Question

If A - B = 60° then sin square A + sin square B - SinAsinB =

Answer 1

Given :-

• (A - B) = 60° .

To Find :-

• sin²A + sin²B - sinA*sinB = ?

SOLUTION :-

→ sin²A + sin²B - sinA*sinB

Adding & subtracting sinA*sinB ,

→ sin²A + sin²B - sinA*sinB - sinA*sinB + sinA*sinB

→ sin²A + sin²B - 2sinA*sinB + sinA*sinB

Comparing it with a² + b² - 2ab = (a - b)² ,

→ (sinA - sinB)² + sinA*sinB

using sinC - sinD = 2sin(C-D/2)*cos(C+D/2)

→ [ 2sin(A - B/2)*cos(A+B/2) ]² + sinA*sinB

Putting (A - B)= 60° ,

→ [ 2 * sin30° * cos(A+B/2) ]² + sinA*sinB

→ (2 * 1/2 * cos(A+B/2) )² + sinA*sinB

→ cos²(A+B/2) + sinA*sinB

Now,

→ cos²(A+B/2) + (2*sinA*sinB)/2

using 2*sinA*sinB = cos(A-B) - cos(A+B)

→ cos²(A+B/2) + (1/2) [ cos(A-B) - cos(A+B) ]

→ cos²(A+B/2) + (1/2) [ cos60° - cos(A+B) ]

→ cos²(A+B/2) + (1/2) [ (1/2) - cos(A+B) ]

→ cos²(A+B/2) - cos(A+B)/2 + (1/4)

Now, using cosA = 2cos²(A/2) - 1 , we get,

→ cos²(A+B/2) - (1/2) {2cos²(A+B/2) - 1} + (1/4)

→ cos²(A+B/2) - cos²(A+B/2) - (1/2) + (1/4)

→ (1/4) - (1/2)

→ (1 - 2)/4

→ (-1)/4 (Ans).

(Nice Question).

Answer 2

$\huge{\fbox{\fbox{\bigstar{\mathfrak{\red{Answer}}}}}}$

Sin²A + sin²B - sinA × sinB

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Adding and subtracting

sin²A + sin²B - sinA × sinB + sinA× sinB

sin²A + sin²B - 2sinA × sinB + sinA × sinB

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Comparing it with a² + b² -2ab = ( a - b² )

( sinA - sinB )² + sinA × sinB

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using sinC - sinD = 2sin(C- D/2) × cos( C+ D/2 )

[2sin (A -B/2) × cos(A +B/2)² + sinA × sinB

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Let's put (A - B ) = 60°

➠ [2 × sin30° × cos(A + B/2) ]² + sinA × sinB

➠ (2 × 1/2 × cos(A +B/2) )² + sinA × sinB

➠ cos²(A + B/2) sinA × sinB

➠cos²(A +B/2) + (2 × sinA × sinB )/2

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using 2×sinA×sinB = cos(A-B) - cos (A+B)

➠cos²(A+B/2) + (1/2) [cos(A-B) - cos(A+B)]

➠cos²(A + B/2) + (1/2) [cos60° & cos(A+B)]

➠cos²(A+B/2) +(1/2) [(1/2) - cos(A+B) ]

➠ cos²(A+B/2) - cos(A +B)/2 + (1/4)

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using cosA = 2cos²(A/2) - 1,

➩cos²(A +B/2) -(1/2) {2cos²(A+B/2) - 1} +(1/4)

➩cos²(A+B/2) - cos²(A+B/2) - (1/2) + (1/4)

➩(1/4) - (1/2)

➩(1-2)/4

➩(-1)/4

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