Question

(cosx+cosy)^2 + ( sinx-siny)^2=?​

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

Step-by-step explanation:

Applying our trigonometric knowledge,we know that,

2sinAsinB=cos(A-B)-cos (A+B)…(1)

And,

2cosAcosB=cos(A+B)+cos (A-B)…(2)

Using (1) & (2),we obtain,

(cosx+cosy) ^2+ (sinx+siny) ^2=cosx^2+sinx^2+2cosxcosy+siny^2+cosy^2+2cosxcosy

=1+1+2cosxcosy+2sinxsiny[Since,sinA^2+cosA^2=1]

=2+cos(x+y)+cos(x-y)+cos(x-y)-cos(x+y)

=2+2cos(x-y)

=2+2[{2cos (x-y)^2–1}][N.B-cos2A=2(cosA^2)–1]

=2–2+{4cos(x-y/2)^2}

={4cos(x-y/2)^2}【proved】.