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1.	\operatorname { cos } 2 A \operatorname { cos } 2 B + \operatorname { sin } ^ { 2 } ( A - B ) - \operatorname { sin } ^ { 2 } ( A + B ) = \operatorname { cos } 2 ( A + B )
Answer» 1 answer·Mathematics Answer i) sin²(A + B) - sin²(A - B) = {sin(A + B) + sin(A - B)}{sin(A + B) - sin(A - B)}[Application of a² - b² = (a + b)(a - b)] ii) Applying identity, sin(A + B) + sin(A - B) = 2sin(A)cos(B)and sin(A + B) - sin(A - B) = 2cos(A)sin(B) sin²(A + B) - sin²(A - B) = {2sin(A)cos(B)}{2cos(A)sin(B)} = {2sin(A)cos(A)}{2sin(B)cos(B)} ==> sin²(A + B) - sin²(A - B) = sin(2A)sin(2B) ---------- (1) iii) Left side is: cos(2A)cos(2B) + sin²(A - B) - sin²(A + B) = cos(2A)cos(2B) - {sin²(A + B) - sin²(A - B)} = cos(2A)cos(2B) - sin(2A)sin(2B) [Substituting from eqn. (1) above] = cos(2A + 2B) = Right side Thus, cos(2A)*cos(2B) + sin²(A - B) - sin²(A + B) = cos(2A + 2B) [Proved]

\operatorname { cos } 2 A \operatorname { cos } 2 B + \operatorname { sin } ^ { 2 } ( A - B ) - \operatorname { sin } ^ { 2 } ( A + B ) = \operatorname { cos } 2 ( A + B )

Answer»

1 answer·Mathematics

Answer

i) sin²(A + B) - sin²(A - B) = {sin(A + B) + sin(A - B)}*{sin(A + B) - sin(A - B)}[Application of a² - b² = (a + b)(a - b)]

ii) Applying identity, sin(A + B) + sin(A - B) = 2*sin(A)*cos(B)and sin(A + B) - sin(A - B) = 2*cos(A)*sin(B)

sin²(A + B) - sin²(A - B) = {2*sin(A)*cos(B)}*{2*cos(A)*sin(B)} = {2*sin(A)*cos(A)}*{2*sin(B)*cos(B)}

==> sin²(A + B) - sin²(A - B) = sin(2A)*sin(2B) ---------- (1)

iii) Left side is: cos(2A)*cos(2B) + sin²(A - B) - sin²(A + B)

= cos(2A)*cos(2B) - {sin²(A + B) - sin²(A - B)}

= cos(2A)*cos(2B) - sin(2A)*sin(2B) [Substituting from eqn. (1) above]

= cos(2A + 2B) = Right side

Thus, cos(2A)*cos(2B) + sin²(A - B) - sin²(A + B) = cos(2A + 2B) [Proved]