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First Derivative Test for Local Maximum

Prove that: s...

Question

Prove that: sinα + sin (α + 2π/3) + sin (α + 4π/3) = 0
Kindly proceed by using sin (A+B) = sinAcosB + cosAsinB (in sin (α + 2π/3) and sin (α + 4π/3)) in first step. :)

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Solution

$We know that : \sin (α + \frac{2 π}{3}) = \sin α \times \cos (\frac{2 π}{3}) + \cos α \times \sin (\frac{2 π}{3}) \Rightarrow \sin (α + \frac{2 π}{3}) = \sin α \times \cos (π - \frac{π}{3}) + \cos α \times \sin (π - \frac{π}{3}) \Rightarrow \sin (α + \frac{2 π}{3}) = \sin α [- \cos (\frac{π}{3})] + \cos α \times \sin (\frac{π}{3}) \Rightarrow \sin (α + \frac{2 π}{3}) = \sin α \times (- \frac{1}{2}) + \cos α \times \frac{\sqrt{3}}{2} \Rightarrow \sin (α + \frac{2 π}{3}) = - \frac{\sin α}{2} + \frac{\sqrt{3}}{2} \cos α \sin (α + \frac{4 π}{3}) = \sin α \times \cos (\frac{4 π}{3}) + \cos α \times \sin (\frac{4 π}{3}) \Rightarrow \sin (α + \frac{4 π}{3}) = \sin α \times \cos (π + \frac{π}{3}) + \cos α \times \sin (π + \frac{π}{3}) \Rightarrow \sin (α + \frac{4 π}{3}) = \sin α [- \cos (\frac{π}{3})] + \cos α \times [- \sin (\frac{π}{3})] \Rightarrow \sin (α + \frac{4 π}{3}) = \sin α \times (- \frac{1}{2}) + \cos α \times (- \frac{\sqrt{3}}{2}) \Rightarrow \sin (α + \frac{4 π}{3}) = - \frac{\sin α}{2} - \frac{\sqrt{3}}{2} \cos α \sin α + \sin (α + \frac{2 π}{3}) + \sin (α + \frac{4 π}{3}) = \sin α - \frac{\sin α}{2} + \frac{\sqrt{3}}{2} \cos α - \frac{\sin α}{2} - \frac{\sqrt{3}}{2} \cos α = 0$

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tan α = \frac{1}{7}, sin β = \frac{1}{\sqrt{10}},

then using the formula

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(α + 2 β)

sin α, {sin}^{2} α, 1, {sin}^{4} α

{sin}^{5} α

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α

lies in the interval-

s i n α = \frac{1}{\sqrt{5}} a n d s i n β = \frac{3}{5}, t h e n β - α

lies in the interval

β

α

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c o s α + c o s (α + β) + c o s (α + β + γ) = 0

s i n α + s i n (α + β) + s i n (α + β + γ) = 0

β = γ = 2 π / 3

$I f \frac{2 s i n α}{1 + c o s α + s i n α} = y, t h e n \frac{1 - c o s α + s i n α}{1 + s i n α} i s e q u a l t o$

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