last step is wrong (Question:656414.0)

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Ramesh Kumar

· started a discussion

· 1 Months ago

last step is wrong (Question:656414.0)

sin [(A+B+C)/2]= sin (180/2) = sin 90 = 1

Question:

If (A + B + C) = 90°, then what is the value of sin (A/2) sin [(180 – B – C)/2] + cos (A/2) sin (B + C)/2?

Options:

1/2

1/√2

√3/2

Solution:

A+B+C=90°

$\begin{matrix} s i n (\frac{A}{2}) . s i n (\frac{180 - B - C}{2}) + c o s (\frac{A}{2}) . s i n \frac{(B + C)}{2} \\ = s i n (\frac{A}{2}) . s i n (90 - \frac{(B + C)}{2}) + c o s (\frac{A}{2}) . s i n \frac{(B + C)}{2} \\ = s i n (\frac{A}{2}) . c o s (\frac{(B + C)}{2}) + c o s (\frac{A}{2}) . s i n \frac{(B + C)}{2} \\ = s i n (\frac{A}{2} + \frac{B + C}{2}) = s i n (\frac{90}{2}) = s i n 45 ° \\ = \frac{1}{\sqrt{2}} \end{matrix}$

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