Chương 6: CUNG VÀ GÓC LƯỢNG GIÁC. CÔNG THỨC LƯỢNG GIÁC - Hỏi đáp

Áp dụng công thức biến tích thành tổng:

\(cos\left(a+b\right).cos\left(a-b\right)=\dfrac{1}{2}\left(cos2a+cos2b\right)\)

\(=\dfrac{1}{2}\left(2cos^2a-1+1-2sin^2b\right)=\dfrac{1}{2}\left(2cos^2a-2sin^2b\right)\)

\(=cos^2a-sin^2b\)

\(cos\left(\dfrac{\pi}{4}+a\right).cos\left(\dfrac{\pi}{4}-a\right)+\dfrac{1}{2}sin^2a=\dfrac{1}{2}\left(cos\dfrac{\pi}{2}+cos2a\right)+\dfrac{1}{2}sin^2a\)

\(=\dfrac{1}{2}cos2a+\dfrac{1}{2}sin^2a=\dfrac{1}{2}\left(cos^2a-sin^2a\right)+\dfrac{1}{2}sin^2a\)

\(=\dfrac{1}{2}cos^2a\)

\(\dfrac{cosa+cos5a+cos3a}{sina+sin5a+sin3a}=\dfrac{2cos3a.cos2a+cos3a}{2sin3a.cos2a+sin3a}\)

\(=\dfrac{cos3a\left(2cos2a+1\right)}{sin3a\left(2cos2a+1\right)}=\dfrac{cos3a}{sin3a}=cot3a\)

\(\left(\dfrac{cosa}{sinb}+\dfrac{sina}{cosb}\right)\left(\dfrac{1-cos4b}{cos\left(a-b\right)}\right)=\dfrac{\left(cosa.cosb+sina.sinb\right)}{sinb.cosb}.\dfrac{2sin^22b}{cos\left(a-b\right)}\)

\(=\dfrac{cos\left(a-b\right)}{\dfrac{1}{2}sin2b}.\dfrac{2sin^22b}{cos\left(a-b\right)}=4sin2b\)

\(\dfrac{1+cos2a-sin2a}{1+cos2a+sin2a}=\dfrac{2cos^2a-2sina.cosa}{2cos^2a+2sinacosa}\)

\(=\dfrac{2cosa\left(cosa-sina\right)}{2cosa\left(cosa+sina\right)}=\dfrac{cosa-sina}{cosa+sina}=\dfrac{\sqrt{2}sin\left(\dfrac{\pi}{4}-a\right)}{\sqrt{2}cos\left(\dfrac{\pi}{4}-a\right)}=tan\left(\dfrac{\pi}{4}-a\right)\)

\(\dfrac{1+cos2a-cosa}{sin2a-sina}=\dfrac{2cos^2a-cosa}{2sina.cosa-sina}=\dfrac{cosa\left(2cosa-1\right)}{sina\left(2cosa-1\right)}=\dfrac{cosa}{sina}=cota\)

i

với (cosx khác 0)

VT: \(\dfrac{cosx+sinx}{cosx^3}=\dfrac{\dfrac{cosx}{cosx}+\dfrac{sinx}{cosx}}{\dfrac{cosx^3}{cosx}}=\dfrac{1+tanx}{cosx^2}\)

VP:

\(tanx^3+tanx^2+tanx+1=\left(tanx+1\right)\left(tanx^2+1\right)\\ =\left(tanx+1\right).\dfrac{1}{cosx^2+1}\)

Vậy VT=VP

\(\sqrt{9+tan^4A}\ge\sqrt{2\sqrt{9.tan^4A}}=\sqrt{6}.tanA\) , chứng minh tương tự

\(\Rightarrow\sqrt{9+tan^4A}+\sqrt{9+tan^4B}+\sqrt{9+tan^4C}\ge\sqrt{6}\left(tanA+tanB+tanC\right)\)

lại có trong tam giác ABC:

\(A+B+C=180^0\Rightarrow A+B=180^0-C\Rightarrow tan\left(A+B\right)=tan\left(180^0-C\right)\)

\(\Rightarrow\dfrac{tanA+tanB}{1-tanA.tanB}=tan\left(180^0-C\right)=-tanC\)

\(\Rightarrow tanA+tanB=-tanC+tanA.tanB.tanC\)

\(\Rightarrow tanA+tanB+tanC=tanA.tanB.tanC\)

Do ABC là tam giác nhọn \(\Rightarrow tanA,tanB,tanC>0\)

Áp dụng BĐT Cauchy: \(tanA+tanB+tanC\ge3\sqrt[3]{tanA.tanB.tanC}\)

\(\Rightarrow\dfrac{\left(tanA+tanB+tanC\right)^3}{27}\ge tanA.tanB.tanC=tanA+tanB+tanC\)

\(\Rightarrow\left(tanA+tanB+tanC\right)^2\ge27\) \(\Rightarrow tanA+tanB+tanC\ge3\sqrt{3}\)

\(\Rightarrow\sqrt{6}\left(tanA+tanB+tanC\right)\ge\sqrt{6}.3\sqrt{3}=9\sqrt{2}\)

\(\Rightarrow\sqrt{9+tan^4A}+\sqrt{9+tan^4B}+\sqrt{9+tan^4C}\ge9\sqrt{2}\)

Dấu "=" xảy ra khi tam giác ABC đều

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