Sin^a-Sin^b=1 then the value of Cos^(a+b)=?
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There is a mistake in the given exercise.
given sin²a + sin² b = 1
sin²b = cos² a => cos a = +- sin b
same way, sin²a = cos²b => cos b = +- sin a
cos² (a+b) = (cos a sin b + sin a cos b)²
= cos²a sin²b + sin²a cos²b + 2 sin a sin b cos a cos b
= sin⁴b + cos⁴b +- 2 sin²b cos²b
= (sin² b + cos²b)² OR (sin²b - cos²b)²
= 1 or cos²2b
So a+b = 0 or 2π
OR, a+b = 2b or 180° - 2b or 360 - 2b
ie., a = b = π/4 => cos²(a+b) = 0
or, a + 3b = 180°
or a + 3b = 360°
given sin²a + sin² b = 1
sin²b = cos² a => cos a = +- sin b
same way, sin²a = cos²b => cos b = +- sin a
cos² (a+b) = (cos a sin b + sin a cos b)²
= cos²a sin²b + sin²a cos²b + 2 sin a sin b cos a cos b
= sin⁴b + cos⁴b +- 2 sin²b cos²b
= (sin² b + cos²b)² OR (sin²b - cos²b)²
= 1 or cos²2b
So a+b = 0 or 2π
OR, a+b = 2b or 180° - 2b or 360 - 2b
ie., a = b = π/4 => cos²(a+b) = 0
or, a + 3b = 180°
or a + 3b = 360°
Anonymous:
Thanks
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