Question

Prove that sin² α + cos² (α + β) + 2 sin α sin β cos (α + β) is independent of α.

Answer 1

we have to prove that sin² α + cos² (α + β) + 2 sin α sin β cos (α + β) is independent of α.

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

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

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

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

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

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

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

= $cos^2\beta$

we see that solution of sin² α + cos² (α + β) + 2 sin α sin β cos (α + β) doesn't appear $\alpha$ hence, given expression is independent of $\alpha$

Answer 2

HELLO DEAR,

GIVEN:-
$\bold{sin^2\alpha+\{cos(\alpha+\beta)\}^2+2sin\alpha sin\beta cos(\alpha+\beta)}$

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

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

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

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

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

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

=> $\bold{cos^2\beta}$

we see that solution of sin² α + cos² (α + β) + 2 sin α sin β cos (α + β) doesn't appear $\alpha$ hence, given expression is independent of $\alpha$

I HOPE IT'S HELP YOU DEAR,
THANKS