Question

if cos3A+cos5A+cos7A+cod15A=4cosmAcosnAcosPA​

Answer 1

Answer:

Answer:

\cos 3 A + \cos 5 A + \cos 7 A + \cos 15 A = 4 \cos 4 A \cos 5 A \cos 6 Acos3A+cos5A+cos7A+cos15A=4cos4Acos5Acos6A

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 Acos3A+cos5A+cos7A+cos15A=4cos4Acos5Acos6A

Consider L.H.S only,

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

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

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

$$\begin{lgathered}\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

Step-by-step explanation:

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