Question

You are given that = -\frac{7}{25}" alt="sin(A) = -\frac{7}{25}" align="absmiddle" class="latex-formula">, with A in Quadrant IV, and $sin(B) = -\frac{12}{13}$, with B in Quadrant III. Find cos(A+B). Give your answer as a fraction

Answer 1

Answer:

- $\frac{204}{325}$

Step-by-step explanation:

Using the addition formula for cosine

cos(A + B) = cosAcosB - sinAsinB

Given sin A = - $\frac{7}{25}$ = $\frac{opposite}{hypotenuse}$ and in fourth quadrant, then

the triangle is a 7 - 24- 25 right triangle with adjacent side 24

In fourth quadrant cosine ratio is positive , thus

cos A = $\frac{adjacent}{hypotenuse}$ = $\frac{24}{25}$

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Given sin B = - $\frac{12}{13}$ = $\frac{opposite}{hypotenuse}$ and in third quadrant, then

the triangle is a 5- 12- 13 right triangle with adjacent side 5

In third quadrant cosine ratio is negative, thus

cos B = $\frac{adjacent}{hypotenuse}$ = - $\frac{5}{13}$

Thus

cosAcosB - sinAsinB

= ($\frac{24}{25}$ × - $\frac{5}{13}$ ) - (- $\frac{7}{25}$ × - $\frac{12}{13}$ )

= - $\frac{120}{325}$ - $\frac{84}{325}$

= - $\frac{204}{325}$