\left. \begin{array} { l } { \operatorname { sin } 10 ^ { \circ } + \o

1.	\left. \begin{array} { l } { \operatorname { sin } 10 ^ { \circ } + \operatorname { sin } 20 ^ { \circ } + \operatorname { sin } 40 ^ { \circ } + \operatorname { sin } 50 ^ { \circ } = \operatorname { sin } 70 ^ { \circ } + \operatorname { sin } 80 ^ { \circ } } \\ { \operatorname { sin } ( 45 ^ { \circ } + A ) + \operatorname { sin } ( 45 ^ { \circ } - A ) = \sqrt { 2 } \operatorname { cos } A } \end{array} \right.
Answer» L.H.S. = 2sin15cos5+2sin45cos5 [ using sin C+sin D= 2sin C+D/2 cos C-D/2 for sin10+sin20 & sin40+sin50] = 2cos5 (sin15+sin45) = 2cos5 (2sin30cos15) [ using sin C+sin D= 2sin C+D/2 cos C-D/2 ] = 2cos5 (2 x 1/2 x cos15) = 2cos5 cos15 R.H.S. = sin70+sin80 = 2sin75cos5 [ using sin C+sin D= 2sin C+D/2 cos C-D/2 ] sin75 = sin(90-15) = cos 15 L.H.S = 2cos5 cos15 R.H.S. = 2cos15 cos5 [ since, sin75 = cos15 ]

\left. \begin{array} { l } { \operatorname { sin } 10 ^ { \circ } + \operatorname { sin } 20 ^ { \circ } + \operatorname { sin } 40 ^ { \circ } + \operatorname { sin } 50 ^ { \circ } = \operatorname { sin } 70 ^ { \circ } + \operatorname { sin } 80 ^ { \circ } } \\ { \operatorname { sin } ( 45 ^ { \circ } + A ) + \operatorname { sin } ( 45 ^ { \circ } - A ) = \sqrt { 2 } \operatorname { cos } A } \end{array} \right.

Answer»

L.H.S. = 2sin15cos5+2sin45cos5 [ using sin C+sin D= 2sin C+D/2 cos C-D/2 for sin10+sin20 & sin40+sin50]

= 2cos5 (sin15+sin45)

= 2cos5 (2sin30cos15) [ using sin C+sin D= 2sin C+D/2 cos C-D/2 ]

= 2cos5 (2 x 1/2 x cos15)

= 2cos5 cos15

R.H.S. = sin70+sin80

= 2sin75cos5 [ using sin C+sin D= 2sin C+D/2 cos C-D/2 ]

sin75 = sin(90-15) = cos 15

L.H.S = 2cos5 cos15

R.H.S. = 2cos15 cos5 [ since, sin75 = cos15 ]