Question

Prove that cos² (α - β) + cos² β + 2 cos (α - β) cos α cos β is independent of β.

Answer 1

we have to prove : cos² (α - β) + cos² β - 2 cos (α - β) cos α cos β is independent of β.

cos² (α - β) = $(cos\alpha.cos\beta+sin\alpha.sin\beta)^2$

=$cos^2\alpha.cos^2\beta+sin^2\alpha.sin^2\beta+2cos\alpha cos\beta sin\alpha sin\beta$

now, cos² (α - β) + cos² β - 2 cos (α - β) cos α cos β

=$cos^2\alpha.cos^2\beta+sin^2\alpha.sin^2\beta+2cos\alpha cos\beta sin\alpha sin\beta+cos^2\beta-2(cos\alpha.cos\beta+sin\alpha.sin\beta)cos\alpha.cos\beta$

=$cos^2\alpha.cos^2\beta+sin^2\alpha.sin^2\beta+2cos\alpha cos\beta sin\alpha sin\beta+cos^2\beta-2cos\alpha.cos\beta.sin\alpha.sin\beta -2cos^2\alpha.cos^2\beta$

=$cos^2\beta+sin^2\alpha.sin^2\beta-cos^2\alpha.cos^2\beta$

= $cos^2\beta(1-cos^2\alpha)+sin^2\alpha.sin^2\beta$

= $cos^2\beta sin^2\alpha+sin^2\alpha.sin^2\beta$

= $sin^2\alpha(sin^2\beta+cos^2\beta)$

= $sin^2\alpha$

here it is clear that solution of cos² (α - β) + cos² β - 2 cos (α - β) cos α cos β independent of $\beta$

Answer 2

HELLO DEAR,




GIVEN:-
cos² (α - β) = $\bold{(cos\alpha.cos\beta+sin\alpha.sin\beta)^2}$

=> $\bold{cos^2\alpha.cos^2\beta+sin^2\alpha.sin^2\beta+2cos\alpha cos\beta sin\alpha sin\beta}$


now,
cos² (α - β) + cos² β - 2 cos (α - β) cos α cos β

=> $\bold{cos^2\alpha.cos^2\beta+sin^2\alpha.sin^2\beta+2cos\alpha cos\beta sin\alpha sin\beta+cos^2\beta-2(cos\alpha.cos\beta+sin\alpha.sin\beta)cos\alpha.cos\beta}$

=> $\bold{cos^2\alpha.cos^2\beta+sin^2\alpha.sin^2\beta+2cos\alpha cos\beta sin\alpha sin\beta+cos^2\beta-2cos\alpha.cos\beta.sin\alpha.sin\beta -2cos^2\alpha.cos^2\beta}$

=> $\bold{cos^2\beta+sin^2\alpha.sin^2\beta-cos^2\alpha.cos^2\beta}$

=> $\bold{cos^2\beta(1-cos^2\alpha)+sin^2\alpha.sin^2\beta}$

=> $\bold{cos^2\beta sin^2\alpha+sin^2\alpha.sin^2\beta}$

=> $\bold{sin^2\alpha(sin^2\beta+cos^2\beta)}$

=> $\bold{sin^2\alpha}$


hence, it is clear that solution of cos² (α - β) + cos² β - 2 cos (α - β) cos α cos β independent of $\beta$



I HOPE IT'S HELP YOU DEAR,
THANKS