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\(M=sinx.cosx+\dfrac{sin^2x}{1+cotx}+\dfrac{cos^2x}{1+tanx}\)
\(=sinx.cosx+\dfrac{sin^2x}{\dfrac{cosx+sinx}{sinx}}+\dfrac{cos^2x}{\dfrac{cosx+sinx}{cosx}}\)
\(=sinx.cosx+\dfrac{sin^3x+cos^3x}{cosx+sinx}\)
\(=sinx.cosx+\dfrac{\left(sinx+cosx\right)\left(sin^2x+cos^2x-sinx.cosx\right)}{cosx+sinx}\)
\(=sinx.cosx+sin^2x+cos^2x-sinx.cosx\)
\(=sin^2x+cos^2x=1\)
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\(cosx=\sqrt{1-\dfrac{7}{16}}=\dfrac{3}{4}\)
\(tanx=\dfrac{\sqrt{7}}{4}:\dfrac{3}{4}=\dfrac{\sqrt{7}}{3}\)
\(cotx=1:\dfrac{\sqrt{7}}{3}=\dfrac{3}{\sqrt{7}}=\dfrac{3\sqrt{7}}{7}\)
\(M=\left(\dfrac{3}{7}\sqrt{7}+\dfrac{1}{3}\sqrt{7}\right):\left(\dfrac{3}{7}\sqrt{7}-\dfrac{1}{3}\sqrt{7}\right)\)
\(=\dfrac{16}{21}:\dfrac{2}{21}=8\)
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a) sin = đối / huyền => sinx < 1 => sinx - 1 < 0
b) cos = kề / huyền => cosx < 1 => 1 - cosx > 0
c) sinx - cosx = sinx - sin(90-x)
Nếu x > 90-x hay x > 45 thì sinx - sin(90-x) > 0 hay sinx - cosx > 0
Nếu x < 90-x hay x < 45 thì sinx - sin(90-x) < 0 hay sinx - cosx < 0
d) Tương tự câu c)
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Đinh Phương Linh tham khảo nha:
2sin²2x + sin7x - 1 = sinx <=> sin7x - sinx - cos4x = 0
<=> 2cos4x.sin3x - cos4x = 0 <=> cos4x.(2sin3x -1) = 0
<=> [ cos4x = 0
----- [ sin3x = 1/2
<=> [ x = pi/8 + kpi/4
----- [ x = pi/18 + 2kpi/3
----- [ x = 5pi/18 + 2kpi