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I have a problem that says : express sin11pi/12+sin5pi/12 as a product and find the exact value of the product I have no clue aout this problem I was wondering could you work this out and show me how u did it please
We’ll divide these into terms (1) sin(11π/12) and (2) sin(5π/12).
1] Now for the term (1), sin(11π/12), we know that it is the same as
sin(6π/12 + 5π/12) = sin(π/2 + 5π/12), which is equal to sin(π/2)cos(5π/12) + co(π/2)sin(5π/12).
Since sin(π/2) = √3/2 and cos(π/2) = ˝, this becomes
(A) sin(11π/12) = √3cos(5π/12)/2 + sin(5π/12)/2.
2] For term (2), sin(5π/12), we know that sin(5π/12) =
sin(6π/12 - π/12) = sin(π/2 - π/12) =
sin(π/2)cos(π/12) – cos(π/2)sin(π/12)
As was just stated, we know the sin() and cos() of π/2,
so this is the same as (B) = √3cos(π/12)/2 – sin(π/12)/2.
Combining (A) and (B) we get √3cos(5π/12)
since in both terms they are over 2.
The sin()’s disappear since one is positive and one is negative.
This looks like a double angle formula.
It is known that cos˛(Θ) = (1 + cos(2Θ))/2. Using this with Θ=5π/12,
what we have is √3cos(5π/12) = √3*(±√(1 + cos(5π/6))).
Now cos(5π/6) = -cos(π/6), which we know as -√3/2.
Putting that back in (1 + cos(2Θ))/2 leads to
±√3( 1 - √3/2) = ±√3((2 - √3)/2 = ±(2√3 – 3)/2.
This could also be written as √3 – 3/2.
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