Question

You are given that sin(A)=−20/29, with A in Quadrant III, and cos(B)=12/13, with B in Quadrant I. Find sin(A+B). Give your answer as a fraction.

Answer 1

Answer:

$sin(A+B)=-\dfrac{345}{377}$

Step-by-step explanation:

Given that:

$sin(A)=-\dfrac{20}{29}\\cos(B)=\dfrac{12}{13}$

A is in 3rd quadrant and B is in 1st quadrant.

To find: sin(A+B)

Solution:

We know the Formula:

1. $sin(A+B) = sinAcosB+cosAsinB$

2. $sin^{2} \theta+cos^{2} \theta=1$

Now, let us find the values of cosA and sinB.

$sin^{2} A+cos^{2} A=1\\\Rightarrow (\frac{-20}{29})^2+cos^{2} A=1\\\Rightarrow cos^{2} A=1- \dfrac{400}{941}\\\Rightarrow cos^{2} A=\dfrac{941-400}{941}\\\Rightarrow cos^{2} A=\dfrac{441}{941}\\\Rightarrow cos A=\pm \dfrac{21}{29}$

A is in 3rd quadrant, so cosA will be negative,

$\therefore cos A=-\dfrac{21}{29}$

$sin^{2} B+cos^{2} B=1\\\Rightarrow sin^{2} A+(\frac{12}{13})^2=1\\\Rightarrow sin^{2} B=1- \dfrac{144}{169}\\\Rightarrow sin^{2} B=\dfrac{169-144}{169}\\\Rightarrow sin^{2} B=\dfrac{25}{169}\\\Rightarrow sinB=\pm \dfrac{5}{13}$

B is in 1st quadrant, sin B will be positive.

$sinB =\dfrac{5}{13}$

Now, using the formula:

$sin(A+B) = sinAcosB+cosAsinB\\\Rightarrow -\dfrac{20}{29} \times \dfrac{12}{13}-\dfrac{21}{29}\times \dfrac{5}{13}\\\Rightarrow -\dfrac{20\times 12+21\times 5}{29\times 13} \\\Rightarrow -\dfrac{240+105}{29\times 13} \\\Rightarrow -\dfrac{345}{377}$

$sin(A+B)=-\dfrac{345}{377}$