Question

Prove that sin²b=sin²a+sin²(a-b)-2sinacosb×sin(a-b)​

Answer 1

Step-by-step explanation:

Sin^2 B = sin^2 A + sin^2 (A - B) - 2sinA.cosB.sin(A - B)

Where;

Sin(A - B) = sinA.cosB - cosA.sinB,

We have;

sin^2 A + sin^2 (A - B) - 2sinA.cosB.sin(A - B)

= sin^2 A + [sinA.cosB - cosA.sinB]^2 - 2sinA.cosB [sinA.cosB - cosA.sinB]

= sin^2 A + sin^2 A.cos^2 B - 2sinA.cosB.cosA.sinB + cos^2 A.sin^2 B - 2sin^2 A.cos^2 B + 2sinA.cosB.cosA.sinB

= sin^2 A + sin^2 A.cos^2 B + cos^2 A.sin^2 B - 2sin^2 A.cos^2 B

= sin^2 A - sin^2 A.cos^2 B + cos^2 A.sin^2 B

= sin^2 A [1 - cos^2 B] + cos^2 A.sin^2 B

Where;

Sin^2 B + cos^2 B = 1

1 - cos^2 B = sin^2 B, we have;

= sin^2 A.sin^2 B + cos^2 A.sin^2 B

= sin^2 B [sin^2 A + cos^2 A]

= sin^2 B [ 1 ]

= sin^2 B.

Hence, proved

Answer 2

Answer:

Step-by-step explanation:

sin^2 b = sin^2 a + sin^2 (a - b) - 2sin a x cos b x sin(a - b)

sin(a - b) = sin a x cos b - cos a x sin b

sin^2 a + sin^2 (a - b) - 2sin a x cos b x sin(a - b)

= sin^2 a + [sin a x cos b - cos a x sin b]^2 - 2sin a x cos b [sina x cos b - cosa x               sin b ]

= sin^2 a + sin^2 a x cos^2 b - 2sin a x cos b x cos a x sin b + cos^2 a x sin^2 b - 2sin^2 a x cos^2 b + 2sin a x cos b x cos a x sin b

= sin^2 a + sin^2 a x cos^2 b + cos^2 a x sin^2 b - 2sin^2 a x cos^2 b

= sin^2 a - sin^2 a x cos^2 b + cos^2 a x sin^2 b

= sin^2 a[1 - cos^2 b] + cos^2 a x sin^2 b

sin^2 b + cos^2 b= 1

1 - cos^2 b = sin^2 b

= sin^2 a x sin^2 b + cos^2 a x sin^2 b

= sin^2 b [sin^2 a + cos^2 a]

= sin^2 b [ 1 ]

= sin^2 b

Hope it helps you  :)