If -90- show that the maximum value of cos-cos- is 1-2

Let y= cosα cosβ then, y=1/2(2cosα cosβ ) =1/2[cos(α +β )+cos(α β)] =1/2[cos90^∘+cos(α β)] [∵α +β=90^∘] =1/2[0+cos(α β)] =1/2cos(α β) ⇒ y=1/2cos(α β) ...

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Byju's Answer

Standard XII

Mathematics

Basic Trigonometric Identities

If α+β=90∘, s...

Question

If $α + β = 90^{\circ}$ , show that the maximum value of $c o s α c o s β i s \frac{1}{2}$

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Solution

Let $y = c o s α c o s β$ then,
$y = \frac{1}{2} (2 c o s α c o s β) = \frac{1}{2} [c o s (α + β) + c o s (α - β)] = \frac{1}{2} [c o s 90^{\circ} + c o s (α - β)] [∵ α + β = 90^{\circ}] = \frac{1}{2} [0 + c o s (α - β)] = \frac{1}{2} c o s (α - β) \Rightarrow y = \frac{1}{2} c o s (α - β)$

Now,
$- 1 \leq c o s (α - β) \leq 1 \Rightarrow y = \frac{- 1}{2} \leq \frac{1}{2} c o s (α - β) \leq \frac{1}{2} \Rightarrow \frac{- 1}{2} \leq \frac{1}{2} c o s (α - β) \leq \frac{1}{2} \Rightarrow \frac{- 1}{2} \leq y \leq \frac{1}{2} \Rightarrow \frac{- 1}{2} \leq c o s α c o s β \leq \frac{1}{2}$
Hence, the maximum values of $c o s α c o s β$ is $\frac{1}{2}$

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If α+β=90∘, show that the maximum value of cosα cosβis12

If $α + β = 90^{\circ}$ , show that the maximum value of $c o s α c o s β i s \frac{1}{2}$