Question

(cosa+cosb)^2+(sina+sinb)^2=4cos^2(A-B/2)

Answer 1

this is soulation of your questions

Answer 2

Solution:

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)]

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

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

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

=$RHS$

Therefore,

(cosA+cosB)²+(sinA+sinB)²

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

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