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Use cos a cos b=1/2 [cos (a + b) + cos (a-b)]to derive cos x + cos y= cos 2 (x+y/2) cos (x-y/2) .

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The trigonometric identity cos x + cos y = 2 cos (x + y/2) cos (x - y/2), is derived using the trigonometric identity cos a cos b=1/2 [cos (a + b) + cos (a-b)].

In the question, we are asked to derive the trigonometric identity, cos x + cos y = 2 cos (x + y/2) cos (x - y/2), using the trigonometric identity cos a cos b=1/2 [cos (a + b) + cos (a-b)].

We are given the trigonometric identity cos a cos b=1/2 [cos (a + b) + cos (a-b)].

Substituting a = x + y/2 and b = x - y/2 in this, we get:

cos (x + y/2) cos (x - y/2) = 1/2[cos (x + y/2 + x - y/2) + cos (x + y/2 - x - y/2)],

or, cos (x + y/2) cos (x - y/2) = 1/2[ cos (2x/2) + cos (2y/2) ],

or, 2 cos (x + y/2) cos (x - y/2) = cos x + cos y, which on inter-changing the sides, gives us:

cos x + cos y = 2 cos (x + y/2) cos (x - y/2), which is the required trigonometric identity.

Thus, the trigonometric identity cos x + cos y = 2 cos (x + y/2) cos (x - y/2), is derived using the trigonometric identity cos a cos b=1/2 [cos (a + b) + cos (a-b)].

Learn more about deriving trigonometric identities at

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The provided question is incorrect. The correct question is:

"Use cos a cos b=1/2 [cos (a + b) + cos (a-b)]to derive cos x + cos y = 2 cos (x + y/2) cos (x - y/2) ."

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