(cos2a-cos2b)² + (sin2a+sin2b)²=4sin²(a+b)​

LHS = (cosA+cosB)²+(sinA+sinB)²

= (cos²A+2cosAcosB+cos²B)+

(sin²A+2sinAsinB+sin²B)

/* By algebraic identity:

(x+y)² = x²+2xy+y² */

Rearranging the terms, we get

= (cos²A+sin²A)+(cos²B+sin²B)+2(cosAcosB+sinAsinB)

= 1+1+2cos(A-B)

we know that,

i) cos²x + sin²x = 1

ii) cosxcosy+sinxsiny = sin(x-y)

= 2+2cos(A-B)

= 2[1+cos(A-B)]

\boxed {1+cos\theta = 2cos^{2}\frac{\theta}{2}}

1+cosθ=2cos

2

2

θ

= 2\times 2cos^{2}\frac{(A-B)}{2}2×2cos

2

2

(A−B)

=4cos^{2}\frac{(A-B)}{2}4cos

2

2

(A−B)

=RHSRHS

Therefore,

(cosA+cosB)²+(sinA+sinB)²

=4cos^{2}\frac{(A-B)}{2}4cos

2

2

(A−B)

i found two solutions,   n = 2 • ± √3 = ± 3.4641

step-by-step explanation: