Cho biết \(\sin\alpha+\cos\alpha=\dfrac{7}{5},\left(0^o< \alpha< 90^o\right)\)
Tính \(\tan\alpha\)?
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Câu 1:
Ta có: \(\cos\left(90^0-\alpha\right)=\sin\alpha\)
\(\Leftrightarrow\sin\alpha=1:\sqrt{\dfrac{1^2+2^2}{1}}=1:\sqrt{5}=\dfrac{\sqrt{5}}{5}\)
Câu 2:
a) \(\cos\alpha=\sqrt{1-\sin^2\alpha}=\sqrt{1-\dfrac{16}{25}}=\dfrac{3}{5}\)
\(\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}=\dfrac{4}{5}:\dfrac{3}{5}=\dfrac{4}{3}\)
( sin a + cos a )^2 = (7/5)^2
=> sin^2 a + cos^2a + 2.sina . cos a = 49/25
=> 1 + 2.sin a . cos a = 49/25
=> 2.sin a + cos a = 49/25 - 1 = 24 / 25
( sin a - cos a )^2 = sin ^2 a + cos ^2a - 2. sin a . cos a = 1 - 24/25 = 1/25
=> sin a - cos a = 1/5 (2)
TA có sina + cos a = 7/5 (1)
Từ (1) và (1) => 2 sina = 8/5 => sin a = 8/5 : 2 = 8/10 = 4/5
=> cos a = sin a - 1/5 = 4/5 - 1/5 = 3/5
tan a = \(\frac{sina}{cosa}=\frac{\frac{4}{5}}{\frac{3}{5}}=\frac{4}{5}\cdot\frac{5}{3}=\frac{4}{3}\)
a)\(sin^2\left(180^o-\alpha\right)+tan^2\left(180-\alpha\right).tan^2\left(270^o+\alpha\right)\)\(+sin\left(90^o+\alpha\right)cos\left(\alpha-360^o\right)\)
\(=sin^2\alpha+tan^2\alpha.cot^2\alpha+cos\alpha cos\alpha\)
\(=sin^2\alpha+cos^2\alpha+\left(tan\alpha cot\alpha\right)^2=1+1=2\).
\(\dfrac{cos\left(\alpha-180^o\right)}{sin\left(180^o-\alpha\right)}+\dfrac{tan\left(\alpha-180^o\right)cos\left(180^o+\alpha\right)sin\left(270^o+\alpha\right)}{tan\left(270^o+\alpha\right)}\)
\(=\dfrac{cos\left(180^o-\alpha\right)}{sin\left(180^o-\alpha\right)}+\dfrac{-tan\left(180^o-\alpha\right).cos\alpha.sin\left(90^o+\alpha\right)}{-tan\left(90^o+\alpha\right)}\)
\(=tan\left(180^o-\alpha\right)+\dfrac{tan\alpha.cos\alpha.cos\alpha}{cot\alpha}\)
\(=-tan\alpha+tan^2\alpha cos^2\alpha\)
\(=tan\alpha\left(-1+tan\alpha cos^2\alpha\right)\)
\(=tan\alpha\left(sin\alpha cos\alpha-1\right)\).
