Chứng minh các biểu thức sau không phụ thuộc vào các góc nhọn \(\alpha\)
a) \(C=\cos^4\alpha+\sin^2\alpha.\cos^2\alpha+\sin^2\alpha\)
b) \(D=\sin^2\alpha.\sin^2\beta+\sin^2\alpha.\cos^2\beta+\cos^2\alpha\)
c) E=\(\sin^6\alpha+\sin^6\beta+3.\sin^2\alpha.\cos^2\alpha\)
d) \(M=\frac{\left(\cos\alpha-\sin\alpha\right)^2-\left(\cos\alpha+\sin\alpha\right)^2}{\cos\alpha.\sin\alpha}\)
Tính a) sin^4α - cos^4α , biết cos2α=3/5
b) cos(α-β) biết sinα - sinβ = 1/3 và cosα - cosβ = 1/2
1. Cho \(2\cos\left(\alpha+\beta\right)=\cos\alpha\cos\left(\pi+\beta\right)\)
Tính \(A=\dfrac{1}{2\sin^2\alpha+3\cos^2\alpha}+\dfrac{1}{2\sin^2\beta+3\cos^2\beta}\)
2. Rút gọn: a) \(A=4\cos\dfrac{2x}{3}\cos\dfrac{\pi+2x}{3}\cos\dfrac{\pi-2x}{3}\)
b) \(B=\dfrac{\sin\left(a-b\right).\sin\left(a+b\right)}{\cos^2a.\sin^2b}-\tan^2a.\cot^2b\)
3. Chứng minh rằng: Nếu \(2\tan a=\tan\left(a+b\right)\) thì:
a) \(\sin b=\sin a.\cos\left(a+b\right)\)
b) \(3\sin b=\sin\left(2a+b\right)\)
1.
\(2cos\left(a+b\right)=cosa.cos\left(\pi+b\right)\)
\(\Leftrightarrow2cosa.cosb-2sina.sinb=-cosa.cosb\)
\(\Leftrightarrow2sina.sinb=3cosa.cosb\Rightarrow4sin^2a.sin^2b=9cos^2a.cos^2b\)
\(\Rightarrow4\left(1-cos^2a\right)\left(1-cos^2b\right)=9cos^2a.cos^2b\)
\(\Leftrightarrow4-4\left(cos^2a+cos^2b\right)=5cos^2a.cos^2b\)
\(A=\dfrac{1}{cos^2a+2\left(sin^2a+cos^2a\right)}+\dfrac{1}{cos^2b+2\left(sin^2b+cos^2b\right)}\)
\(=\dfrac{1}{2+cos^2a}+\dfrac{1}{2+cos^2b}=\dfrac{4+cos^2a+cos^2b}{4+2\left(cos^2a+cos^2b\right)+cos^2a.cos^2b}\)
\(=\dfrac{4+cos^2a+cos^2b}{4+2\left(cos^2a+cos^2b\right)+\dfrac{4}{5}-\dfrac{4}{5}\left(cos^2a+cos^2b\right)}=\dfrac{4+cos^2a+cos^2b}{\dfrac{24}{5}+\dfrac{6}{5}\left(cos^2a+cos^2b\right)}=\dfrac{5}{6}\)
2.
\(A=2cos\dfrac{2x}{3}\left(cos\dfrac{2\pi}{3}+cos\dfrac{4x}{3}\right)=2cos\dfrac{2x}{3}\left(cos\dfrac{4x}{3}-\dfrac{1}{2}\right)\)
\(=2cos\dfrac{2x}{3}.cos\dfrac{4x}{3}-cos\dfrac{2x}{3}\)
\(=cos3x+cos\dfrac{2x}{3}-cos\dfrac{2x}{3}\)
\(=cos3x\)
\(B=\dfrac{cos2b-cos2a}{cos^2a.sin^2b}-tan^2a.cot^2b=\dfrac{1-2sin^2b-\left(1-2sin^2a\right)}{cos^2a.sin^2b}-tan^2a.cot^2b\)
\(=\dfrac{2sin^2a-2sin^2b}{cos^2a.sin^2b}-tan^2a.cot^2b=2tan^2a\left(1+cot^2b\right)-2\left(1+tan^2a\right)-tan^2a.cot^2b\)
\(=2tan^2a+2tan^2a.cot^2b-2-2tan^2a-tan^2a.cot^2b\)
\(=tan^2a.cot^2b-2\)
3.
