Question

40.What is the minimum value of SinA if 0≤ A ≤ 90°.:​

Answer 1

Answer:

Sin¥ . Cos ¥ = 1/2 (sin 2¥)

Max value of sine is 1

So

1/2 × 1

=1/2

2nd approach

f(¥) = Sin¥ . Cos ¥

f ' (¥) = sin ¥ (- sin ¥) + Cos¥ . Cos¥

= Cos^2 ¥ - sin ^2 ¥

for max/ min

f ' (¥)=0

Cos^2 ¥ - sin ^2 ¥ = 0

Cos^2 ¥ = sin ^2 ¥

Cos ¥ = sin ¥

¥ = 45°

f '’ (¥) = -2 Sin¥ . Cos ¥ -2 Sin¥ . Cos ¥

= -4 Sin¥ . Cos ¥

For ¥ =45°

f '’ (¥) < 0

So its point of maxima

f(¥) = Sin¥ . Cos ¥

f (45°) = 1/2

Answer 2

Answer:

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Step-by-step explanation:

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