Question

suppose A is in quadrant 2 and B is in Quadrant 2

Answer 1

Solution:

Given that:

$\cos(A)=-\frac{9}{41},\text{ }\sin(B)=\frac{63}{65}$

To evaluate sin (A+B) and sin (A-B),

Step 1: Express the compound angle formula.

According to the compound angle formula,

$\begin{gathered} \sin(A+B)=\sin(A)\cos(B)+\sin(B)\cos(A) \\ \sin(A-B)=\sin(A)\cos(B)-\sin(B)\cos(A) \end{gathered}$

Step 2: Evaluate cos (B) and sin (A).

According to the Pythagorean identities,

$\cos^2\theta+sin^2\theta=1$

Thus,

$\begin{gathered} \sin^(A)=\sqrt{1-\cos^2A}\text{ =}\sqrt{1-(-\frac{9}{41})^2} \\ =\sqrt{1-\frac{81}{1681}} \\ =\sqrt{\frac{1600}{1681}} \\ \Rightarrow\sin^(A)=\frac{40}{41} \end{gathered}$

Similarly,

$\begin{gathered} \cos(B)=\sqrt{1-\sin^2B}\text{ =}\sqrt{1-(\frac{63}{65})^2} \\ =\sqrt{1-\frac{3969}{4225}} \\ =\sqrt{\frac{256}{4225}} \\ \Rightarrow\cos(B)=-\frac{16}{65}\text{ \lparen second quadrant\rparen} \end{gathered}$

Step 3: Evaluate sin (A+B).

Recall,

$\begin{gathered} \sin(A+B)=\sin(A)\cos(B)+\sin(B)\cos(A) \\ =(\frac{40}{41}\times-\frac{16}{65})+(\frac{63}{65}\times-\frac{9}{41}) \\ =-\frac{1207}{2665} \end{gathered}$

Step 4: Evaluate sin(A-B).

Recall,

$\begin{gathered} \sin(A-B)=\sin(A)\cos(B)-\sin(B)\cos(A) \\ =(\frac{40}{41}\times-\frac{16}{65})-(\frac{63}{65}\times-\frac{9}{41}) \\ =-\frac{73}{2665} \end{gathered}$

Hence, we have

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