\(\dfrac{2sina}{cosa}=\dfrac{sin\left(a+b\right)}{cos\left(a+b\right)}\Leftrightarrow2sina.cos\left(a+b\right)=cosa.sin\left(a+b\right)\)
\(\Leftrightarrow sina.cos\left(a+b\right)=sin\left(a+b\right).cosa-cos\left(a+b\right)sina\)
\(\Leftrightarrow sina.cos\left(a+b\right)=sin\left(a+b-a\right)\)
\(\Leftrightarrow sina.cos\left(a+b\right)=sinb\)
b.
\(\dfrac{2sina}{cosa}=\dfrac{sin\left(a+b\right)}{cos\left(a+b\right)}\Leftrightarrow2sina.cos\left(a+b\right)=cosa.sin\left(a+b\right)\)
\(\Leftrightarrow sin\left(2a+b\right)+sin\left(-b\right)=\dfrac{1}{2}sin\left(2a+b\right)+\dfrac{1}{2}sinb\)
\(\Leftrightarrow\dfrac{1}{2}sin\left(2a+b\right)=\dfrac{3}{2}sinb\)
\(\Leftrightarrow sin\left(2a+b\right)=3sinb\)
Giúp mình vs chiều phải nộp bài rồi
a)C= \(4\cos^2\alpha-3\sin^2\alpha.cos=\frac{4}{7}\)
b)\(\cos^2\alpha+\cos^2\beta+\cos^2\alpha.\sin^2\beta+\sin^2\alpha\)
c)2\(\left(\sin\alpha-\cos\alpha\right)^2-\left(\sin\alpha+\cos\alpha\right)^2+\left(\sin\alpha.\cos\alpha\right)\)
d)\(\left(\tan\alpha-\cot\alpha\right)^2-\left(\sin\alpha+\cot\alpha\right)^2\)
Bạn không ghi rõ yêu cầu đề bài thì làm sao mà làm?
1) Cho: \(\tan\alpha=\frac{1}{2}\). Tính \(\frac{\cos\alpha+\sin\alpha}{\cos\alpha-\sin\alpha}\)
2) Cho: \(\cos\beta=2\sin\beta.\) Hãy tính: \(\sin\beta.\cos\beta\)
3)Chứng minh hệ thức:
a/ \(\frac{1+\cos\alpha}{\sin\alpha}=\frac{\sin\alpha}{1-\cos\alpha}\)
b/ \(\cot^2\alpha-\cos^2\alpha=\cot^2\alpha.\cos\alpha\)
1. \(\frac{cos\alpha+sin\alpha}{cos\alpha-sin\alpha}=\frac{1+\frac{sin\alpha}{cos\alpha}}{1-\frac{sin\alpha}{cos\alpha}}=\frac{1+\frac{1}{2}}{1-\frac{1}{2}}=3\)
2. \(cos\beta=2sin\beta\Rightarrow cos^2\beta=4sin^2\beta\). Do \(cos^2\beta+sin^2\beta=1\Rightarrow5sin^2\beta=1\Rightarrow sin\beta=\frac{1}{\sqrt{5}}\)
\(\Rightarrow cos\beta=\frac{2}{\sqrt{5}}\). Vậy \(sin\beta.cos\beta=\frac{2}{5}\)
3. a. Nhân chéo ra được hệ thức \(sin^2\alpha+cos^2\alpha=1\)
b. Chú ý \(cot^2\alpha=\frac{cos^2\alpha}{sin^2\alpha}\)
1.Cho các góc\(\alpha,\beta\)nhọn và \(\alpha< \beta\). Chứng minh \(\sin\left(\beta-\alpha\right)=\sin\beta\cos\alpha-\cos\beta\sin\alpha\)
2.Cho các góc \(\alpha,\beta\)nhọn và \(\alpha< \beta\).Chứng minh \(\cos\left(\beta-\alpha\right)=\cos\beta\cos\alpha+\sin\beta\sin\alpha\)
