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Lời giải:
Ta có: \(\left\{\begin{matrix} \sin a+\cos a=\frac{7}{5}\\ \sin ^2a+\cos ^2a=1\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} \sin a+\cos a=\frac{7}{5}\\ (\sin a+\cos a)^2-2\sin a\cos a=1\end{matrix}\right.\)
\(\Rightarrow \left\{\begin{matrix} \sin a+\cos a=\frac{7}{5}\\ (\frac{7}{5})^2-2\sin a\cos a=1\end{matrix}\right.\)
\(\Rightarrow \left\{\begin{matrix} \sin a+\cos a=\frac{7}{5}\\ \sin a\cos a=\frac{12}{25}\end{matrix}\right.\)
\(\Rightarrow \sin a(\frac{7}{5}-\sin a)=\frac{12}{25}\)
\(\Leftrightarrow \sin ^2a-\frac{7}{5}\sin a+\frac{12}{25}=0\)
\(\Leftrightarrow (\sin a-\frac{4}{5})(\sin a-\frac{3}{5})=0\Rightarrow \left[\begin{matrix} \sin a=\frac{4}{5}\\ \sin a=\frac{3}{5}\end{matrix}\right.\)
Nếu \(\sin a=\frac{4}{5}\Rightarrow \cos a=\frac{3}{5}\Rightarrow \tan a=\frac{\sin a}{\cos a}=\frac{4}{3}\)
Nếu \(\sin a=\frac{3}{5}\rightarrow \cos a=\frac{4}{5}\Rightarrow \tan a=\frac{\sin a}{\cos a}=\frac{3}{4}\)
bài này bn có thể biến đổi sao cho bt được giá trị của tổng và tích giữa \(sinx;cosx\) như cô Akai rồi sử dụng viét đảo để giải tiếp nha
a, ta có \(\tan\alpha=\frac{\sin\alpha}{\cos\alpha}\)
\(\frac{1}{3}\)= \(\frac{\sin\alpha}{\cos\alpha}\)
\(\cos\alpha\)= 3 \(\sin\alpha\)
ta có \(\frac{\cos\alpha+\sin\alpha}{\cos\alpha-\sin\alpha}\)= \(\frac{3\sin\alpha+\sin\alpha}{3\sin\alpha-\sin\alpha}\)= \(\frac{4\sin\alpha}{2\sin\alpha}\)= \(2\)
#mã mã#
a) ta có : \(A=tan1.tan2.tan3...tan89\)
\(=\left(tan1.tan89\right).\left(tan2.tan88\right).\left(tan3.tan87\right)...\left(tan44.tan46\right).tan45\)
\(=\left(tan1.tan\left(90-1\right)\right).\left(tan2.tan\left(90-2\right)\right).\left(tan3.tan\left(90-3\right)\right)...\left(tan44.tan\left(90-44\right)\right).tan45\)
\(=\left(tan1.cot1\right).\left(tan2.cot2\right).\left(tan3.cot3\right)...\left(tan44.cot44\right).tan45\) \(=tan45=1\)b) ta có \(B=\dfrac{sin\alpha+2cos\alpha}{3sin\alpha-4cos\alpha}=\dfrac{\dfrac{sin\alpha}{cos\alpha}+\dfrac{2cos\alpha}{cos\alpha}}{\dfrac{3sin\alpha}{cos\alpha}-\dfrac{4cos\alpha}{cos\alpha}}\)
\(=\dfrac{tan\alpha+2}{3tan\alpha-4}=\dfrac{\dfrac{1}{2}+2}{\dfrac{3}{2}-4}=-1\)
ta có \(D=\dfrac{2sin^2\alpha-3cos^2\alpha}{4cos^2\alpha-5sin^2\alpha}=\dfrac{\dfrac{2sin^2\alpha}{cos^2\alpha}-\dfrac{3cos^2\alpha}{cos^2\alpha}}{\dfrac{4cos^2\alpha}{cos^2\alpha}-\dfrac{5sin^2\alpha}{cos^2\alpha}}\)
\(=\dfrac{2tan^2\alpha-3}{4-5tan^2\alpha}=\dfrac{2\left(\dfrac{1}{2}\right)^2-3}{4-5\left(\dfrac{1}{2}\right)^2}=\dfrac{-10}{11}\)
Lời giải:
\(M=\frac{\frac{\sin a}{\cos a}+1}{\frac{\sin a}{\cos a}-1}=\frac{\tan a+1}{\tan a-1}=\frac{\frac{3}{5}+1}{\frac{3}{5}-1}=-4\)
\(N = \frac{\frac{\sin a\cos a}{\cos ^2a}}{\frac{\sin ^2a-\cos ^2a}{\cos ^2a}}=\frac{\frac{\sin a}{\cos a}}{(\frac{\sin a}{\cos a})^2-1}=\frac{\tan a}{\tan ^2a-1}=\frac{\frac{3}{5}}{\frac{3^2}{5^2}-1}=\frac{-15}{16}\)
