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 #2
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Dec 12, 2018
 #8
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+9

General Solution - Trigonometry

Sin Ø  - Sin 2Ø = Sin 4Ø - Sin 3Ø

 

sin(ϕ)sin(2ϕ)=sin(4ϕ)sin(3ϕ)orsin(3ϕ)+sin(ϕ)=sin(4ϕ)+sin(2ϕ)

 

Formula:

sin(3ϕ)+sin(ϕ)=2sin(2ϕ)cos(ϕ)=sin(4ϕ)+sin(2ϕ)=2sin(3ϕ)cos(ϕ)2sin(2ϕ)cos(ϕ)=2sin(3ϕ)cos(ϕ)sin(2ϕ)cos(ϕ)=sin(3ϕ)cos(ϕ)sin(3ϕ)cos(ϕ)sin(2ϕ)cos(ϕ)=0cos(ϕ)(sin(3ϕ)sin(2ϕ))=0

 

Formula:

cos(ϕ)(sin(3ϕ)sin(2ϕ)=2cos(52ϕ)sin(12ϕ))=0cos(ϕ)2cos(52ϕ)sin(12ϕ)=0|:2cos(ϕ)cos(52ϕ)sin(12ϕ)=0

 

General Solution:

1.cos(ϕ)=0ϕ=2nπ±arccos(0)ϕ=2nπ±π2ϕ=(2n1)π2

2.cos(52ϕ)=052ϕ=2nπ±arccos(0)52ϕ=2nπ±π2|25ϕ=45nπ±π5ϕ=(2n1)π5

3.sin(12ϕ)=012ϕ=nπ+(1)narcsin(0)12ϕ=nπ+(1)nπ|2ϕ=2nπ+(1)n2πϕ=2nπ

 

laugh

Dec 12, 2018
 #1
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0
Dec 12, 2018
 #1
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0
Dec 12, 2018
 #3
avatar+6251 
+1
Dec 12, 2018
 #6
avatar+26396 
+9
Dec 12, 2018
 #4
avatar+26396 
+9

Hello CPhill,

"how was that method derived":

 

1.)

This is the Legendre Theorem from 1808:

Source: Legendre's Theorem - The Prime Factorization of Factorials https://www.cut-the-knot.org/blue/LegendresTheorem.shtml

 

 

2.)

There is a second nice Theorem from Legendre to solve this problem:

Source: Legendre's Theorem - The Prime Factorization of Factorials https://www.cut-the-knot.org/blue/LegendresTheorem.shtml

 

Example:

The prime factorization of 109!  So n=109

What is the exponent of 3?  So p=3
109 in base 3:10910=110013

 

The exponent of 3 is:

The exponent of 3==n109(sum of all the digits in the expansion of n in base p1+1+0+0+1)3=p1The exponent of 3=10932The exponent of 3=1062The exponent of 3=53

 

laugh

Dec 12, 2018

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