Question

Value of

$( \dfrac{1 + sin \dfrac{2x}{9} + cos \dfrac{2x}{9} }{1 + sin \dfrac{2x}{9} - cos \dfrac{2x}{9} } ) ^{3} is$
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Answer 1

we have to find the value of [{1 + sin(2π/9) + i cos(2π/9)}/{1 + sin(2π/9) - i cos(2π/9)}]³

solution : sin(2π/9) = sin(π/2 - 5π/18) = cos(5π/18)

sin(2π/9) = sin(π/2 - 5π/18) = cos(5π/18) similarly, cos(2π/9) = sin(5π/18)

so, [ {1 + sin(2π/9) + i cos(2π/9)}/{1 + sin(2π/9) - i cos(2π/9)} ]³

= [{1 + cos(5π/18) + i sin(5π/18)}/{1 + cos(5π/18) - i sin(5π/18)} ]³

we know, 1 + cos2A = 2cos²A

we know, 1 + cos2A = 2cos²A and sin2A = 2sinA cosA

we know, 1 + cos2A = 2cos²A and sin2A = 2sinA cosA so 1 + cos(5π/18) = 2cos²(5π/36)

we know, 1 + cos2A = 2cos²A and sin2A = 2sinA cosA so 1 + cos(5π/18) = 2cos²(5π/36) sin(5π/18) = 2sin(5π/36) cos(5π/36)

= [{2cos²(5π/36) + 2i sin(5π/36) cos(5π/36)}/{2cos²(5π/36) - 2i sin(5π/36) cos(5π/36)}]³

= [{(cos(5π/36) + i sin(5π/36)}/{cos(5π/36) - i sin(5π/36)}]³

we know, if (sinA + icosA)/(sinA - icosA) = (sinA + icosA)²

= [{cos(5π/36) + i sin(5π/36)}²]³

= [cos(5π/36) + i sin(5π/36)]⁶

we know, (cosA + isinA)ⁿ = cosnA + i sinnA

= cos(5π/6) + i sin(5π/6)

= -√3/2 + i × 1/2

= i/2 - √3/2

Therefore the value of [{1 + sin(2π/9) + i cos(2π/9)}/{1 + sin(2π/9) - i cos(2π/9)} ]³ = -i/2 + √3/2