Question :
If 2 cos α = x + (1/x) and 2 cos β = y + (1/y), show that
(i) (x/y) + (y/x) = 2 cos (α - β)
(ii) xy - (1/xy) = 2i sin (α + β)
(iii) (xm/yn) - (yn/xm) = 2i sin (mα - nβ)
(iv) (xmyn) + 1/(xmyn) = 2 cos (mα + nβ)
Solution :
2 cos α = x + (1/x)
x2 + 1 = (2 cos α) x
x2 - (2 cos α) x + 1 = 0
Solving for x, we get
= [-b + √(b2 - 4ac)] / 2a
= [(2 cos α) + √((2 cos α)2 - 4(1)] / 2(1)
= [(2 cos α) + √-4 (1 - cos2α) / 2
= (2 cos α + i 2 sin α) / 2
x = cos α + i sin α
2 cos β = y + (1/y)
y2 + 1 = (2 cos β) y
y2 - (2 cos β) y + 1 = 0
Solving for y, we get
= [-b + √(b2 - 4ac)] / 2a
= [(2 cos β) + √((2 cos β)2 - 4(1)] / 2(1)
= [(2 cos β) + √-4 (1 - cos2β) / 2
= (2 cos β + i 2 sin β) / 2
y = cos β + i sin β
(i) (x/y) + (y/x) = 2 cos (α - β)
xy-1 = (cos α + i sin α)(cos (-β) + i sin (-β))
(x/y) = cos (α - β) + i sin (α - β)
(y/x) = cos (α - β) - i sin (α - β)
By adding, we get
(x/y) + (y/x) = 2 cos (α - β)
(ii) xy - (1/xy) = 2i sin (α + β)
xy = (cos α + i sin α)(cos β + i sin β)
= cos (α + β) + i sin (α + β)
1/xy = = cos (α + β) - i sin (α + β)
xy - (1/xy)
= [cos (α + β) + i sin (α + β)] - [cos (α + β) - i sin (α + β)]
= -2i sin (α + β)
(iii) (xm/yn) - (yn/xm) = 2i sin (mα - nβ)
x = cos α + i sin α
xm = (cos α + i sin α)m
xm = (cos mα + i sin mα)
y = cos β + i sin β
yn = (cos β + i sin β)n
yn = (cos nβ + i sin nβ)
(xm/yn) = cos (mα - nβ) + i sin (mα - nβ)
(yn/xm) = cos (mα - nβ) - i sin (mα - nβ)
(xm/yn) - (yn/xm) = -2i sin (mα - nβ)
(iv) (xmyn) + 1/(xmyn) = 2 cos (mα + nβ)
xm = (cos mα + i sin mα)
yn = (cos nβ + i sin nβ)
xmyn = cos (mα + nβ) + i sin (mα + nβ)
1/(xmyn) = cos (mα + nβ) - i sin (mα + nβ)
(xmyn) + 1/(xmyn) = 2cos (mα + nβ)
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