3.Cho các góc \(\alpha,\beta\)nhọn. Chứng minh \(\sin\left(\alpha+\beta\right)=\sin\alpha\cos\beta+\sin\beta\cos\alpha\)
4.Cho các góc \(\alpha,\beta\)nhọn. Chứng minh \(\cos\left(\alpha+\beta\right)=\cos\alpha\cos\beta-\sin\alpha\sin\beta\)
1.Cho \(\alpha,\beta\left(\alpha\ne\beta\right)\in\left(0;\dfrac{\pi}{2}\right)\)và thỏa mãn điều kiện \(\dfrac{cosx-cos\alpha}{cosx-cos\beta}=\dfrac{sin^2\alpha cos\beta}{sin^2\beta cos\alpha}\)
(giả sử \(x\) xác định). Chứng minh\(tan^2\dfrac{x}{2}=tan^2\dfrac{\alpha}{2}tan^2\dfrac{\beta}{2}\)
2. Giải hệ phương trình \(\left\{{}\begin{matrix}xy-2y-3=\sqrt{y-x-1}+\sqrt{y-3x+5}\\\left(1-y\right)\sqrt{2x-y}+2\left(x-1\right)=\left(2x-y-1\right)\sqrt{y}\end{matrix}\right.\)
3. Cho ba số thực dương a, b, c thỏa mãn \(\dfrac{1}{a+2}+\dfrac{1}{b+3}+\dfrac{1}{c+4}=1\). Tìm Min của biểu thức \(P=a+b+c+\dfrac{4}{\sqrt[3]{a\left(b+1\right)\left(c+2\right)}}+3\)
4. Tìm m để hệ bất phương trình \(\left\{{}\begin{matrix}x^2-5x+9\le\left|x-6\right|\\x^2+2x-3m^2+4\left|m\right|-4\le0\end{matrix}\right.\)
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 \(\left(x;y\right)=\left(3;5\right)\)
Trong trường hợp nào dưới đây \(cos\alpha = cos\beta \) và \(sin\alpha = - sin\beta \).
\(\begin{array}{l}A.\;\beta = - \alpha \\B.\;\beta = \pi - \alpha \\C.\;\beta = \pi + \alpha \\D.\;\beta = \frac{\pi }{2} + \alpha \end{array}\)
+) Xét \(\beta = - \alpha \), khi đó:
\(\begin{array}{l}cos\beta = cos\left( {-{\rm{ }}\alpha } \right) = cos\alpha ;\\sin\beta = sin\left( {-{\rm{ }}\alpha } \right) = -sin\alpha \Leftrightarrow sin\alpha = -sin\beta .\end{array}\)
Do đó A thỏa mãn.
Đáp án: A
Cho \(\alpha-\beta=\frac{\pi}{3}\). Tính giá trị bthuc
a) \(A=\left(cos\alpha+cos\beta\right)^2+\left(sin\alpha+sin\beta\right)^2\)
b) \(B=\left(cos\alpha+sin\beta\right)^2+\left(cos\beta-sin\alpha\right)^2\)
\(A=cos^2a+cos^2b+2cosa.cosb+sin^2a+sin^2b+2sina.sinb\)
\(=2+2\left(cosa.cosb+sina.sinb\right)\)
\(=2+2.cos\left(a-b\right)=2+2.cos\frac{\pi}{3}=3\)
\(B=cos^2a+sin^2b+2cosa.sinb+cos^2b+sin^2a-2sina.cosb\)
\(=2-2\left(sina.cosb-cosa.sinb\right)\)
\(=2-2sin\left(a-b\right)=2-2sin\frac{\pi}{3}=2-\sqrt{3}\)
