Math, asked by T0UGH, 1 month ago

answer it fast....as soon as possible..​

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Answered by Anonymous
13

(cosα - cosβ)² + (sinα - sinβ)² = 4sin²(α-β/2)

  • Let's prove from LHS

→ (cosα - cosβ)² + (sinα - sinβ)²

  • Apply identity
  • (a - b)² = + - 2ab

→ cos²α + cos²β - 2cosαcosβ + sin²α + sin²β - 2sinαsinβ

  • Rearranging

→ (cos²α + sin²α) + (cos²β + sin²β) - 2cosαcosβ - 2sinαsinβ

  • cos² ∅ + sin²∅ = 1

→ cos²α + sin²α + cos²β + sin²β - 1 [2cosαcosβ + 2sinαsinβ]

  • Take - 1 as a common
  • 2cosxcosy = cos(x + y) + cos(x - y)
  • 2sinxsiny = cos(x - y) - cos(x + y)

→ 1 + 1 - 1[cos(α + β) + cos(α - β) + cos(α - β) - cos(α + β)]

→ 2 - 1[cos(α + β) - cos(α + β) + cos(α - β) + cos(α - β)]

→ 2 - 1[2cos(α - β)]

  • Take 2 as a common

→ 2 - 2cos(α - β)

→ 2[1 - cos(α - β)]

  • cos2x = 1 - 2sin²x
  • cosx = 1 - 2sin²x/2
  • 1 - cosx = 2sin²x/2

→ 2[2sin²(α - β)/2]

→ 4sin²(α - β)/2 = RHS

  • Hence, LHS = RHS proved
Answered by MichWorldCutiestGirl
128

QuEsTiOn,

  • In the Attachment!

(cos alpha - cos beta)²+ (sin alpha - sin beta)²= 4sin²(alpha - beta / 2)

Let's prove from LHS

→ (cosa - cosß)² + (sina - sinß)²

SoLuTiOn,

• Apply identity • (a - b)² = a² + b² - 2ab

→ cos²a + cos²B - 2cosacosß + sin³a + sin²ß - 2sinasinß

Rearranging

→ (cos²a + sin²a) + (cos²B + sin²B) - 2cosacosß - 2sinasinß

• cos² Ø + sin²0 = 1

→ cos²a + sin²a + cos²ß + sin²3 -1 [2cosacosß +2sinasinß]

Take 1 as a common

  • 2cosxcos y = cos(x + y) + cos(x - y)

  • 2sinxsiny = cos(x - y) - cos(x + y)

1+1-1[cos(a + 3) + cos(a - B) + cos(a - B) -

cos(a + B)]

→ 2-1[cos(a + B) cos(a + B) + cos(a - B) +

cos(a - b)]

→ 2-1[2cos(a - B)]

. Take 2 as a common

→ 2-2cos(a - ß)

→ 2[1- cos(a - B)]

• cos 2x = 1 - 2sin² x • cosx = 1 - 2sinx/2 • 1 - cos x = 2sin² x / 2

2[2sin²(a - B)/2]

4sin²(alpha - beta) / 2 = RHS

FiNaL AnSwEr,

  •  \color{blue} \boxed{ \huge \sf \: LHS \:  RHS  \: proved}

Hope you get your AnSwEr.

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