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Math problem of trigonometric function in senior one.
Hello!

1)

According to the area formula of triangle S=acsinB/2=bcsinA/2.

So asinB=bsinA, so a/b=sinA/sinB.

In addition, the title tells acosB=bcosA, and a/b=cosA/cosB is obtained.

therefore

sinA/sinB=cosA/cosB

So sinAcosB=cosAsinB

sinAcosB-cosAsinB=sin(A-B)=0

So A=B, then a = B.

According to cosine theorem

c^2=a^2+b^2-2abcosC

c^2=a^2+a^2-2a^2*3/4

c^2=2a^2-3a^2/2

c^2=a^2/2

A= radical number 2*c

According to a+c=2+ radical number 2

So a=2, c= radical number 2, and b=a=2.

So the area of the triangle S=absinC/2=2*2* root number 7/8= root number 7/2.

2)

Y=L-4 radical 7S/7

=2a+c-4 root number 7*absinC/(2*7)

=2a+c-ab/2

=2a+c-a^2/2

=2a+ radical 2a/2-a 2/2

=- 1/2(a2-4a- radical 2a)

=- 1/2[a2-2(2+ radical 2/2)a]

=- 1/2[A 2-2(2+ radical number 2/2)a+(2+ radical number 2/2)2-(2+ radical number 2/2) 2]

=- 1/2[a-(2+ radical number 2/2)]2+(2+ radical number 2/2) 2/2

Therefore, when a=2+ radical number 2/2, y max =(2+ radical number 2/2)2/2 =(4+ 1/2+ radical number 2)/2=9/4+ radical number 2.

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