\operatorname { cos } 2 \theta \cdot \operatorname { cos } \frac { \th

1.	\operatorname { cos } 2 \theta \cdot \operatorname { cos } \frac { \theta } { 2 } - \operatorname { cos } 3 \theta \cdot \operatorname { cos } \frac { 9 \theta } { 2 } = \operatorname { sin } 5 \theta \cdot \operatorname { sin } \frac { 5 \theta } { 2 }
Answer» multiply and divide L.H.S. by 2. u'll get 1/2[2cos2x.cosx/2 - 2cos3x.cos9x/2] Now , 2cosAcosB = cos(A+B) + cos(A-B). so, LHS= 1/2[cos5x/2 + cos3x/2 - [cos15x/2 + cos3x/2] =1/2[cos5x/2 - cos15x/2] by using cosC - cosD= 2.sin(C+D)/2.sin(D-C)/2 therefore, LHS= 1/2[2.sin20x/4.sin10x/4] =sin5x.sin5x/2= RHS.

\operatorname { cos } 2 \theta \cdot \operatorname { cos } \frac { \theta } { 2 } - \operatorname { cos } 3 \theta \cdot \operatorname { cos } \frac { 9 \theta } { 2 } = \operatorname { sin } 5 \theta \cdot \operatorname { sin } \frac { 5 \theta } { 2 }

Answer»

multiply and divide L.H.S. by 2. u'll get 1/2[2cos2x.cosx/2 - 2cos3x.cos9x/2]

Now , 2cosAcosB = cos(A+B) + cos(A-B).

so, LHS= 1/2[cos5x/2 + cos3x/2 - [cos15x/2 + cos3x/2]

=1/2[cos5x/2 - cos15x/2]

by using cosC - cosD= 2.sin(C+D)/2.sin(D-C)/2

therefore, LHS= 1/2[2.sin20x/4.sin10x/4]

=sin5x.sin5x/2= RHS.