Prove that: (i)sin2 B = sin2 A + sin^2 (A-B) – 2sin A cos B sin (A

Prove that: (i)sin2 B = sin2 A + sin^2 (A-B) – 2sin A cos B sin (A – B)

Prove that: (i)sin2 B = sin2 A + sin^2 (A-B) – 2sin A cos B sin (A – B)

(ii)

RD Sharma Solutions for Class 11 Maths Chapter 7 – Values of Trigonometric Functions at Sum or Difference of Angles image- 12

Solution:

(i)

$si{{n}^{2}}B=si{{n}^{2}}A+si{{n}^{2}}\left( A-B \right)2sin\text{}A\text{}cosB\text{}sin\left(A\text{}-\text{}B\right)$

Let us consider RHS:

$si{{n}^{2}}A\text{}+\text{}si{{n}^{2}}\left(A\text{}-B\right)\text{}\text{}2\text{}sin\text{}A\text{}cos\text{}B\text{}sin\text{}\left(A\text{}-\text{}B\right)$

$si{{n}^{2}}A\text{ }+\text{ }sin\text{ }\left( A\text{}-B \right)\text{}\left[sin\text{}\left(A\text{}-B \right)\text{}\text{}2\text{}sin\text{}A\text{}cos\text{}B\right]$

Since,

$sin\text{ }\left( A-\text{ }B \right)\text{}=\text{}sin\text{}A\text{}cos\text{}B\text{}-\text{}cos\text{}A\text{ }sin\text{ }B$

Therefore,

$si{{n}^{2}}A+sin\left(A-B\right)\left[sinAcosB-cosAsinB-2sinAcos\text{}B\right]$

$si{{n}^{2}}A\text{ }+\text{ }sin\text{ }\left( A\text{}-B\right)\text{}\left[-sin\text{}A\text{}cos\text{}B\text{}-\text{}cos\text{}A\text{}sin\text{}B \right]$

$si{{n}^{2}}A\text{}-\text{}sin\text{}\left(A\text{}-B\right)\text{}\left[sin\text{}A\text{}cos\text{}B\text{}+\text{}cos\text{}A\text{}sin\text{}B\right]$