Question

suppose sin(A)= ¼. use the trig identity sin²(A) + cos²(A) =1 to find cos(A) in quadrant II. round to nearest ten-thousandth a. -0.8572b. 0.1397c. 0.4630d. -0.9682

Answer 1

The value of cos (A) on the second quadrant is -0.9682

Here, we want to find the value of cos (A) in the second quadrant using the given relation

We proceed as follows;

$\begin{gathered} \sin ^2A+cos^2\text{ A = 1} \\ \\ \cos ^2A=1-sin^2A \\ \\ \sin ^2A=(sinA)^2 \\ \\ \text{Insert value of 1/4} \\ \\ \sin ^2A\text{ = (}\frac{1}{4})^2 \\ \\ \sin ^2A\text{ = }\frac{1}{16} \\ \\ \text{cos A = }\sqrt[]{\cos ^2A} \\ \\ \cos \text{ A = }\sqrt[]{(1-\sin ^2A)} \\ \\ \cos \text{ A = }\sqrt[]{1-\frac{1}{16}} \\ \\ \cos \text{ A =}\sqrt[]{\frac{15}{16}} \\ \\ \cos \text{ A = 0.9682} \end{gathered}$

However, the value of cosine on the second quadrant is negative (COSINE value is only positive on the first and the fourth axes)

Thus, in this case our answer will be negative which makes it -0.9682