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Question
Prove that cos(30° – A) cos(30° + A) + cos(45° – A) cos(45° + A) = `cos 2"A" + 1/4`
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Solution
cos(30° – A) cos(30° + A) + cos(45° – A) . cos(45° + A)
= cos(30° + A) cos(30° – A) + cos(45° + A) cos(45° – A)
= `1/2`[cos(30° + A + 30° – A) + cos(30° + A – (30° + A ))] + `1/2`[cos(45° + A + 45° – A) + cos(45° + A – (450 + A))
= `1/2`[cos 60° + cos(30° + A – 30° + A)] + `1/2`[cos 90° + cos(45° + A – 45° + A)]
= `1/2`[cos 60° + cos 2A] + `1/2`[cos 90° + 2A]
= `1/2 cos 60^circ + 1/2 cos 2"A" + 1/ cos 90^cir + 1/2 cos 2"A"`
= `1/2 xx 1/2 + cos 2"A" +1/2 xx 0`
= `1/4 + cos 2"A"`
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