Question

prove that 2sin7A*cosA=sin9A+sin5A

Answer 1

Answer:

tan 5A

Explanation:

Here, the given equation is

\dfrac{sin A + sin 5A + sin 9 A}{cos A + cos 5A + cos 9A}

we know the formula of sin C + sin D and sin C + sin D = sin \dfrac{C+D}{2} cos \dfrac{C-D}{2} \\and\\cos C + cos D = cos \dfrac{C+D}{2} cos \dfrac{C-D}{2} \\

apply the formula in the given equation

\dfrac{sin A + sin 5A + sin 9 A}{cos A + cos 5A + cos 9A}[tex]=\dfrac{sin A + sin 9A + sin 5A}{cos A + cos 9A + cos 5A}\\\\=\dfrac{sin\frac{(9A+A)}{2}cos\frac{(9A-A)}{2} + sin5A }{cos\frac{(9A+A)}{2}cos\frac{(9A-A)}{2} + cos5A } \\=\dfrac{sin5A cos 8A+sin 5A}{cos5Acos8A+cos5A} \\=\dfrac{sin5A(cos8A+1)}{cos5A(cos8A+1)} \\=\dfrac{sin5A}{cos5A} \\=tan5A

Hence proved.

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

$\blacksquare \:$FORMULA TO BE IMPLEMENTED

$\displaystyle \: 2 \: sinA \: cos \: B = sin \: {( A + B ) } + \: sin \: {( A - B ) }$

$\blacksquare \:$CALCULATION

$2 \: sin7A \: cosA$

$\displaystyle \: \: = \: sin \: {(7A + A) } \: + \: sin \: {(7A - A) } \: \:$

$= sin \: 8A \: + sin \: 6A$