Question

Prove:-
[sin(B+C)] × [sin(B-C)] = sin²(B) - sin²(C)​

Answer 1

Given to prove :-

sin ( B + C ) × sin ( B - C ) = sin²B - sin²C

Formulae to know :-

sin ( B + C ) = sinBcosC + sinC cosB

sin ( B - C ) = sinB cosC - sinC cosB

( a + b ) ( a - b ) = a² - b²

sin²A = 1- cos²A

Solution :-

sin ( B + C ) × sin ( B - C )

= (sinBcosC + sinC cosB) ×( sinB cosC - sinC cosB)

It is in form of( a + b ) ( a - b ) = a² - b²

So,

= (sinBcosC + sinC cosB) ×( sinB cosC - sinC cosB)

= sin²B cos²C - sin²C cos²B

As we are observing RHS The RHS is in interms of sinB , sinC

So, lets convert interms of sinB, sinC

= sin²B cos²C - sin²C cos²B

= sin²B ( 1 - sin²C ) - sin²C ( 1 - sin²B )

= sin²B - sin²B sin²C - sin²C + sin²Bsin²C

Keep like terms together

= sin²B - sin²B sin²C + sin²B sin²C - sin²C

+ sin²Bsin²C - sin²Bsin²C get cancelled

= sin²B - sin²C

So,

sin ( B + C ) × sin ( B - C ) = sin²B - sin²C

Hence proved !

Know more:-

Trigonmetric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

csc²θ - cot²θ = 1

Trigonmetric relations

sinθ = 1/cscθ

cosθ = 1 /secθ

tanθ = 1/cotθ

tanθ = sinθ/cosθ

cotθ = cosθ/sinθ

Trigonmetric ratios

sinθ = opp/hyp

cosθ = adj/hyp

tanθ = opp/adj

cotθ = adj/opp

cscθ = hyp/opp

secθ = hyp/adj