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Question
If cos θ = `1/2 ("a" + 1/"a")`, show that cos 3θ = `1/2 ("a"^3 + 1/"a"^3)`
Solution
cos θ = `1/2 ("a" + 1/"a")`
cos 3θ = 4 cos3θ – 3 cos θ
= `4[1/2("a" + 1/"a")]^3 - 3[1/2("a" + 1/"a")]`
= `4 xx 1/8("a" + 1/"a")^3 - 3/2("a" + 1/"a")`
= `1/2("a" + 1/"a")^3 - 3/2("a" + 1/"a")`
= `1/2["a"^3 + 3"a"^2(1/"a") + 3"a"(1/"a")^2 + 1/"a"^3] - 3/2(a" + 1/"a")`
= `1/2["a"^3 + 3"a" + 3/"a" + 1/"a"^3] - 3/2"a" - 3/(2"a")`
= `1/2 "a"^3 + 3/2"a" + 3/(2"a") + 1/(2"a"^3) - 3/2"a" - 3/(2"a")`
cos 3θ = `1/2"a"^3 + 1/(2"a"^3)`
= `1/2("a"^3 + 1/"a"^3)`
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