The angles $$ A $$ and $$ B $$ are such that $$ \sin A=\frac{3}{5} $$ and $$ \cos B=-\frac{4}{5} $$, where $$ A $$ and $$ B $$ are in the same quadrant. Find the value of each of the following without using a calculator. $$ \begin{array}{ll} ext { (i) } an (A+B) & ext { (ii) } \cos (A+B) \ ext { (iii) } \sin (A-B) & \end{array} $$

Question

The angles $$ A $$ and $$ B $$ are such that $$ \sin A=\frac{3}{5} $$ and $$ \cos B=-\frac{4}{5} $$, where $$ A $$ and $$ B $$ are in the same quadrant. Find the value of each of the following without using a calculator. $$ \begin{array}{ll}	ext { (i) } 	an (A+B) & 	ext { (ii) } \cos (A+B) \ 	ext { (iii) } \sin (A-B) & \end{array} $$

UpStudy AI · Accepted Answer

Given that both angles $$ A $$ and $$ B $$ are in the same quadrant and:

$$
\sin A = \frac{3}{5}, \quad \cos B = -\frac{4}{5}
$$

### Determining the Quadrant:
- $$ \sin A = \frac{3}{5} > 0 $$ implies that angle $$ A $$ is either in the **First** or **Second** quadrant.
- $$ \cos B = -\frac{4}{5} < 0 $$ implies that angle $$ B $$ is either in the **Second** or **Third** quadrant.

Since $$ A $$ and $$ B $$ are in the **same quadrant**, both must be in the **Second** quadrant.

### Finding Remaining Trigonometric Values:
For angle $$ A $$ in the Second Quadrant:
$$
\cos A = -\sqrt{1 - \sin^2 A} = -\sqrt{1 - \left(\frac{3}{5}\right)^2} = -\frac{4}{5}
$$

For angle $$ B $$ in the Second Quadrant:
$$
\sin B = \sqrt{1 - \cos^2 B} = \sqrt{1 - \left(-\frac{4}{5}\right)^2} = \frac{3}{5}
$$

### (i) $$ \tan(A + B) $$:

Using the tangent addition formula:
$$
\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}
$$

First, calculate $$ \tan A $$ and $$ \tan B $$:
$$
\tan A = \frac{\sin A}{\cos A} = \frac{\frac{3}{5}}{-\frac{4}{5}} = -\frac{3}{4}
$$
$$
\tan B = \frac{\sin B}{\cos B} = \frac{\frac{3}{5}}{-\frac{4}{5}} = -\frac{3}{4}
$$

Now, substitute into the addition formula:
$$
\tan(A + B) = \frac{-\frac{3}{4} - \frac{3}{4}}{1 - \left(-\frac{3}{4}\right)\left(-\frac{3}{4}\right)} = \frac{-\frac{6}{4}}{1 - \frac{9}{16}} = \frac{-\frac{3}{2}}{\frac{7}{16}} = -\frac{24}{7}
$$

### (ii) $$ \cos(A + B) $$:

Using the cosine addition formula:
$$
\cos(A + B) = \cos A \cos B - \sin A \sin B
$$
$$
\cos(A + B) = \left(-\frac{4}{5}\right)\left(-\frac{4}{5}\right) - \left(\frac{3}{5}\right)\left(\frac{3}{5}\right) = \frac{16}{25} - \frac{9}{25} = \frac{7}{25}
$$

### (iii) $$ \sin(A - B) $$:

Using the sine subtraction formula:
$$
\sin(A - B) = \sin A \cos B - \cos A \sin B
$$
$$
\sin(A - B) = \left(\frac{3}{5}\right)\left(-\frac{4}{5}\right) - \left(-\frac{4}{5}\right)\left(\frac{3}{5}\right) = -\frac{12}{25} + \frac{12}{25} = 0
$$

### Summary of Results:
$$
\begin{aligned}
&\text{(i)} \quad \tan(A + B) = -\frac{24}{7} \
&\text{(ii)} \quad \cos(A + B) = \frac{7}{25} \
&\text{(iii)} \quad \sin(A - B) = 0 \
\end{aligned}
$$

**Answer:**

(i) $$ \tan(A+B) = -\dfrac{24}{7} $$

(ii) $$ \cos(A+B) = \dfrac{7}{25} $$

(iii) $$ \sin(A-B) = 0 $$
$$ \tan(A+B) = -\dfrac{24}{7} $$, $$ \cos(A+B) = \dfrac{7}{25} $$, $$ \sin(A-B) = 0 $$.

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