Answers: sin(A+B)=[tex]\frac{-2\sqrt{2}-\sqrt{42}}{10}[/tex], cos(A+B)=[tex]\frac{2\sqrt{2}-\sqrt{42}}{10}[/tex], A+B=Quadrant 3, sin(2A)=[tex]\frac{4\sqrt{21}}{25}[/tex]
NOTES:
sin A = [tex]\frac{2}{5}[/tex] ⇒ cos A = [tex]\frac{\sqrt{21}}{5}[/tex]
since tan B = 1 and is in Quadrant 3, then
sin B = [tex]-\frac{\sqrt{2}}{2}[/tex] and cos B = [tex]-\frac{\sqrt{2}}{2}[/tex]
SIN (A + B):
sin (A + B) = (sin A * cos B) + (cos A * sin B)
= ([tex]\frac{2}{5}[/tex])([tex]-\frac{\sqrt{2}}{2}[/tex]) + ([tex]\frac{\sqrt{21}}{5}[/tex])([tex]-\frac{\sqrt{2}}{2}[/tex])
= [tex]-\frac{2\sqrt{2}}{10}[/tex] + [tex]-\frac{\sqrt{42}}{10}[/tex]
= [tex]\frac{-2\sqrt{2}-\sqrt{42}}{10}[/tex]
COS (A + B):
cos (A + B) = (cos A * cos B) - (sin A * sin B)
= ([tex]\frac{\sqrt{21}}{5}[/tex])([tex]-\frac{\sqrt{2}}{2}[/tex]) - ([tex]\frac{2}{5}[/tex])([tex]-\frac{\sqrt{2}}{2}[/tex])
= [tex]-\frac{\sqrt{42}}{10}[/tex] - [tex]-\frac{2\sqrt{2}}{10}[/tex]
= [tex]\frac{2\sqrt{2}-\sqrt{42}}{10}[/tex]
SIN 2A:
sin (A + A) = 2 (sin A * cos A)
= 2 ([tex]\frac{2}{5}[/tex])([tex]\frac{\sqrt{21}}{5}[/tex])
= [tex]\frac{4\sqrt{21}}{25}[/tex]
A + B:
sin A = [tex]\frac{2}{5}[/tex]
A = sin⁻¹[tex](\frac{2}{5})[/tex]
A = 23.57°
tan B = 1
B = tan⁻¹(1)
B = 45°
= 45° + 180° in Quadrant 3
= 225°
A + B = 23.6° + 225°
= 248.6° which lies in Quadrant 3
