sin(A-B)sin(A+B)=cos^2B-cos^2A
Answers
Answered by
64
sin(A+B) sin(A-B)
= (sinA cosB + cosA sinB) (sinA cosB - cosA sinB)
= (sinA cosB)^2 - (cosA sinB)^2
= sin²A cos²B - cos²A sin²B
= (1 - cos²A) cos²B - cos²A (1 - cos²B)
= cos²B - cos²A cos²B - cos²A + cos²A cos²B
= cos²B - cos²A
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Plz mark the answer as brainly answer
= (sinA cosB + cosA sinB) (sinA cosB - cosA sinB)
= (sinA cosB)^2 - (cosA sinB)^2
= sin²A cos²B - cos²A sin²B
= (1 - cos²A) cos²B - cos²A (1 - cos²B)
= cos²B - cos²A cos²B - cos²A + cos²A cos²B
= cos²B - cos²A
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Plz mark the answer as brainly answer
Zantastic:
can you try to answer the previous question?
Answered by
32
hello users ,,,,,,,,
we have to show that ,,,
Sin(A-B)sin(A+B)=cos²B-cos²A
solution :-
we know that ,
sin (A + B ) = sin A cos B + cos A sin B
and
sin (A-B ) = sin A cos B - cos A sin B
here ,
taking LHS
Sin(A-B)sin(A+B)
= { (sin A cos B + cos A sin B ) (sin A cos B - cos A sin B) }
= (sin A cos B)² - (cos A sin B)²
= sin² A cos² B - cos² A sin² B
now converting sin into cos
we get ,
sin² A cos² B - cos² A sin² B
= ( 1 - cos²A ) cos²B - cos²A (1 - cos²B)
= cos²B - cos²A cos²B - cos²A + cos²A cos²B
= (cos²B - cos²A) + (cos²A cos²B - cos²A cos²B )
= cos²B - cos²A = RHS
:(): hope it helps :():
we have to show that ,,,
Sin(A-B)sin(A+B)=cos²B-cos²A
solution :-
we know that ,
sin (A + B ) = sin A cos B + cos A sin B
and
sin (A-B ) = sin A cos B - cos A sin B
here ,
taking LHS
Sin(A-B)sin(A+B)
= { (sin A cos B + cos A sin B ) (sin A cos B - cos A sin B) }
= (sin A cos B)² - (cos A sin B)²
= sin² A cos² B - cos² A sin² B
now converting sin into cos
we get ,
sin² A cos² B - cos² A sin² B
= ( 1 - cos²A ) cos²B - cos²A (1 - cos²B)
= cos²B - cos²A cos²B - cos²A + cos²A cos²B
= (cos²B - cos²A) + (cos²A cos²B - cos²A cos²B )
= cos²B - cos²A = RHS
:(): hope it helps :():
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