Trig Identities (1 Viewer)

[Lukybear]

Help with this Q plz.

Prove:
sin^2a . cos^2b - cos^2a . sin^2b = sin^2a + sin^2b

 

[lyounamu]

Help with this Q plz.

Prove:
sin^2a . cos^2b - cos^2a . sin^2b = sin^2a + sin^2b

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]

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

 

[Drongoski]

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 ..

 

[Lukybear]

Yea sorry it is.