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If a + b =π/2 , show that maximum value of cosAcosB is 1/2
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Answer 1

Answer:

Cosacosb=(1/2)*(cos(a+b)+cos(a-b))

since a+b is 90 degrees cos(a+b) is 0

so cosacosb=1/2*cos(a-b)

max value of cos function is 1 when a=b=45 degrees I.e a-b=0

hence max value is 1/2

Hope it's help you.....

Answer 2

$\huge\bigstar\mathfrak\pink{\underline{\underline{SOLUTION}}}$

a+b= π/2= (90°)

a= 90°-B........(1)

A/q,

$= > \frac{1}{2} (2cosA \:cosB) \\ \\ = > \frac{1}{2} [2cos(90 - B)cosB]\: \: \: \: \: \: [from \: (1)]\\ \\ = > \frac{1}{2} (2sinB \: cosB) \\ \\ = > \frac{1}{2} (sinB). \: \: \: \: \: \: (sin2 \theta = 2sin \theta \: cos \theta)$

Maximum value of sin i1 from [-1,1], putting in above

$= > \frac{1}{2} (1) \\ \\ = > \frac{1}{2} \: \: \: \: [hence \: proved]$

Hope it helps ☺️