Question

If A+B+C =0° then prove that
sin A +SinB -sinc =-4 cosA/2×cosB/2× sinC/2​

Answer 1

Step-by-step explanation:

=B=60

then A=60

B=60

then

cos(A-B)=cosAcosB+sinAsinB

cos(60-60)=cos60cos60+sin60sin60

cos0=cso60cos60+sin60sin60

(we know that

\begin{gathered}cos0 = 1 \\ cos60 = 1 \div 2 \\ sin60 = \sqrt{3} \div 2 \\ from \: the \: trigonometry \: table\end{gathered}

cos0=1

cos60=1÷2

sin60=

3

÷2

fromthetrigonometrytable

\begin{gathered}1 = {1 \div 2 \times 1 \div 2} + { \sqrt{3 \div } \div 2 \times \sqrt{3} \div 2} \\ 1 = 1 \div 4 + 3 \div 4 \\ 1 = 4 \div 4 \\ 1 = 1\end{gathered}

1=1÷2×1÷2+

3÷

÷2×

3

÷2

1=1÷4+3÷4

1=4÷4

1=1