Toán Lớp 11: Giúp em với 1.Sin^2 2x-cos^2 8x=sin(10x+17π/2) 2.sin(2x+17π/2)-3cos(x-15π/2)=1+2sinx 3.2sin^3 x-sinx=2cos^3x-cosx+cos2x

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Toán Lớp 11:  Giúp em với 1.Sin^2 2x-cos^2 8x=sin(10x+17π/2) 2.sin(2x+17π/2)-3cos(x-15π/2)=1+2sinx 3.2sin^3 x-sinx=2cos^3x-cosx+cos2x

dongoctuan · Answer

Giải đáp:       Lời giải và giải thích chi tiết:   1. $sin(10x +  dfrac{17&pi;}{2}) = sin(10x +  dfrac{&pi;}{2} + 8&pi;)$   $ = sin(10x +  dfrac{&pi;}{2}) = cos10x$   $ PT &hArr; 2sin&sup2;2x - 2cos&sup2;8x = 2cos10x$   $ &hArr; (1 - cos4x) - (1 + cos16x) = 2cos10x$   $ &hArr; (cos16x + cos4x) + 2cos10x = 0$   $ &hArr; 2cos10xcos6x + 2cos10x = 0$   $ &hArr; 2cos10x(cos6x + 1) = 0$   - TH1 $: cos10x = 0 &hArr; 10x = (2k + 1) dfrac{&pi;}{2} &hArr; x = (2k + 1) dfrac{&pi;}{20} $   - TH2 $: cos6x = - 1 &hArr; 6x = (2k +1)&pi; &hArr; x = (2k + 1) dfrac{&pi;}{6} $   2. Tương tự c&acirc;u 1$: sin(2x +  dfrac{17&pi;}{2}) = cos2x$   $ cos(x -  dfrac{15&pi;}{2}) = cos(x +  dfrac{&pi;}{2} - 8&pi;)$   $ = cos(x +  dfrac{&pi;}{2}) = - sinx$   $ PT &hArr; cos2x + 3sinx = 1 + 2sinx$   $ &hArr; (1 - cos2x) - sinx = 0$   $ &hArr; 2sin&sup2;x - sinx = 0$   $ &hArr; sinx(2sinx - 1) = 0$   - TH1 $: sinx = 0 &hArr; x = k&pi;$   - TH2 $: sinx =  dfrac{1}{2} &hArr; x =  dfrac{&pi;}{6} + 2k&pi;;  x =  dfrac{5&pi;}{6} + 2k&pi; $   3. $PT &hArr; - sinx(1 - 2sin&sup2;x) - cosx(2cos&sup2;x - 1) - cos2x = 0$   $ &hArr; - sinxcos2x - cosxcos2x - cos2x = 0$   $ &hArr; - cos2x(sinx + cosx + 1) = 0$   - TH1 $: cos2x = 0 &hArr; 2x = (2k + 1) dfrac{&pi;}{2} &hArr; x = (2k + 1) dfrac{&pi;}{4} $   - TH2 $: sinx + cosx = - 1 $   $ &hArr;  sqrt{2}sin(x +  dfrac{&pi;}{4}) = - 1 &hArr; sin(x +  dfrac{&pi;}{4}) = -  dfrac{ sqrt{2}}{2}$   $  x +  dfrac{&pi;}{4} = -  dfrac{&pi;}{4} + k2&pi; &hArr; x = -  dfrac{&pi;}{2} + k2&pi; $   $  x +  dfrac{&pi;}{4} = -  dfrac{3&pi;}{4} + k2&pi; &hArr; x = - &pi; + k2&pi; $