Question

Prove that: sin A + sin 3 A + sin 5 A + sin 7A / cos A+ cos 3A + cos 5A +cos 7A =tan 4A​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{\dfrac{sin(A)+sin(3A)+sin(5A)+sin(7A)}{cos(A)+cos(3A)+cos(5A)+cos(7A)}}$

$\sf{=\dfrac{sin(A)+sin(7A)+sin(3A)+sin(5A)}{cos(A)+cos(7A)+cos(3A)+cos(5A)}}$

$\sf{=\dfrac{2sin\bigg(\dfrac{7A+A}{2}\bigg)cos\bigg(\dfrac{7A-A}{2}\bigg)+2sin\bigg(\dfrac{5A+3A}{2}\bigg)cos\bigg(\dfrac{5A-3A}{2}\bigg)}{2cos\bigg(\dfrac{7A+A}{2}\bigg)cos\bigg(\dfrac{7A-A}{2}\bigg)+2cos\bigg(\dfrac{5A+3A}{2}\bigg)cos\bigg(\dfrac{5A-3A}{2}\bigg)}}$

$\sf{=\dfrac{2sin(4A)cos(3A)+2sin(4A)cos(A)}{2cos(4A)cos(3A)+2cos(4A)cos(A)}}$

$\sf{=\dfrac{2sin(4A)\{cos(3A)+cos(A)\}}{2cos(4A)\{cos(3A)+cos(A)\}}}$

$\sf{=\dfrac{sin(4A)}{cos(4A)}}$

$\sf{=tan(4A)}$