In a triangle ABC,if B = 3C,then the values o | vedclass

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(B) Given $B = 3C$. By the sine rule,$\frac{b}{\sin B} = \frac{c}{\sin C} = k$,so $b = k \sin 3C$ and $c = k \sin C$.First,consider $\sqrt{\frac{b + c

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$\cos C, \sin \frac{A}{2}$

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(B) Given $B = 3C$. By the sine rule,$\frac{b}{\sin B} = \frac{c}{\sin C} = k$,so $b = k \sin 3C$ and $c = k \sin C$.First,consider $\sqrt{\frac{b + c}{4c}} = \sqrt{\frac{\sin 3C + \sin C}{4 \sin C}}$.Using the sum-to-product formula $\sin X + \sin Y = 2 \sin \frac{X+Y}{2} \cos \frac{X-Y}{2}$,we get $\sin 3C + \sin C = 2 \sin 2C \cos C$.So,$\sqrt{\frac{2 \sin 2C \cos C}{4 \sin C}} = \sqrt{\frac{2 (2 \sin C \cos C) \cos C}{4 \sin C}} = \sqrt{\cos^2 C} = \cos C$.Next,consider $\frac{b - c}{2c} = \frac{\sin 3C - \sin C}{2 \sin C}$.Using the difference-to-product formula $\sin X - \sin Y = 2 \cos \frac{X+Y}{2} \sin \frac{X-Y}{2}$,we get $\sin 3C - \sin C = 2 \cos 2C \sin C$.So,$\frac{2 \cos 2C \sin C}{2 \sin C} = \cos 2C$.Since $A + B + C = 180^{\circ}$ and $B = 3C$,we have $A + 4C = 180^{\circ}$,so $2C = 90^{\circ} - \frac{A}{2}$.Thus,$\cos 2C = \cos(90^{\circ} - \frac{A}{2}) = \sin \frac{A}{2}$.Therefore,the values are $\cos C$ and $\sin \frac{A}{2}$.

In a triangle $ABC$,if $B = 3C$,then the values of $\sqrt{\frac{b + c}{4c}}$ and $\frac{b - c}{2c}$ are

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