Prove sin 5θ

N

nasi112

Full Member
Joined
Aug 23, 2020
Messages
617
Prove [MATH]\frac{\sin 5\theta}{\sin \theta} = 16 \cos^{4} \theta - 12 \cos^{2} \theta + 1[/MATH]
I tried all the methods I know. I also watched some videos on youtube but I failed to solve it. I can prove things with [MATH]2\theta[/MATH] easily but [MATH]5\theta[/MATH] never seen like this! If any one can help?
 
Try writing [MATH]\sin(5\theta) = \sin(2\theta+3\theta)[/MATH], or maybe [MATH]\sin(5\theta) = \sin(4\theta+\theta)[/MATH]. Do you have formulas for triple or quadruple angles? If not derive that first, similarly.

Also, it will be helpful if you show your work on at least one method, so we can see if you are starting out in a reasonable way and making some silly error, or just giving up too soon. It may not be obvious that you are near the goal until you reach it!
 
Dr Peterson gave you great advice.

Another method is using deMoive's theorem. You will get that cos(5θ) + i*sin(5θ) = (cos(θ) + i*sin(θ))5. You can multiply out the right hand side hopefully using the binomial theorem and equate the real part to cos(θ) and the imaginary part to sin(θ). That will give you the numerator you need and simply divide that by sin(θ) to get your result.
 
i agree with Jomo

you can use complex numbers to solve this problem

it will be more fun, easier, and straight forward. you don't need to remember any identities or formulas such as double angles, triple angles,...etc

you just need a little knowledge of how de Moivre’s theorem works

try it with binomial theorem, and see how powerful it is
 
The question, of course, is what methods you know. If you haven't learned about the connection of complex numbers to trigonometry, then you aren't ready for this powerful method. This is one reason we ask for information about your background, and for examples of the work you tried, to get clues about what you are capable of and what kinds of hints to give.
 
I studied complex numbers. I will try both methods. I am not familiar with [MATH]3\theta[/MATH], so I will do the other. [MATH]\sin 5\theta = \sin(4\theta + \theta) = \sin 4\theta \cos \theta + \sin \theta \cos 4\theta = 2\sin 2\theta \cos 2\theta \cos \theta + \sin \theta (\cos^2 2\theta - \sin^2 2\theta) = 4 \sin \theta \cos^2 \theta (\cos^2 \theta - \sin^2 \theta) + \sin \theta \cos^2 2\theta - \sin \theta \sin^2 2\theta = 4 \sin \theta \cos^4 \theta - 4 \sin^3 \theta \cos^2 \theta + \sin \theta (\cos^2 \theta - \sin^2 \theta)^2 - \sin \theta (2 \sin \theta \cos \theta)^2 = 4 \sin \theta \cos^4 \theta - 4 \sin^3 \theta \cos^2 \theta + \sin \theta \cos^4 \theta - 2\cos^2 \theta \sin^3 \theta + \sin^5 \theta - 4\sin^3 \theta \cos^2 \theta[/MATH]
[MATH]\frac{\sin 5\theta}{\sin \theta} = 4\cos^4 \theta - 4\sin^2 \theta \cos^2 \theta + \cos^4 \theta - 2\cos^2 \theta \sin^2 \theta + \sin^4 \theta - 4\sin^2 \theta \cos^2 \theta = 4\cos^4 \theta - (4\cos^2 \theta - 4\cos^4 \theta) + \cos^4 \theta - (2\cos^2 \theta - 2\cos^4) + \sin^4 \theta - (4\cos^2 \theta - 4\cos^4 \theta) = 4\cos^4 \theta - 4\cos^2 \theta + 4\cos^4 \theta + \cos^4 \theta - 2\cos^2 \theta + 2\cos^4 + \sin^4 \theta - 4\cos^2 \theta + 4\cos^4 \theta = 15\cos^4 \theta - 10\cos^2 \theta + \sin^4 \theta = 15\cos^4 \theta - 10\cos^2 \theta + (1 - \cos^2 \theta)^2 = 15\cos^4 \theta - 10\cos^2 \theta + 1 - 2\cos^2 \theta + \cos^4 \theta = 16\cos^4 \theta - 12\cos^2 \theta + 1[/MATH]
Thank you so much Dr Peterson. Your way works pretty good:) I want to do the second method which Jomo and Joshuaa told. How to start?
 
[MATH](a + b)^5 = {5 \choose 0} a^5 + {5 \choose 1} a^4 b + {5 \choose 2} a^3 b^2+ {5 \choose 3} a^2 b^3 + {5 \choose 4} a b^4 + {5 \choose 5} b^5 = a^5 + 5a^4 b + 10a^3 b^2+ 10a^2 b^3 + 5a b^4 + b^5[/MATH]
 
[MATH](\cos \theta + i\sin \theta)^5 = {5 \choose 0}\cos^5 \theta + {5 \choose 1} \cos^4 \theta (i\sin \theta) + {5 \choose 2} \cos^3 \theta (i\sin \theta)^2 + {5 \choose 3} \cos^2 \theta (i\sin \theta)^3 + {5 \choose 4} \cos \theta (i\sin \theta)^4 + {5 \choose 5} (i\sin \theta)^5 = \cos^5 \theta + 5\cos^4 \theta (i\sin \theta) - 10\cos^3 \theta (\sin^2 \theta) - 10\cos^2 \theta (i\sin^3 \theta) + 5\cos \theta (\sin^4 \theta) + (i\sin^5 \theta)[/MATH]
Now, I take the imaginary part

[MATH]\sin 5\theta = 5\cos^4 \theta \sin \theta - 10\cos^2 \theta \sin^3 \theta + \sin^5 \theta[/MATH]
[MATH]\frac{\sin 5\theta}{\sin \theta} = 5\cos^4 \theta - 10\cos^2 \theta \sin^2 \theta + \sin^4 \theta = 5\cos^4 \theta - 10\cos^2 \theta + 10\cos^4 \theta + 1 - 2\cos^2 \theta + \cos^4 \theta = 16\cos^4 \theta - 12\cos^2 \theta + 1[/MATH]
Thank you Dr. Peterson, Jomo and Joshuaa. It worked again:)
 
You must log in or register to reply here.
Top