suppose sin(a)+cos(b)=-0.2 while sin(b)+cos(a)=1.6

reposted here:

https://web2.0calc.com/questions/suppose-sin-a-cos-b-0-2-while-sin-b-cos-a-1-6_1

suppose sin(a)+cos(b)=-0.2 while sin(b)+cos(a)=1.6 

What is the value ofsin (a+b)

 

\(sin(a)+cos(b)=-0.2 \qquad \qquad and \qquad\qquad sin(b)+cos(a)=1.6\\ (sin(a)+cos(b))^2=(-0.2)^2 \qquad and \qquad \qquad (sin(b)+cos(a))^2=(1.6)^2\\ sin^2(a)+cos^2(b)+2sinacosb=0.04 \quad and \quad sin^2(b)+cos^2(a)+2sinbcosa=2.56\\ sin^2(a)+cos^2(b)+2sinacosb=0.04 \;\;(1)\;\; and \quad sin^2(b)+cos^2(a)+2sinbcosa=2.56 \;\;(2)\\ (1)+(2)\\ sin^2(a)+cos^2(a)+sin^2(b)+cos^2(b)+2sinacosb+2sinbcosa=0.04+2.56\\ 1+1+2(sinacosb+sinbcosa)=3\\ 2(sinacosb+sinbcosa)=1\\ sin(a+b)=0.5\\\)

You need to check it.

 

 

LaTex:

sin(a)+cos(b)=-0.2 \qquad \qquad and  \qquad\qquad sin(b)+cos(a)=1.6\\

(sin(a)+cos(b))^2=(-0.2)^2 \qquad and \qquad \qquad (sin(b)+cos(a))^2=(1.6)^2\\
sin^2(a)+cos^2(b)+2sinacosb=0.04 \quad and \quad  sin^2(b)+cos^2(a)+2sinbcosa=2.56\\
sin^2(a)+cos^2(b)+2sinacosb=0.04 \;\;(1)\;\; and \quad  sin^2(b)+cos^2(a)+2sinbcosa=2.56 \;\;(2)\\
(1)+(2)\\
sin^2(a)+cos^2(a)+sin^2(b)+cos^2(b)+2sinacosb+2sinbcosa=0.04+2.56\\
1+1+2(sinacosb+sinbcosa)=3\\
2(sinacosb+sinbcosa)=1\\
sin(a+b)=0.5\\