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Do \(\alpha\in\left(\frac{\pi}{2};\frac{3\pi}{4}\right)\Rightarrow sin\alpha>0;cos\alpha< 0;tan\alpha< 0\)
\(\frac{tana}{cota}=\frac{\sqrt{5}-1}{\sqrt{5}+1}\Leftrightarrow tan^2a=\frac{\sqrt{5}-1}{\sqrt{5}+1}=\frac{\left(\sqrt{5}-1\right)^2}{4}\Rightarrow tana=\frac{1-\sqrt{5}}{2}\Rightarrow cota=\frac{-1-\sqrt{5}}{2}\)
\(1+tan^2a=\frac{1}{cos^2a}\Rightarrow cos^2a=\frac{1}{1+tan^2a}=\frac{5+\sqrt{5}}{10}\)
\(\Rightarrow sin^2a=1-cos^2a=\frac{5-\sqrt{5}}{10}\)
\(sin2a=2sina.cosa=2tana.cos^2a=-\frac{2\sqrt{5}}{5}\)
Thay vào ta được:
\(P=...\)
Bạn tự thay số và bấm máy
1.
\(\Leftrightarrow3x=k\pi\Leftrightarrow x=\frac{k\pi}{3}\)
2.
\(\Leftrightarrow cos5x=0\Leftrightarrow5x=\frac{\pi}{2}+k\pi\Leftrightarrow x=\frac{\pi}{10}+\frac{k\pi}{5}\)
4.
\(cos3x+cosx+cos2x=0\)
\(\Leftrightarrow2cos2x.cosx+cos2x=0\)
\(\Leftrightarrow cos2x\left(2cosx+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\cosx=-\frac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+\frac{k\pi}{2}\\x=\pm\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
5.
\(sin6x+sin2x+sin4x=0\)
\(\Leftrightarrow2sin4x.cos2x+sin4x=0\)
\(\Leftrightarrow sin4x\left(2cos2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin4x=0\\cos2x=-\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{k\pi}{4}\\x=\pm\frac{\pi}{3}+k\pi\end{matrix}\right.\)
6. ĐKXĐ; ...
\(\Leftrightarrow tanx+tan2x=1-tanx.tan2x\)
\(\Leftrightarrow\frac{tanx+tan2x}{1-tanx.tan2x}=1\)
\(\Leftrightarrow tan3x=1\)
\(\Leftrightarrow x=\frac{\pi}{12}+\frac{k\pi}{3}\)
1.a) \(4cos\dfrac{\alpha}{2}.cos\dfrac{\beta}{2}.cos\dfrac{f}{2}\)
\(=\dfrac{1}{2}.4\left[cos\left(\dfrac{\alpha-\beta}{2}\right)+cos\left(\dfrac{\alpha+\beta}{2}\right)\right].cos\dfrac{f}{2}\)
\(=2.cos\left(\dfrac{\alpha-\beta}{2}\right)cos\dfrac{f}{2}+2.cos\left(\dfrac{\alpha+\beta}{2}\right).cos\dfrac{f}{2}\)
\(=cos\left(\dfrac{\alpha-\left(\beta+f\right)}{2}\right)+cos\left(\dfrac{\alpha-\beta+f}{2}\right)+cos\left(\dfrac{\alpha+\beta-f}{2}\right)+cos\left(\dfrac{\alpha+\beta+f}{2}\right)\)
\(=cos\left(\dfrac{2\alpha-\pi}{2}\right)+cos\left(\dfrac{\pi-2\beta}{2}\right)+cos\left(\dfrac{\pi-2f}{2}\right)+cos\left(\dfrac{\pi}{2}\right)\)
