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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 11: Tìm nghiệm x thuộc(-90°,360°) của pt: sin^2(x+45°)-sin^2(x-30°)=sin15°.cos(2x+15°)

Tháng Bảy 16, 2022 Toán học 1 Comment 0 views

Toán Lớp 11: Tìm nghiệm x thuộc(-90°,360°) của pt: sin^2(x+45°)-sin^2(x-30°)=sin15°.cos(2x+15°)

Comments ( 1 )

  1. quynhthanhnhu
    quynhthanhnhu
    16 Tháng Bảy, 2022 at 7:42 sáng
    Reply
    Chứng minh: ${\sin ^2}a – {\sin ^2}b = \sin \left( {a + b} \right)\sin \left( {a – b} \right)$
    $\begin{array}{l} \sin \left( {a + b} \right)\sin \left( {a – b} \right)\\  = \left( {\sin a\cos b + \sin b\cos a} \right)\left( {\sin a\cos b – \sin b\cos a} \right)\\  = {\left( {\sin a\cos b} \right)^2} – {\left( {\cos a\sin b} \right)^2}\\  = {\sin ^2}a\left( {1 – {{\sin }^2}b} \right) – \left( {1 – {{\sin }^2}a} \right){\sin ^2}b\\  = {\sin ^2}a – {\sin ^2}b\left( {dpcm} \right) \end{array}$
    $\begin{array}{l}
    {\sin ^2}\left( {x + {{45}^o}} \right) – {\sin ^2}\left( {x – {{30}^o}} \right) = \sin {15^o}\cos \left( {2x + {{15}^o}} \right)\\
     \Leftrightarrow \sin \left( {x + {{45}^o} + x – {{30}^o}} \right)\sin \left( {x + {{45}^o} – x + {{30}^o}} \right) = \sin {15^o}\cos \left( {2x + {{15}^o}} \right)\\
     \Leftrightarrow \sin \left( {2x + {{15}^o}} \right)\sin {75^o} = \sin {15^o}\cos {\left( {2x + 15} \right)^o}\\
     \Leftrightarrow \sin \left( {2x + {{15}^o}} \right)\cos {15^o} – \sin {15^o}\cos {\left( {2x + 15} \right)^o} = 0\\
     \Leftrightarrow \sin \left( {2x + {{15}^o} – {{15}^o}} \right) = 0\\
     \Leftrightarrow \sin 2x = 0\\
     \Leftrightarrow 2x = k{180^o}\\
     \Leftrightarrow x = k{90^o}\\
    x \in \left( { – {{90}^o};360} \right) \Rightarrow k = 0,1,2,3\\
     \Rightarrow x = {0^o},{90^o},{180^o},{270^o}
    \end{array}$

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222-9+11+12:2*14+14 = ? ( )

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