Find Least Value of Alpha in Following Trigonometric equation?

sami123

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May 8, 2019
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3
Given question is :
sin(5α+θ)=cos(θ−3α)

We are to find the least positive value of α for which above equation holds.
The way I did is as,
sin5αcosθ+cos5αsinθ=cosθcos3α+sinθsin3α

Now for this to be true
sin5α=cos3αsin⁡5α=cos⁡3α
and
cos5α=sin3αcos⁡5α=sin⁡3α

How do I find the value of α that satisfies the above criteria?
 

Subhotosh Khan

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Jun 18, 2007
Messages
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Given question is :
sin(5α+θ)=cos(θ−3α)

We are to find the least positive value of α for which above equation holds.
The way I did is as,
sin5αcosθ+cos5αsinθ=cosθcos3α+sinθsin3α

Now for this to be true
sin5α=cos3αsin⁡5α=cos⁡3α
and
cos5α=sin3αcos⁡5α=sin⁡3α

How do I find the value of α that satisfies the above criteria?
Hint:

sin(Θ) = cos[(2*n+1)/2 * π - Θ]..................... n = 0, 1, 2, 3, .......
 

ksdhart2

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Mar 25, 2016
Messages
1,194
I follow your work up until:

\(\displaystyle \sin(5\alpha) \cos(\theta) + \cos(5\alpha) \sin(\theta) = \cos(\theta) \cos(3\alpha) + \sin(\theta) \sin(3\alpha)\)

But after this, everything goes off the rails. How did you get from the above, true statement to:

\(\displaystyle \sin(5\alpha) = \cos(3\alpha) \sin(⁡5\alpha) = cos(⁡3\alpha)\)

This compound equality has no solutions, meaning something went very very wrong in your workings... :confused:
 

Dr.Peterson

Elite Member
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Nov 12, 2017
Messages
4,433
It appears that each expression at that point was repeated twice, as I have seen sometimes in copying from certain formats.

It's really just
sin(5α)=cos(3α)​
cos(5α)=sin(3α)​

That becomes
cos(π/2 - 5α + 2kπ)=cos(3α)​
cos(5α)=sin(π/2 - 3α + 2kπ)​
 

Dr.Peterson

Elite Member
Joined
Nov 12, 2017
Messages
4,433
I hope you recognized my typo; I forgot to change one sin to cos:
That becomes
cos(π/2 - 5α + 2kπ) = cos(3α)​
cos(5α) = cos(π/2 - 3α + 2kπ)​
Also, I put the "+ 2kπ" there, at Khan's suggestion; it could have been left until the next step, namely removal of the trig functions.
 
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