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Trig Identities (1 Viewer)
- Thread starter Lukybear
- Start date
lyounamu
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sin^2a x cos^2b - (1-sin^2a) x (1-cos^2b) = sin^2a*cos^2b - (1 - sin^2a - cos^2b + sin^2a*cos^2b) = sin^2a - 1 + cos^2b = sin^2a -cos^2b
I think it is sin^2a - sin^2b instead of sin^2a + sin^2b
annabackwards
<3 Prophet 9
My answer was sin^2a - sin^2b as well
LHS = sin^2a(1 - sin^2b) - sin^2b(1 - sin^2a)
= sin^2a - sin^2a.sin^2b - sin^2b + sin^2b.sin^2a
= sin^2a - sin^2b
LHS = sin^2a(1 - sin^2b) - sin^2b(1 - sin^2a)
= sin^2a - sin^2a.sin^2b - sin^2b + sin^2b.sin^2a
= sin^2a - sin^2b
I think you are right lyounamu.
LHS = (1 - sin2 a) - cos2 a (1 - cos2 b)
= cos^b - cos^2a . cos^2b - cos^2 a + cos^a cos^2 b
= cos^2b - cos^2 a
= (1 - sin^2 b) - (1 - sin^2 a)
= sin^2 a - sin^2 b
PS: annabackwards' is better .. I suspected I should have converted cos^2 a to sin^2 a etc but ..
LHS = (1 - sin2 a) - cos2 a (1 - cos2 b)
= cos^b - cos^2a . cos^2b - cos^2 a + cos^a cos^2 b
= cos^2b - cos^2 a
= (1 - sin^2 b) - (1 - sin^2 a)
= sin^2 a - sin^2 b
PS: annabackwards' is better .. I suspected I should have converted cos^2 a to sin^2 a etc but ..
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