a/ Có \(\tan\alpha=\frac{1}{3}\Rightarrow\frac{\sin\alpha}{\cos\alpha}=\frac{1}{3}\Leftrightarrow\cos\alpha=3\sin\alpha\)
Thay vào biểu thức có:
\(\frac{3\sin\alpha+\sin\alpha}{3\sin\alpha-\sin\alpha}=\frac{4\sin\alpha}{2\sin\alpha}=2\)
b/ Có \(\sin\alpha+\cos\alpha=\frac{7}{5}\Rightarrow\sin\alpha=\frac{7}{5}-\cos\alpha\) (1)
Có \(\sin^2\alpha+\cos^2\alpha=1\) (2)
Thay (1) vào (2) rồi tự thay số vào giải PTB2 để tìm cos và sin
Có \(\tan\alpha=\frac{\sin\alpha}{\cos\alpha}\)
Thay vào là OK
$\begin{cases}sinα+cosα=\dfrac{7}{5}\\sin^2α+cos^2α=1\\\end{cases}$
`<=>` $\begin{cases}sinα+cosα=\dfrac{7}{5}\\(sinα+cosα)^2-2sinαcosα=1\\\end{cases}$
`<=>` $\begin{cases}sinα+cosα=\dfrac{7}{5}\\sinα.cosα=\dfrac{12}{25}\\\end{cases}$
`<=>` \(\left\{{}\begin{matrix}\left[{}\begin{matrix}sinα=\dfrac{4}{5}\\cosα=\dfrac{3}{5}\end{matrix}\right.\\\left[{}\begin{matrix}sinα=\dfrac{3}{5}\\cosα=\dfrac{4}{5}\end{matrix}\right.\end{matrix}\right.\)
`=>` \(\left[{}\begin{matrix}tanα=\dfrac{3}{4}\\tanα=\dfrac{4}{3}\end{matrix}\right.\)
Vậy...
Ta có: \(\left(\sin\alpha+\cos\alpha\right)^2=\dfrac{49}{25}\)
\(\Leftrightarrow2\cdot\sin\alpha\cdot\cos\alpha=\dfrac{49}{25}-1=\dfrac{24}{25}\)
Ta có: \(\left(\sin\alpha-\cos\alpha\right)^2\)
\(=\sin^2\alpha+\cos^2\alpha-\dfrac{24}{25}\)
\(=1-\dfrac{24}{25}=\dfrac{1}{25}\)
\(\Leftrightarrow\sin\alpha-\cos\alpha=\dfrac{1}{5}\)
mà \(\sin\alpha+\cos\alpha=\dfrac{7}{5}\)
nên \(2\cdot\sin\alpha=\dfrac{8}{5}\)
hay \(\sin\alpha=\dfrac{4}{5}\)
\(\Leftrightarrow\cos\alpha=\dfrac{7}{5}-\dfrac{4}{5}=\dfrac{3}{5}\)
\(\Leftrightarrow\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}=\dfrac{4}{5}:\dfrac{3}{5}=\dfrac{4}{3}\)
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 \(\tan\alpha=\frac{\sin\alpha}{\cos\alpha}\))
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) \(\cos\alpha=\sqrt{1-\sin^2\alpha}=\sqrt{1-\dfrac{16}{25}}=\dfrac{3}{5}\)
\(\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}=\dfrac{4}{5}:\dfrac{3}{5}=\dfrac{4}{3}\)
\(\dfrac{1}{cos^2\alpha}=1+tan^2\alpha=1+\left(\dfrac{7}{24}\right)^2=\dfrac{625}{576}\)
\(\Rightarrow cos^2\alpha=\dfrac{576}{625}\)
\(cot\alpha=\dfrac{1}{tan\alpha}=\dfrac{24}{7}\)
\(1+tan^2\alpha=\dfrac{1}{cos^2\alpha}\Rightarrow cos^2\alpha=\dfrac{576}{625}\Rightarrow cos\alpha=\dfrac{24}{25}\)
\(1+cot^2\alpha=\dfrac{1}{sin^2\alpha}\Rightarrow sin^2\alpha=\dfrac{49}{625}\Rightarrow cos\alpha=\dfrac{7}{25}\)
Đặt sina=a; cosa=b
Theo đề, ta có: \(\left\{{}\begin{matrix}a+b=1.4\\a^2+b^2=1\end{matrix}\right.\Leftrightarrow ab=\dfrac{1.4^2-1}{2}=0.48\)
=>a,b là các nghiệm của pt là:
\(x^2-1.4x+0.48=0\)
=>x=0,6 hoặc x=0,8
=>(a,b)=(0,6;0,8) hoặc (a,b)=(0,8;0,6)
TH1: a=0,6; b=0,8
tan a=a/b=3/4
TH2: a=0,8; b=0,6
tan a=a/b=4/3