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Sagot :
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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