\(=cos\left(-\dfrac{\pi}{2}+\alpha\right)+cos\left(\dfrac{\pi}{2}-\beta\right)+cos\left(\dfrac{\pi}{2}-f\right)\)
\(=sin\alpha+sin\beta+sinf\) (đpcm)
a2) \(1+4sin\dfrac{\alpha}{2}.sin\dfrac{\beta}{2}.sin\dfrac{f}{2}\)
\(=1+2\left[cos\left(\dfrac{\alpha-\beta}{2}\right)-cos\left(\dfrac{\alpha+\beta}{2}\right)\right].sin\dfrac{f}{2}\)
\(=1+2.cos\left(\dfrac{\alpha-\beta}{2}\right).sin\dfrac{f}{2}-2.cos\left(\dfrac{\alpha+\beta}{2}\right).sin\dfrac{f}{2}\)
\(=1+sin\left(\dfrac{f-\alpha+\beta}{2}\right)+sin\left(\dfrac{a-\beta+f}{2}\right)-sin\left(\dfrac{f-\left(\alpha+\beta\right)}{2}\right)-sin\left(\dfrac{\alpha+\beta+f}{2}\right)\)
\(=1+sin\left(\dfrac{\pi-2\alpha}{2}\right)+sin\left(\dfrac{\pi-2\beta}{2}\right)-sin\left(\dfrac{2f-\pi}{2}\right)-sin\left(\dfrac{\pi}{2}\right)\)
\(=sin\left(\dfrac{\pi}{2}-\alpha\right)+sin\left(\dfrac{\pi}{2}-\beta\right)+sin\left(\dfrac{\pi}{2}-f\right)\)
\(=cos\alpha+cos\beta+cosf\) (đpcm)
1) đặc : \(f\left(x\right)=y=cot4x\)
điều kiện xác định : \(sin4x\ne0\Leftrightarrow4x\ne k\pi\Leftrightarrow x\ne\dfrac{k\pi}{4}\)
\(\Rightarrow x\in D\) thì \(-x\in D\)
ta có : \(f\left(-x\right)=cot\left(-4x\right)=-cot4x=-f\left(x\right)\)
\(\Rightarrow\) hàm này là hàm lẽ
2) đặc : \(f\left(x\right)=y=\left|cotx\right|\)
điều kiện xác định : \(sinx\ne0\Leftrightarrow x\ne k\pi\)
\(\Rightarrow x\in D\) thì \(-x\in D\)
ta có : \(f\left(-x\right)=\left|cot\left(-x\right)\right|=\left|-cotx\right|=\left|cotx\right|=f\left(x\right)\)
\(\Rightarrow\) hàm này là hàm chẳn
3) đặc : \(f\left(x\right)=y=1-sin^2x=cos^2x\)
điều kiện xác định : \(D=R\)
\(\Rightarrow x\in D\) thì \(-x\in D\)
ta có : \(f\left(-x\right)=cos^2\left(-x\right)=cos^2x=f\left(x\right)\)
\(\Rightarrow\) hàm này là hàm chẳn
4) đặc : \(f\left(x\right)=y=sin\left(x+\dfrac{\pi}{4}\right)=\dfrac{sinx+cosx}{\sqrt{2}}\)
điều kiện xác định : \(D=R\)
\(\Rightarrow x\in D\) thì \(-x\in D\)
ta có : \(f\left(-x\right)=\dfrac{sin\left(-x\right)+cos\left(-x\right)}{\sqrt{2}}=\dfrac{-sinx+cosx}{\sqrt{2}}\ne f\left(x\right);-f\left(x\right)\)
\(\Rightarrow\) hàm này là hàm không chẳn không lẽ
mấy bài còn lại bn làm tương tự cho quen nha
2.
