# Answer to Question #11789 in Trigonometry for tanya jaiswal

Question #11789

if α+β = 90° find the maximum and minimum values of sinα.sinβ

Expert's answer

sinα.sinβ = (cos(α-β) - cos(α+β)) / 2

If α+β = 90°,

sinα.sinβ = (cos(α-β)

- cos(90°)) / 2 = (cos(α-β) - 0) / 2 = cos(α-β) / 2.

So, max(cos(α-β)) = 1,

if α = β = 45°. min(cos(α-β)) = 0, if α = 90°,

β = 0°, or β = 90°, α = 0°.

If α+β = 90°,

sinα.sinβ = (cos(α-β)

- cos(90°)) / 2 = (cos(α-β) - 0) / 2 = cos(α-β) / 2.

So, max(cos(α-β)) = 1,

if α = β = 45°. min(cos(α-β)) = 0, if α = 90°,

β = 0°, or β = 90°, α = 0°.

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