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Sample Question 9:
Step 1: Consider the equation \(\sin(x+y)\leq \sin(x)+\sin(y)\). In the interval [0, π], sine is nonnegative.
Step 2: We can rewrite the original equation using the addition formula for sine, which is \(\sin(x + y) = \sin x \cos y + \sin y \cos x\).
Step 3: By doing so, we get \(\sin x \cos y + \sin y \cos x \le \sin x + \sin y\).
Step 4: We can see that equality is obtained when \(\cos x = \cos y = 1\), which is the maximum value of cosine. Therefore, the largest subset of values of \(y\) for which the equation is satisfied is all \(y\) in the interval [0, π].
Final Answer: \(\textbf{(E) } 0\le y\le \pi\).
Which of the following describes the largest subset of values of \(y\) within the closed interval \([0,\pi]\) for which \(\sin(x+y)\leq \sin(x)+\sin(y)\)for every \(x\) between \(0\) and \(\pi\), inclusive?
\(\textbf{(A) } y=0 \qquad \textbf{(B) } 0\leq y\leq \frac{\pi}{4} \qquad \textbf{(C) } 0\leq y\leq \frac{\pi}{2} \qquad \textbf{(D) } 0\leq y\leq \frac{3\pi}{4} \qquad \textbf{(E) } 0\leq y\leq \pi\)
Answer Keys
Question 9: ESolutions
Question 9Step 1: Consider the equation \(\sin(x+y)\leq \sin(x)+\sin(y)\). In the interval [0, π], sine is nonnegative.
Step 2: We can rewrite the original equation using the addition formula for sine, which is \(\sin(x + y) = \sin x \cos y + \sin y \cos x\).
Step 3: By doing so, we get \(\sin x \cos y + \sin y \cos x \le \sin x + \sin y\).
Step 4: We can see that equality is obtained when \(\cos x = \cos y = 1\), which is the maximum value of cosine. Therefore, the largest subset of values of \(y\) for which the equation is satisfied is all \(y\) in the interval [0, π].
Final Answer: \(\textbf{(E) } 0\le y\le \pi\).