{({100}^{\circ }-{20}^{\circ })}{2}=2cos{60}^{\circ }cos{40}^{\circ }=2\times \frac{1}{2}cos{40}^{\circ }[\because cos{60}^{\circ }=\frac{1}{2}]=cos{40}^{\circ }=RHS$.

(iii) $sin{50}^{\circ }+sin{10}^{\circ }=cos{20}^{\circ }LHS=sin{50}^{\circ }+sin{10}^{\circ }[\because sinC+sinD=2sin\frac{C+D}{2}cos\frac{C-D}{2}]sin{50}^{\circ }+sin{10}^{\circ }=2sin\frac{{60}^{\circ }}{w}cos{20}^{\circ }=2sin{30}^{\circ }cos{20}^{\circ }=2\times \frac{1}{2}cos{20}^{\circ }=cos{20}^{\circ }=RHS[\because sin{30}^{\circ }\frac{1}{2}]$

(iv) $sin{23}^{\circ }+sin{37}^{\circ }=cos{7}^{\circ }LHS=sin{23}^{\circ }+sin{37}^{\circ }=2sin\left(\frac{{23}^{\circ }+{37}^{\circ }}{2}\right)cos\left(\frac{{23}^{\circ }+{37}^{\circ }}{2}\right)[\because sinC+sinD=2sin\frac{C+D}{2}cos\frac{C-D}{2}]=2sin\left(30{)}^{\circ }cos\right(-7{)}^{\circ }=2\times \frac{1}{2}cos7=cos{7}^{\circ }=RHS[\because cos(-\theta )=cos\theta ,sin{30}^{\circ }=\frac{1}{2}]$

(v) $sin{105}^{\circ }+cos{105}^{\circ }=cos{45}^{\circ }LHS=sin{105}^{\circ }+cos{105}^{\circ }=sin{105}^{\circ }+cos({90}^{\circ }+{15}^{\circ })=sin{105}^{\circ }-sin{15}^{\circ }=2sin\left(\frac{{105}^{\circ }-{15}^{\circ }}{2}\right)cos\left(\frac{{105}^{\circ }+{15}^{\circ }}{2}\right)=2sin{45}^{\circ }cos{60}^{\circ }=2\frac{1}{\sqrt{2}}\times \frac{1}{2}=\frac{1}{\sqrt{2}}=cos{45}^{\circ }=RHS$

(vi) $sin{40}^{\circ }+sin{20}^{\circ }=cos{10}^{\circ }LHS=sin{40}^{\circ }+sin{20}^{\circ }=2sin\left(\frac{{40}^{\circ }+{20}^{\circ }}{2}\right)cos\left(\frac{{40}^{\circ }-{20}^{\circ }}{2}\right)[\because sinC+sinD=2sin\frac{C+D}{2}cos\frac{C-D}{2}]=2sin{30}^{\circ }cos{10}^{\circ }=2\times \frac{1}{2}cos{10}^{\circ }=cos{10}^{\circ }=RHS[\because sin{30}^{\circ }=\frac{1}{2}]$