Question

In triangle ABC, Prove that

${sin}^{2} A + {sin}^{2} B + {sin}^{2} C \leqslant \frac{9}{4}$
​

Answer 1

Answer:

mark me brainliest

Step-by-step explanation:

cos2A=1-2sin^2A

<br>

sin^2A=(1-cos2A)/2

<br>

=(1-cos2A)/2+(1-cos2B)/2+sin^2C

<br>

=1+sin^2C-1/2(cos2A+cos2B)

<br>

=1+sin^2C-1/2(2cos(A_B)*cos(A-B))

<br>

=1+sin^2C-cos(A+B)*cos(A-B)

<br>

A+B+C=pi

<br>

A+B=pi-C

<br>

cos(A+B)=cos(pi-C)

<br>

=-cosC

<br>

=1+(1-cos^2C)+cosCcos(A-B)

<br>

=2-cos^2C+cosCcos(A-B)

<br>

-1<=costheta<=1

<br>

-1<=cos(A-B)<=1

<br> LHS<br>

sin^2A+sin^2B+sin^2C<=2-cos^2C+cosC

<br>

=2-(cos^2C-cosC+1/4-1/4)

<br>

=2-(cosC-1/2)^2+1/4

<br>

=2+1/4-(cosC-1/2)^2

<br>

sin^2A+sin^2B+sin^2C<=9/4