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Answers
(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 = a² + b² - 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
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,
Hope you get your AnSwEr.