ĐK: \(2x-y\ge0;y\ge0;y-x-1\ge0;y-3x+5\ge0\)
\(\left\{{}\begin{matrix}xy-2y-3=\sqrt{y-x-1}+\sqrt{y-3x+5}\left(1\right)\\\left(1-y\right)\sqrt{2x-y}+2\left(x-1\right)=\left(2x-y-1\right)\sqrt{y}\left(2\right)\end{matrix}\right.\)
\(\left(2\right)\Leftrightarrow\left(1-y\right)\sqrt{2x-y}+y-1+2x-y-1-\left(2x-y-1\right)\sqrt{y}=0\)
\(\Leftrightarrow\left(1-y\right)\left(\sqrt{2x-y}-1\right)+\left(2x-y-1\right)\left(1-\sqrt{y}\right)=0\)
\(\Leftrightarrow\left(1-\sqrt{y}\right)\left(\sqrt{2x-y}-1\right)\left(1+\sqrt{y}\right)+\left(\sqrt{2x-y}-1\right)\left(1-\sqrt{y}\right)\left(\sqrt{2x-y}+1\right)=0\)
\(\Leftrightarrow\left(1-\sqrt{y}\right)\left(\sqrt{2x-y}-1\right)\left(\sqrt{y}+\sqrt{2x-y}+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=1\\y=2x-1\end{matrix}\right.\) (Vì \(\sqrt{y}+\sqrt{2x-y}+2>0\))
Nếu \(y=1\), khi đó:
\(\left(1\right)\Leftrightarrow x-5=\sqrt{-x}+\sqrt{-3x+6}\)
Phương trình này vô nghiệm
Nếu \(y=2x-1\), khi đó:
\(\left(1\right)\Leftrightarrow2x^2-5x-1=\sqrt{x-2}+\sqrt{4-x}\) (Điều kiện: \(2\le x\le4\))
\(\Leftrightarrow2x\left(x-3\right)+x-3+1-\sqrt{x-2}+1-\sqrt{4-x}=0\)
\(\Leftrightarrow\left(x-3\right)\left(\dfrac{1}{1+\sqrt{4-x}}-\dfrac{1}{1+\sqrt{x-2}}+2x+1\right)=0\)
Ta thấy: \(1+\sqrt{x-2}\ge1\Rightarrow-\dfrac{1}{1+\sqrt{x-2}}\ge-1\Rightarrow1-\dfrac{1}{1+\sqrt{x-2}}\ge0\)
Lại có: \(\dfrac{1}{1+\sqrt{4-x}}>0\); \(2x>0\)
\(\Rightarrow\dfrac{1}{1+\sqrt{4-x}}-\dfrac{1}{1+\sqrt{x-2}}+2x+1>0\)
Nên phương trình \(\left(1\right)\) tương đương \(x-3=0\Leftrightarrow x=3\Rightarrow y=5\)
Ta thấy \(\left(x;y\right)=\left(3;5\right)\) thỏa mãn điều kiện ban đầu.
Vậy hệ phương trình đã cho có nghiệm \(\left(x;y\right)=\left(3;5\right)\)
Theo giả thiết ta có 3 góc: \(\alpha;\beta=\alpha+\dfrac{\pi}{3};\gamma=\alpha+\dfrac{2\pi}{3}\).
