1976 AHSME Problems/Problem 17

Problem 17

If $\theta$ is an acute angle, and $\sin 2\theta=a$ , then $\sin\theta+\cos\theta$ equals

$\textbf{(A) }\sqrt{a+1}\qquad \textbf{(B) }(\sqrt{2}-1)a+1\qquad \textbf{(C) }\sqrt{a+1}-\sqrt{a^2-a}\qquad\\ \textbf{(D) }\sqrt{a+1}+\sqrt{a^2-a}\qquad \textbf{(E) }\sqrt{a+1}+a^2-a$

Solution

Let $x = \sin\theta+\cos\theta$ , so we want to find $x$ . First, square the expression to get $x^2 = \sin^2 \theta + 2\sin\theta\cos\theta + \cos^2 \theta$ . Recall that $\sin^2 \theta + \cos^2 \theta = 1$ and $\sin 2\theta = 2\sin\theta\cos\theta$ . Plugging these in, the equation simplifies to $x^2 = 1 + \sin 2\theta$ . Given that $\sin 2\theta=a$ , the equation becomes $x^2=1+a$ . Take the square root of both sides to get $x = \sqrt{1+a}$ .

Hence, the answer is $\boxed{\textbf{(A) }\sqrt{a+1}}$ ~jiang147369

1976 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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1976 AHSME Problems/Problem 17

Problem 17

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