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Algebra Difficulty 3.3 AMC 10/12 Find the answer

Among the following four propositions, the true one is((  ))
A: There exists x(0,π)x \in (0,\pi) such that sinx=tanx\sin x = \tan x
B: The negation of "For any xRx \in \mathbb{R}, x2+x+1>0x^2 + x + 1 > 0" is "There exists x0Rx_0 \in \mathbb{R} such that x02+x0+1<0x_0^2 + x_0 + 1 < 0"
C: For all θR\theta \in \mathbb{R}, the function f(x)=sin(2x+θ)f(x) = \sin(2x + \theta) is not an even function
D: In triangle ABCABC, "sinA+sinB=cosA+cosB\sin A + \sin B = \cos A + \cos B" is a necessary and sufficient condition for "C=π2C = \frac{\pi}{2}"

This was a multiple-choice question, but the options didn't survive into the source we have. The answer given is D, and the solution below works it through.

Solution

Solution:
A. If sinx=tanx\sin x = \tan x, then sinx=tanx=sinxcosx\sin x = \tan x = \frac{\sin x}{\cos x},
Since x(0,π)x \in (0,\pi), we have sinx0\sin x \neq 0, which implies 1=1cosx1 = \frac{1}{\cos x}, i.e., cosx=1\cos x = 1,
But since x(0,π)x \in (0,\pi), cosx=1\cos x = 1 does not hold. Therefore, the statement that there exists x(0,π)x \in (0,\pi) such that sinx=tanx\sin x = \tan x is false. Hence, option A is incorrect.
B. The negation of "For any xRx \in \mathbb{R}, x2+x+1>0x^2 + x + 1 > 0" is "There exists x0Rx_0 \in \mathbb{R} such that x02+x0+10x_0^2 + x_0 + 1 \leqslant 0". Therefore, option B is incorrect.
C. When θ=π2\theta = \frac{\pi}{2}, f(x)=sin(2x+θ)=sin(2x+π2)=cos2xf(x) = \sin(2x + \theta) = \sin(2x + \frac{\pi}{2}) = \cos 2x is an even function. Hence, option C is incorrect.
D. In triangle ABCABC, if C=π2C = \frac{\pi}{2}, then A+B=π2A + B = \frac{\pi}{2},
Thus, from sinA+sinB=sin(π2B)+sin(π2A)=cosB+cosA\sin A + \sin B = \sin(\frac{\pi}{2} - B) + \sin(\frac{\pi}{2} - A) = \cos B + \cos A, the necessity holds;
Since sinA+sinB=cosA+cosB\sin A + \sin B = \cos A + \cos B,
We have sinAcosA=cosBsinB\sin A - \cos A = \cos B - \sin B,
Squaring both sides gives sin2A2sinAcosA+cos2A=sin2B2sinBcosB+cos2B\sin^2 A - 2\sin A\cos A + \cos^2 A = \sin^2 B - 2\sin B\cos B + \cos^2 B,
Thus, 12sinAcosA=12sinBcosB1 - 2\sin A\cos A = 1 - 2\sin B\cos B,
Therefore, sin2A=sin2B\sin 2A = \sin 2B,
This implies 2A=2B2A = 2B or 2A=π2B2A = \pi - 2B,
i.e., A=BA = B or A+B=π2A + B = \frac{\pi}{2},
When A=BA = B, "sinA+sinB=cosA+cosB\sin A + \sin B = \cos A + \cos B" is equivalent to 2sinA=2cosA2\sin A = 2\cos A,
Thus, tanA=1\tan A = 1, i.e., A=B=π4A = B = \frac{\pi}{4}, and in this case, C=π2C = \frac{\pi}{2},
In summary, C=π2C = \frac{\pi}{2} always holds, which means the sufficiency is established,
Overall, in triangle ABCABC, "sinA+sinB=cosA+cosB\sin A + \sin B = \cos A + \cos B" is a necessary and sufficient condition for "C=π2C = \frac{\pi}{2}". Therefore, option D is correct.
Hence, the choice is: D\boxed{D}

A. Judged based on the properties of trigonometric functions.
B. Judged based on the negation of a universal proposition being an existential proposition.
C. Judged based on the odd-even properties of trigonometric functions.
D. Judged based on the definitions of necessary and sufficient conditions, using the squaring method.
This question mainly examines the judgment of the truthfulness of propositions, involving a wide range of knowledge points and requiring strong comprehensive skills to test students' calculation and reasoning abilities.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.