Question

If A is a quadrant 1 angle, such that sinA=2/5 and B is a quadrant 3 angle, such that tanB=1, find sin(A+B) and cos(A+B) and the quadrant in with A+B lies in. Find sin2A.
*Please Show Work*

Answer 1

Answers: sin(A+B)=$\frac{-2\sqrt{2}-\sqrt{42}}{10}$, cos(A+B)=$\frac{2\sqrt{2}-\sqrt{42}}{10}$, A+B=Quadrant 3, sin(2A)=$\frac{4\sqrt{21}}{25}$

NOTES:

sin A = $\frac{2}{5}$      ⇒    cos A = $\frac{\sqrt{21}}{5}$

since tan B = 1 and is in Quadrant 3, then

sin B = $-\frac{\sqrt{2}}{2}$    and  cos B = $-\frac{\sqrt{2}}{2}$

SIN (A + B):

sin (A + B) = (sin A * cos B) + (cos A * sin B)

                 = ($\frac{2}{5}$)($-\frac{\sqrt{2}}{2}$) + ($\frac{\sqrt{21}}{5}$)($-\frac{\sqrt{2}}{2}$)

                  = $-\frac{2\sqrt{2}}{10}$ + $-\frac{\sqrt{42}}{10}$

                  = $\frac{-2\sqrt{2}-\sqrt{42}}{10}$

COS (A + B):

cos (A + B) = (cos A * cos B) - (sin A * sin B)

                  = ($\frac{\sqrt{21}}{5}$)($-\frac{\sqrt{2}}{2}$) - ($\frac{2}{5}$)($-\frac{\sqrt{2}}{2}$)

                  = $-\frac{\sqrt{42}}{10}$ - $-\frac{2\sqrt{2}}{10}$

                  = $\frac{2\sqrt{2}-\sqrt{42}}{10}$

SIN 2A:

sin (A + A) = 2 (sin A * cos A)

                 = 2 ($\frac{2}{5}$)($\frac{\sqrt{21}}{5}$)

                 = $\frac{4\sqrt{21}}{25}$

A + B:

sin A = $\frac{2}{5}$

    A = sin⁻¹$(\frac{2}{5})$

    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