Question

prove that cos 3A +cos 5A + cos7A + cos 15A = 4cos4Acos5Acos6A

Answer 1

2cos[5A+3A/2]cos[5A-3A/2]+2cos[15A+7A/2]cos[15A-7A/2]
=2cos4AcosA+2cos11Acos4A
=2cos4A[cosA+cos11A]
=2cos4A[2cos[11A+A/2]Cos[11A-A/2]
=2cos4A2cos5Acos6A
=4cos4Acos5Acos6A

Answer 2

Answer:

$\cos 3 A + \cos 5 A + \cos 7 A + \cos 15 A = 4 \cos 4 A \cos 5 A \cos 6 A$

Hence proved

Solution:

Given,  

$\cos 3 A + \cos 5 A + \cos 7 A + \cos 15 A = 4 \cos 4 A \cos 5 A \cos 6 A$

Consider L.H.S only,

$\begin{array} { l } { \cos 3 A + \cos 5 A + \cos 7 A + \cos 15 A } \\\\ { = 2 \cos \frac { 3 A + 15 A } { 2 } \cos \frac { 15 A - 3 A } { 2 } + 2 \cos \frac { 7 A + 5 A } { 2 } \cos \frac { 7 A - 5 A } { 2 } } \end{array}$

$\begin{array} { c } { \text { [since, } \cos X + \cos Y = 2 \cos \frac { X + Y } { 2 } \cos \frac { X - Y } { 2 } ] } \\\\ { = 2 \cos \frac { 18 A } { 2 } \cos \frac { 12 A } { 2 } + 2 \cos \frac { 12 A } { 2 } \cos \frac { 2 A } { 2 } } \end{array}$

$\begin{array} { c } { = 2 \cos 9 A \cos 6 A + 2 \cos 6 A \cos A } \\\\ { = 2 \cos 6 A ( \cos 9 A + \cos A ) } \\\\ { = 2 \cos 6 A \left( 2 \cos \frac { 9 A + A } { 2 } \cos \frac { 9 A - A } { 2 } \right) } \end{array}$

$\begin{array} { c } { \left[ \text{ since } , \cos X + \cos Y = 2 \cos \frac { X + Y } { 2 } \cos \frac { X - Y } { 2 } \right] } \\\\ { = 2 \cos 6 A \left( 2 \cos \frac { 10 A } { 2 } \cos \frac { 8 A } { 2 } \right) } \\\\ { = 4 \cos 6 A \cos 5 A \cos 4 A } \end{array}$

$\begin{array} { c } { = 4 \cos 6 A \cos 5 A \cos 4 A = R H S } \\\\ { \therefore L H S = R H S } \\\\ { \cos 3 A + \cos 5 A + \cos 7 A + \cos 15 A = 4 \cos 4 A \cos 5 A \cos 6 A } \end{array}$

Hence proved.