Ta có:
\(tan\alpha.tan\left(\alpha+\dfrac{\pi}{3}\right)+tan\left(\alpha+\dfrac{\pi}{3}\right).tan\left(\alpha+\dfrac{2\pi}{3}\right)+\)\(tan\left(\alpha+\dfrac{2\pi}{3}\right).tan\alpha\)
\(=tan\alpha\left[tan\left(\alpha+\dfrac{\pi}{3}\right)+tan\left(\alpha+\dfrac{2\pi}{3}\right)\right]\)\(+tan\left(a+\dfrac{\pi}{3}\right)tan\left(\alpha+\dfrac{2\pi}{3}\right)\)
\(=tan\alpha\dfrac{sin\left(2\alpha+\pi\right)}{cos\left(\alpha+\dfrac{\pi}{3}\right)cos\left(\alpha+\dfrac{2\pi}{3}\right)}\)\(+\dfrac{sin\left(\alpha+\dfrac{\pi}{3}\right)sin\left(\alpha+\dfrac{2\pi}{3}\right)}{cos\left(\alpha+\dfrac{\pi}{3}\right)cos\left(\alpha+\dfrac{2\pi}{3}\right)}\)
\(=tan\alpha\dfrac{-sin2\alpha}{cos\left(\alpha+\dfrac{\pi}{3}\right)cos\left(\alpha+\dfrac{2\pi}{3}\right)}\)\(+\dfrac{cos\dfrac{\pi}{3}-cos\left(2\alpha+\pi\right)}{2cos\left(\alpha+\dfrac{\pi}{3}\right)cos\left(\alpha+\dfrac{2\pi}{3}\right)}\)
\(=\dfrac{-2sin^2\alpha}{cos\left(\alpha+\dfrac{\pi}{3}\right)cos\left(\alpha+\dfrac{2\pi}{3}\right)}\)\(+\dfrac{\dfrac{1}{2}+cos2\alpha}{2cos\left(\alpha+\dfrac{\pi}{3}\right)cos\left(\alpha+\dfrac{2\pi}{3}\right)}\)
\(=\dfrac{\dfrac{1}{2}-4sin^2\alpha+cos2\alpha}{2cos\left(\alpha+\dfrac{\pi}{3}\right)cos\left(\alpha+\dfrac{2\pi}{3}\right)}\)
\(=\dfrac{\dfrac{1}{2}-4\left(1-cos^2\alpha\right)+2cos^2\alpha-1}{cos\dfrac{\pi}{3}+cos\left(2\alpha+\pi\right)}\)
\(=\dfrac{6cos^2\alpha-\dfrac{9}{2}}{\dfrac{1}{2}-cos2\alpha}\)
\(=\dfrac{3\left(2cos^2\alpha-\dfrac{3}{2}\right)}{\dfrac{1}{2}-\left(2cos^2\alpha-1\right)}=\dfrac{3\left(2cos^2\alpha-\dfrac{3}{2}\right)}{\dfrac{3}{2}-2cos^2\alpha}=-3\).
\(4cos\alpha.cos\beta cos\gamma=4cos\alpha cos\left(\alpha+\dfrac{\pi}{3}\right)cos\left(\alpha+\dfrac{2\pi}{3}\right)\)
\(=4cos\alpha.\dfrac{1}{2}\left(cos\dfrac{\pi}{3}+cos\left(2\alpha+\pi\right)\right)\)
\(=4cos\alpha.\dfrac{1}{2}\left(\dfrac{1}{2}-cos2\alpha\right)\)
\(=cos\alpha-2cos\alpha.cos2\alpha\)
\(=cos\alpha-\left(cos\alpha+cos3\alpha\right)\)
\(=-cos3\alpha\)
\(=cos\left(\pi+3\alpha\right)\)
\(=cos3\left(\dfrac{\pi}{3}+\alpha\right)\)
\(=cos3\beta\) (đpcm).
c: 2(sin^6a+cos^6a)+1
=2[(sin^2a+cos^2a)^3-3*sin^2a*cos^2a]+1
=2-6sin^2acos^2a+1
=3-6*sin^2a*cos^2a
=3(sin^4a+cos^4a)
a:
Sửa đề: =-tana*tanb
\(VT=\left(\dfrac{sina}{cosa}-\dfrac{sinb}{cosb}\right):\left(\dfrac{cosa}{sina}-\dfrac{cosb}{sinb}\right)\)
\(=\dfrac{sina\cdot cosb-sinb\cdot cosa}{cosa\cdot cosb}:\dfrac{cosa\cdot sinb-cosb\cdot sina}{sina\cdot sinb}\)
\(=\dfrac{sin\left(a-b\right)}{cosa\cdot cosb}\cdot\dfrac{sina\cdot sinb}{sin\left(b-a\right)}\)
\(=-tana\cdot tanb\)
=VP