| Nội dung đăng: | 1/ 2√3.sin(x-π/8)cos(x-π/8) + 2cos²(x-π/8) = √3 + 4[sin²x + cos(π/3-x).cos(π/3+x)] ⇔ √3.sin(2x - π/4) + 1 + cos(2x - π/4) = √3 + 2(1 - cos2x) - 1 + 2cos2x ⇔ √3.sin(2x - π/4) + cos(2x - π/4) = √3 ⇔ √3/2.sin(2x - π/4) + 1/2.cos(2x - π/4) = ... |
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Thưởng th.3.2024
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