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  4. In triangle A B C, if a=3, b=5, c=4, then find value of sin (B/2)+ cos (B/2).

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Q. In $\triangle A B C$, if $a=3, b=5, c=4$, then find value of $\sin \frac{B}{2}+\cos \frac{B}{2}$.

Trigonometric Functions

Solution:

Given, $a=3, b=5, c=4$
$\sin \frac{B}{2}+\cos \frac{B}{2}$
$=\sqrt{\frac{(s-a)(s-c)}{c a}}+\sqrt{\frac{s(s-b)}{c a}}$
$=\sqrt{\frac{(6-3)(6-4)}{4 \times 3}}+\sqrt{\frac{6(6-5)}{4 \times 3}}$
$=\sqrt{\frac{3 \times 2}{4 \times 3}}+\sqrt{\frac{6 \times 1}{4 \times 3}}$
$=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}$