Prove that- -cos-cos 60- cos 60- -1-4 for all values of -

|cos θ cos (60 θ) cos (60+θ)| = |cos θ (cos^2 60^∘ sin^2 θ)| {since cos (A B) cos (A+B) = cos^2A sin^2 B } = |cos θ(1/4 sin^2 θ ) | = |cos θ 1/4(1 4 sin^2 θ) ...

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Standard XII

Mathematics

Range of Trigonometric Expressions

Prove that: |...

Question

Prove that: $| c o s θ c o s (60 - θ) c o s (60 + θ) | \leq \frac{1}{4}$ for all values of $θ$

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Solution

$| c o s θ c o s (60 - θ) c o s (60 + θ) | = | c o s θ (c o s^{2} 60^{\circ} - s i n^{2} θ) | {since c o s (A - B) c o s (A + B) = c o s^{2} A - s i n^{2} B} = ∣ ∣ c o s θ (\frac{1}{4} - s i n^{2} θ) ∣ ∣ = ∣ ∣ c o s θ \frac{1}{4} (1 - 4 s i n^{2} θ) ∣ ∣ = ∣ ∣ \frac{1}{4} c o s θ (1 - 4 (1 - c o s^{2} θ)) ∣ ∣ = \frac{1}{4} c o s θ (- 3 + 4 c o s^{2} θ)$

$= ∣ ∣ \frac{1}{4} (4 c o s 3 θ - 3 c o s θ) ∣ ∣ = ∣ ∣ \frac{1}{4} c o s 3 θ ∣ ∣ \leq \frac{1}{4} {since | c o s 3 θ | \leq 1}$

So,

$| c o s θ c o s (60^{\circ} - θ) c o s (60^{\circ} + θ) | \leq \frac{1}{4}$

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