Difference between revisions of "University of South Carolina High School Math Contest/1993 Exam/Problem 13"

Revision as of 20:27, 22 July 2006

Suppose that $x$ and $y$ are numbers such that $\sin(x+y) = 0.3$ and $\sin(x-y) = 0.5$ . Then $\sin(x)\cos(y)=$

$\mathrm{(A) \ }0.1 \qquad \mathrm{(B) \ }0.3 \qquad \mathrm{(C) \ }0.4 \qquad \mathrm{(D) \ }0.5 \qquad \mathrm{(E) \ }0.6$

Expanding $\sin{(x+y)}$ and $\sin{(x-y)}$ , we have: $(1)$ $\sin{x}\cos{y}+\sin{y}\cos{x}=.3$ $(2)$ $\sin{x}\cos{y}-\sin{y}\cos{x}=.5$

$(1)+(2)$ yields $2\sin{x}\cos{y}=.8$ and our answer is $.4$ .

@@ Line 1: / Line 1: @@
 == Problem ==
+Suppose that <math>x</math> and <math>y</math> are numbers such that <math>\sin(x+y) = 0.3</math> and <math>\sin(x-y) = 0.5</math>.  Then <math>\sin(x)\cos(y)=</math>
-<center><math> \mathrm{(A) \ } \qquad \mathrm{(B) \ } \qquad \mathrm{(C) \ } \qquad \mathrm{(D) \ } \qquad \mathrm{(E) \ }  </math></center>
+<center><math> \mathrm{(A) \ }0.1 \qquad \mathrm{(B) \ }0.3 \qquad \mathrm{(C) \ }0.4 \qquad \mathrm{(D) \ }0.5 \qquad \mathrm{(E) \ }0.6  </math></center>
 == Solution ==
+Expanding <math>\sin{(x+y)}</math> and <math>\sin{(x-y)}</math>, we have:
+<math>(1)</math> <math>\sin{x}\cos{y}+\sin{y}\cos{x}=.3</math>
+<math>(2)</math> <math>\sin{x}\cos{y}-\sin{y}\cos{x}=.5</math>
+<math>(1)+(2)</math> yields <math>2\sin{x}\cos{y}=.8</math> and our answer is <math>.4</math>.
 == See also ==
 * [[University of South Carolina High School Math Contest/1993 Exam]]
+[[Category:Intermediate Trigonometry Problems]]