Question

Prove that
$\frac{ \sin( \gamma ) + \sin( 3\gamma ) + \sin( 5\gamma ) + \sin( 7 \gamma ) }{ \cos( \gamma ) + \cos(3 \gamma ) + \cos(5 \gamma ) + \cos(7 \gamma ) } = \tan(4 \gamma )$
​

Answer 1

Step-by-step explanation:

Solution :-

On taking LHS

(Sin ɣ + Sin 3 ɣ + Sin 5ɣ+ Sin 7ɣ) /

(Cos ɣ+Cos 3ɣ+Cos 5 ɣ + Cos 7 ɣ)

It can be rearranged as

=>[ (Sin ɣ + Sin 7 ɣ )+( Sin 3ɣ+ Sin 5ɣ) ]/

[(Cos ɣ+Cos 7ɣ)+(Cos 3ɣ + Cos 5 ɣ)]

=> [ (Sin 7ɣ + Sin ɣ )+( Sin 5ɣ+ Sin 3ɣ) ]/

[(Cos 7ɣ+Cos ɣ)+(Cos 5ɣ + Cos 3ɣ)]

We know that

Sin C + Sin D = 2 Sin (C+D)/2 Cos (C-D)/2

and

Cos C+Cos D = 2 Cos (C+D)/2 Cos(C-D)/2

Now

Sin 7 ɣ + Sin ɣ

=> 2 Sin (7 ɣ+ɣ )/2 Cos (7 ɣ -ɣ)/2

=> 2 Sin (8 ɣ)/2 Cos (6 ɣ)/2

=> 2 Sin 4 ɣ Cos 3 ɣ

and

Sin 5ɣ + Sin 3 ɣ

=> 2 Sin (5 ɣ+3ɣ )/2 Cos (5ɣ -3ɣ)/2

=> 2 Sin (8 ɣ)/2 Cos (2ɣ)/2

=> 2 Sin 4 ɣ Cos ɣ

Now,

(Sin 7 ɣ + Sin ɣ )+(Sin 5ɣ + Sin 3 ɣ )

=> 2 Sin 4 ɣ Cos 3 ɣ+2 Sin 4 ɣ Cos ɣ

=> 2 Sin 4 ɣ (Cos 3ɣ+Cos ɣ) ---------------(1)

and

Cos 7 ɣ + Cos ɣ

=> 2 Cos (7 ɣ+ɣ )/2 Cos (7 ɣ -ɣ)/2

=> 2 Cos (8 ɣ)/2 Cos (6 ɣ)/2

=> 2 Cos 4 ɣ Cos 3 ɣ

and

Cos 5ɣ + Cos 3 ɣ

=> 2 Cos (5 ɣ+3ɣ )/2 Cos (5ɣ -3ɣ)/2

=> 2 Cos (8 ɣ)/2 Cos (2ɣ)/2

=> 2 Cos 4 ɣ Cos ɣ

Now,

(Cos 7 ɣ + Cos ɣ )+(Cos5ɣ + Cos 3 ɣ )

=> 2 Cos 4 ɣ Cos 3 ɣ+2 Cos4 ɣ Cos ɣ

=> 2 Cos 4 ɣ (Cos 3ɣ+Cos ɣ) -----------(2)

Now

From (1)&(2)

We have

=>[ 2 Sin 4 ɣ (Cos 3ɣ+Cos ɣ)] /

[2 Cos 4 ɣ (Cos 3ɣ+Cos ɣ)]

On cancelling (Cos 3ɣ+Cos ɣ) then

=> 2 Sin 4 ɣ / 2 Cos 4 ɣ

=> Sin 4 ɣ / Cos 4 ɣ

=> Tan 4 ɣ

Since Tan A = Sin A / Cos A

=> RHS

LHS = RHS

(Sin ɣ + Sin 3 ɣ + Sin 5ɣ+ Sin 7ɣ) /

(Cos ɣ+Cos 3ɣ+Cos 5ɣ + Cos 7ɣ) =Tan 4ɣ

Hence, Proved.

Used formulae:-

→ SinC + SinD = 2 Sin (C+D)/2 Cos (C-D)/2

→ CosC+CosD = 2 Cos(C+D)/2 Cos(C-D)/2

→ Tan A = Sin A / Cos A

→ ɣ = gamma (Greek alphabet )