§1. Giá trị lượng giác của một góc bất kỳ từ 0 (độ) đến 180 (độ)

\(\dfrac{\pi}{2}< a< \pi\Rightarrow\left\{{}\begin{matrix}sina>0\\cosa< 0\end{matrix}\right.\) \(\Rightarrow tana< 0\)

\(tana-3cota=2\Leftrightarrow tana-\dfrac{3}{tana}=2\)

\(\Leftrightarrow tan^2a-2tana-3=0\Rightarrow\left[{}\begin{matrix}tana=-1\\tana=3>0\left(loại\right)\end{matrix}\right.\)

\(\dfrac{1}{cos^2a}=1+tan^2a\Rightarrow cosa=-\sqrt{\dfrac{1}{1+tan^2a}}=-\dfrac{\sqrt{2}}{2}\)

\(sina=cosa.tana=\dfrac{\sqrt{2}}{2}\)

Chắc là \(0< a< \dfrac{\pi}{2}\)?

\(0< a< \dfrac{\pi}{2}\Rightarrow sina;cosa>0\)

\(\left\{{}\begin{matrix}sina=\sqrt{3}cosa\\sin^2a+cos^2a=1\end{matrix}\right.\) \(\Rightarrow\left(\sqrt{3}cosa\right)^2+cos^2a=1\)

\(\Rightarrow4cos^2a=1\Rightarrow cosa=\dfrac{1}{2}\)

\(\Rightarrow sina=\sqrt{3}cosa=\dfrac{\sqrt{3}}{2}\)

Bạn kiểm tra lại đề, có vẻ như trong 2 cái \(sin^2\) kia phải có 1 cái là \(cos^2\) mới hợp lý

\(P=\dfrac{cos^237+sin^2143+sin26}{1+sin154}=\dfrac{cos^237+sin^2\left(180-37\right)+sin26}{1+sin\left(180-26\right)}\)

\(=\dfrac{cos^237+sin^237+sin26}{1+sin26}=\dfrac{1+sin26}{1+sin26}=1\)

\(\dfrac{A}{2}+\dfrac{B}{2}=\dfrac{\pi}{2}-\dfrac{C}{2}\Rightarrow tan\left(\dfrac{A}{2}+\dfrac{B}{2}\right)=tan\left(\dfrac{\pi}{2}-\dfrac{C}{2}\right)\)

\(\Rightarrow\dfrac{tan\dfrac{A}{2}+tan\dfrac{B}{2}}{1-tan\dfrac{A}{2}tan\dfrac{B}{2}}=cot\dfrac{C}{2}=\dfrac{1}{tan\dfrac{C}{2}}\)

\(\Rightarrow tan\dfrac{A}{2}.tan\dfrac{C}{2}+tan\dfrac{B}{2}tan\dfrac{C}{2}=1-tan\dfrac{A}{2}tan\dfrac{B}{2}\)

\(\Rightarrow tan\dfrac{A}{2}tan\dfrac{B}{2}+tan\dfrac{B}{2}tan\dfrac{C}{2}+tan\dfrac{C}{2}tan\dfrac{A}{2}=1\)

Ta có:

\(tan\dfrac{A}{2}+tan\dfrac{B}{2}+tan\dfrac{C}{2}\ge\sqrt{3\left(tan\dfrac{A}{2}tan\dfrac{B}{2}+tan\dfrac{B}{2}tan\dfrac{C}{2}+tan\dfrac{C}{2}tan\dfrac{A}{2}\right)}=\sqrt{3}\)

Dấu "=" xảy ra khi và chỉ khi \(A=B=C\) hay tam giác ABC đều

- Áp dụng BĐT bunhiacopxki ta có :

\(\left(3sinx+4cosx\right)^2\le\left(3^2+4^2\right)\left(sin^2x+cos^2x\right)=25\)

\(\Leftrightarrow-5\le M\le5\)

P/s : Chắc là đề nhầm :vvv nếu không nhầm thì thêm bớt rồi bunhi xong cộng với cos thêm vào nha

\(y=\sqrt{2}sin\left(2x+\dfrac{\pi}{4}\right)+3\)

Do \(sin\left(2x+\dfrac{\pi}{4}\right)\le1\Rightarrow y\le3+\sqrt{2}\)

\(\Rightarrow a=3;b=1\Rightarrow a+b=\)

\(S=sinx+siny+sin\left(3x+y\right)-sin\left(3x+y\right)-sin\left(x+y\right)\)

\(=sinx+siny-sin\left(x+y\right)\)

\(S^2=\left(sinx+siny-sin\left(x+y\right)\right)^2\le3\left(sin^2x+sin^2y+sin^2\left(x+y\right)\right)\)

\(S^2\le3\left(1-\dfrac{1}{2}\left(cos2x+cos2y\right)+sin^2\left(x+y\right)\right)\)

\(S^2\le3\left[1-cos\left(x+y\right)cos\left(x-y\right)+1-cos^2\left(x-y\right)\right]\)

\(S^2\le3\left[2+\dfrac{1}{4}cos^2\left(x+y\right)-\left[cos\left(x-y\right)-\dfrac{1}{2}cos\left(x+y\right)\right]^2\right]\le3\left[2+\dfrac{1}{4}cos^2\left(x+y\right)\right]\)

\(S^2\le3\left(2+\dfrac{1}{4}\right)=\dfrac{27}{4}\)

\(\Rightarrow S\le\dfrac{3\sqrt{3}}{2}\)

\(\Rightarrow\left\{{}\begin{matrix}a=3\\b=3\\c=2\end{matrix}\right.\)

\(bc.cosA=bc\left(\dfrac{b^2+c^2-a^2}{2bc}\right)=\dfrac{b^2+c^2-a^2}{2}\)

Tương tự: \(ac.cosB=\dfrac{a^2+c^2-b^2}{2}\) ; \(ab.cosC=\dfrac{a^2+b^2-c^2}{2}\)

\(\Rightarrow Q=\dfrac{a^2+b^2+c^2}{2S}\ge\dfrac{\left(a+b+c\right)^2}{6S}=\dfrac{4p^2}{6\sqrt{p\left(p-a\right)\left(p-b\right)\left(p-c\right)}}\)

\(Q\ge\dfrac{2p\sqrt{p}}{3\sqrt{\left(p-a\right)\left(p-b\right)\left(p-c\right)}}\ge\dfrac{2p\sqrt{p}}{3\sqrt{\left(\dfrac{3p-\left(a+b+c\right)}{3}\right)^3}}=\dfrac{2p\sqrt{p}}{3\sqrt{\dfrac{p^3}{27}}}=2\sqrt{3}\)

\(A=\dfrac{4sin^4x-cos^2x\left(1-cos^2x\right)+sin^2x.cos^2x-2cos^2x}{sin^2x}+\dfrac{2}{tan^2x}\)

\(=\dfrac{4sin^4x-sin^2x.cos^2x+sin^2x.cos^2x-2cos^2x}{sin^2x}+2cot^2x\)

\(=4sin^2x-2cot^2x+2cot^2x=4sin^2x\)

\(\Rightarrow\left\{{}\begin{matrix}a=4\\b=2\end{matrix}\right.\)