Help with proving a trigonometric identity

AI Thread Summary
The discussion focuses on proving the trigonometric identity (Sin[A] Sin[2 A] + Sin[3 A] Sin[6 A])/(Sin[A] Cos[2 A] + Sin[3 A] Cos[6 A]) is identical to tan[5A]. The user initially struggles with the problem but suggests using factor formulas. After some exploration, they successfully manipulate the equation by multiplying and applying factor identities, ultimately simplifying it to sin[5A]/cos[5A], confirming it equals tan[5A]. The user expresses satisfaction in solving the identity, highlighting the rewarding nature of mathematical problem-solving.
kenshaw93
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Hi I've got this problem which has really been bothering me.

How are you supposed to prove that:
(Sin[A] Sin[2 A] + Sin[3 A] Sin[6 A])/(Sin[A] Cos[2 A] + Sin[3 A] Cos[6 A]) is identicle to tan[5A].

I am almost sure that I've got to use the factor formulae, but I've had no luck. Maybe someone can help me please :)
Thank you for your time!
 
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kenshaw93 said:
Hi I've got this problem which has really been bothering me.

How are you supposed to prove that:
(Sin[A] Sin[2 A] + Sin[3 A] Sin[6 A])/(Sin[A] Cos[2 A] + Sin[3 A] Cos[6 A]) is identicle to tan[5A].

I am almost sure that I've got to use the factor formulae, but I've had no luck. Maybe someone can help me please :)
Thank you for your time!

Try to "de-factor".
For instance, Sin(A)Sin(2A)=Sin(2A)Sin(A).
Recalling the identity that Cos S + Cos T = ?
 
Ok i found out how to work it out. i first multiply by 2 throughout. then i used the factor formulae until i end up with the terms :

2sin5A sin4A / 2sin4Acos5A which is identicle to sin5A/cos5A which is identicle to tan5A...PROVED!

The satisfaction of proving an identity or equation is rather incredible!
 
yea! Great sense of satisfaction. you are turning 17 this year?
 
haha indeed i am, still young i guess, just scratching the surface of mathematics. The subject really gets you thinking
 
The working out suggests first equating ## \sqrt{i} = x + iy ## and suggests that squaring and equating real and imaginary parts of both sides results in ## \sqrt{i} = \pm (1+i)/ \sqrt{2} ## Squaring both sides results in: $$ i = (x + iy)^2 $$ $$ i = x^2 + 2ixy -y^2 $$ equating real parts gives $$ x^2 - y^2 = 0 $$ $$ (x+y)(x-y) = 0 $$ $$ x = \pm y $$ equating imaginary parts gives: $$ i = 2ixy $$ $$ 2xy = 1 $$ I'm not really sure how to proceed from here.
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