Bạn xem lại biểu thức A. Biểu thức $A$ sau khi rút gọn thì \(A=\frac{-2\sin ^2a}{3\cos 2a}\) vẫn phụ thuộc vào $a$
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Sử dụng công thức: \(\sin (90-a)=\cos a; \cot (90-a)=\tan a\), ta có:
\(B=\tan ^260(\sin ^8a-\cos ^8a)+4\cos 60(\cos ^6a-\sin ^6a)-\cos ^6a(\tan ^2a-1)^3\)
\(=3(\sin ^8a-\cos ^8a)+2(\cos ^6a-\sin ^6a)-\cos ^6a\left(\frac{\sin ^2a}{\cos ^2a}-1\right)^3\)
\(=3(\sin ^8a-\cos ^8a)+2(\cos ^6a-\sin ^6a)-(\sin ^2a-\cos ^2a)^3\)
\(=3(\sin ^2a-\cos ^2a)(\sin ^2a+\cos ^2a)(\sin ^4a+\cos ^4a)+2(\cos ^2a-\sin ^2a)(\cos ^4a+\sin ^2a\cos ^2a+\sin ^4a)-(\sin ^2a-\cos ^2a)^3\)
\(=3(\sin ^2-\cos ^2a)(\sin ^4a+\cos ^4a)-2(\sin ^2a-\cos ^2a)(\cos ^4a+\sin ^2a\cos ^2a+\sin ^4a)-(\sin ^2a-\cos ^2a)^3\)
\(=(\sin ^2a-\cos ^2a)[3(\sin ^4a+\cos ^4a)-2(\cos ^4a+\sin ^2a\cos ^2a+\sin ^4a)-(\sin ^2a-\cos ^2a)^2]\)
\(=(\sin ^2a-\cos ^2a).0=0\). Do đó giá trị của biểu thức không phụ thuộc vào $a$
\(VT=\dfrac{-tan\left(\dfrac{\pi}{2}-a\right)cos\left(2\pi-\dfrac{\pi}{2}+a\right)-sin^3\left(4\pi-\dfrac{\pi}{2}-a\right)}{cos\left(\dfrac{\pi}{2}-a\right)tan\left(2\pi-\dfrac{\pi}{2}+a\right)}\)
\(=\dfrac{-cota.sina+sin^3\left(\dfrac{\pi}{2}+a\right)}{sina.\left(-cota\right)}=\dfrac{-cosa+cos^3a}{-cosa}=1-cos^2a=sin^2a\)
Đặt \(\sin\alpha=x,\cos\alpha=y\)
Ta có hpt:
\(\left\{{}\begin{matrix}x+y=\frac{7}{5}\\x^2+y^2=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x+y=\frac{7}{5}\\xy=\frac{\left(x+y\right)^2-\left(x^2+y^2\right)}{2}=\frac{\left(\frac{7}{5}\right)^2-1}{2}=\frac{12}{25}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\frac{7}{5}-y\\xy=\frac{12}{25}\end{matrix}\right.\)
\(\Rightarrow xy=y\left(\frac{7}{5}-y\right)=\frac{12}{25}\)
\(\Leftrightarrow\frac{7}{5}y-y^2=\frac{12}{25}\Leftrightarrow y^2-\frac{7}{5}y+\frac{12}{25}=0\)
\(\Delta=\frac{49}{25}-4\cdot\frac{12}{25}=\frac{1}{25}>0;\sqrt{\Delta}=\frac{1}{5}\)
phương trình có 2 nghiệm phân biệt:
\(\left\{{}\begin{matrix}y=\frac{\frac{7}{5}+\frac{1}{5}}{2}=\frac{4}{5}\\y=\frac{\frac{7}{5}-\frac{1}{5}}{2}=\frac{3}{5}\end{matrix}\right.\)
Thay vào tìm x ta được các tập nghiệm: \(\left(x,y\right)=\left(\frac{3}{5};\frac{4}{5}\right);\left(\frac{4}{5};\frac{3}{5}\right)\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sin\alpha=\frac{3}{5}\\\cos\alpha=\frac{4}{5}\end{matrix}\right.\\\left\{{}\begin{matrix}\sin\alpha=\frac{4}{5}\\\cos\alpha=\frac{3}{5}\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\tan\alpha=\frac{\frac{3}{5}}{\frac{4}{5}}=\frac{3}{4}\\\tan\alpha=\frac{\frac{4}{5}}{\frac{3}{5}}=\frac{4}{3}\end{matrix}\right.\)
(Áp dụng \(\tan\alpha=\frac{\sin\alpha}{\cos\alpha}\))