Question

Trigonometry proof problem

Answer 1

Prove
\[\frac{ \sin A + \sin 3A + \sin 5A }{ \cos A + \cos3A + \cos5A } = \tan 3A\]

Answer 2

@hartnn
@whpalmer4

Answer 3

did you try
sin x + sin y formula/property ?
sin A+sin 5A =.... ?

Answer 4

cos A +cos 5A =.... ?

Answer 5

SinC + SinD = 2Sin[(C+D)/2]Cos[(C-D)/2]
CosC + CosD = 2Cos[(C+D)/2]Cos[(C-D)/2]

Answer 6

no

Answer 7

SinC + SinD = 2Sin[(C+D)/2]Cos[(C-D)/2]
sin A+sin 5A = 2sin 3A cos 2A
CosC + CosD = 2Cos[(C+D)/2]Cos[(C-D)/2]
cos A+cos 5A = 2cos 3A cos 2A

notice how cos 2A is common ....

Answer 8

\[\frac{ A+5A }{ 2} \]
can become 6A/2 ?

Answer 9

yes,
\(\large \frac{ \sin A + \sin 3A + \sin 5A }{ \cos A + \cos3A + \cos5A }\\ \large =\frac{ \sin 3A + 2\sin 3A \cos 2A }{ 2\cos 3A \cos 2A+ \cos3A } \)
factor out sin 3A from numerator and cos 3A from denominator and you are done.

Answer 10

@L.C hehe

Answer 11

Lol

Answer 12

what ?