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Bài 1:
Áp dụng định lí pytago trong tam giác vuông ABC ta có:
BC2=AC2+AB2
BC2=42+32
BC=\(\sqrt{25}\)=5(cm)
Ta có:
Sin B=\(\dfrac{AC}{BC}=\dfrac{4}{5}=0.8\)
Cos B=\(\dfrac{AB}{BC}=\dfrac{3}{5}=0.6\)
Tag B=\(\dfrac{AC}{AB}=\dfrac{4}{3}\)
Cotg B=\(\dfrac{AB}{AC}=\dfrac{3}{4}=0.75\)
\(A=4\left(1-sin^2x\right)-6sin^2a=4-10sin^2a=4-10.\left(\frac{1}{5}\right)^2=...\)
\(tana+cota=3\Leftrightarrow\frac{sina}{cosa}+\frac{cosa}{sina}=3\Leftrightarrow\frac{sin^2a+cos^2a}{sina.cosa}=3\)
\(\Leftrightarrow\frac{1}{sina.cosa}=3\Leftrightarrow sina.cosa=\frac{1}{3}\)
\(C=cot^2a-cos^2a.cot^2a=cot^2a\left(1-cos^2a\right)=cot^2a.sin^2a\)
\(=\frac{cos^2a}{sin^2a}.sin^2a=cos^2a=1-sin^2a=1-\left(\frac{3}{4}\right)^2=...\)
2. \(\left(\sin a+\cos a\right)^2+\left(\sin a-\cos a\right)^2+2\)
\(=\sin^2a+2.\sin a.\cos a+\cos^2a+\sin^2a\cdot2.\sin a.\cos a+\cos^2a+2\)
\(=2\sin^2a+2\cos^2a+2\)
\(=2\left(\sin^2a+\cos^2a\right)+2\)
\(=2.1+2=4\)
=> biểu thức trên ko phụ thuộc vào a
1. a.) \(\cot a=\dfrac{1}{\tan a}=\dfrac{1}{\sqrt{3}}\)
\(\tan\sqrt{3}=60\Rightarrow a=60^o\)
\(\sin60=\dfrac{\sqrt{3}}{2}\)
\(\cos60=\dfrac{1}{2}\)
b.) \(\cos^2a=1-\left(\dfrac{15}{17}\right)^2=\dfrac{64}{289}\Rightarrow\cos a=\dfrac{8}{17}\)
\(\tan a=\dfrac{\sin a}{\cos a}=\dfrac{\dfrac{15}{17}}{\dfrac{8}{17}}=\dfrac{15}{17}.\dfrac{17}{8}=\dfrac{15}{8}\)
Bài 1:
Ta có:
\(A=\sin ^6a+\cos ^6a+3\sin ^2a\cos ^2a\)
\(=(\sin ^2a)^3+(\cos ^2a)^3+3\sin ^2a\cos ^2a\)
\(=(\sin ^2a+\cos ^2a)(\sin ^4a-\sin ^2a\cos ^2a+\cos ^4a)+3\sin ^2a\cos ^2a\)
\(=\sin ^4a-\sin ^2a\cos ^2a+\cos ^4a+3\sin ^2a\cos ^2a\)
\(=\sin ^4a+2\sin ^2a\cos ^2a+\cos ^4a\)
\(=(\sin ^2a+\cos ^2a)^2=1^2=1\)
Lời giải:
Xét tam giác $ABC$. Gọi cạnh $AB, AC$ là $a,b$ và góc \(\widehat{BAC}=\alpha\)
Kẻ đường cao $BH$ của tam giác $ABC$
Khi đó:
\(S=\frac{BH.AC}{2}\)
Mặt khác, theo công thức lượng giác:
\(\frac{BH}{AB}=\sin \widehat{BAC}=\sin \alpha\Rightarrow BH=\sin \alpha.AB\)
Do đó: \(S=\frac{BH.AC}{2}=\frac{\sin \alpha.AB.AC}{2}=\frac{\sin \alpha.a.b}{2}\) (đpcm)
Bài 1:
Ta có: \(\sin^2\alpha+cos^2\alpha=1\)
=> \(\sin^2\alpha=1-\left(0,4\right)^2\)
=> \(\sin\alpha=\sqrt{0,84}\approx0,9\)
Mặt khác: \(\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{0,9}{0,4}=\frac{9}{4}\)
Mà: \(\tan\alpha\times\cot\alpha=1\)
\(\Rightarrow\cot\alpha=\frac{4}{9}\)
Bài 2: C B A
Vì \(\widehat{A}+\widehat{B}=90^o\)nên hai góc này phụ nhau.
Vậy: \(\cos\widehat{A}=\tan\widehat{B}=\frac{15}{3}\)
a/ \(A=\frac{cot^2a-cos^2a}{cot^2a}-\frac{sina.cosa}{cota}\)
\(=\frac{\frac{cos^2a}{sin^2a}-cos^2a}{\frac{cos^2a}{sin^2a}}-\frac{sina.cosa}{\frac{cosa}{sina}}\)
\(=\left(1-sin^2a\right)-sin^2a=1\)
b/ \(B=\left(cosa-sina\right)^2+\left(cosa+sina\right)^2+cos^4a-sin^4a-2cos^2a\)
\(=cos^2a-2cosa.sina+sin^2a+cos^2a+2cosa.sina+sin^2a+\left(cos^2a+sin^2a\right)\left(cos^2a-sin^2a\right)-2cos^2a\)
\(=2+\left(cos^2a-sin^2a\right)-2cos^2a\)
\(=2-sin^2a-cos^2a=2-1=1\)
Bài 1:
Ta có: \(\tan\alpha+\cot\alpha=3\)
\(\Leftrightarrow\dfrac{\sin\alpha}{\cos\alpha}+\dfrac{\cos\alpha}{\sin\alpha}=3\)
\(\Leftrightarrow\dfrac{\sin^2\alpha}{\cos\alpha.\sin\alpha}+\dfrac{\cos^2\alpha}{\sin\alpha.\cos\alpha}=3\)
\(\Leftrightarrow\dfrac{\sin^2\alpha+\cos^2\alpha}{\sin\alpha.\cos\alpha}=3\)
\(\Leftrightarrow\dfrac{1}{\sin\alpha.\cos\alpha}=3\)
\(\Leftrightarrow\dfrac{1}{\sin\alpha.\cos\alpha}=\dfrac{3\left(\sin\alpha.\cos\alpha\right)}{\sin\alpha.\cos\alpha}\)
\(\Leftrightarrow1=3\left(\sin\alpha.\cos\alpha\right)\)
\(\Leftrightarrow\sin\alpha.\cos\alpha=\dfrac{1}{3}\)
Vậy \(\tan\alpha+\cot\alpha=3\) thì \(\sin\alpha.\cos\alpha=\dfrac{1}{3}\)
A B C H Chứng minh
\(AB^2=AC^2+BC^2-2AC.BC.\cos C\)
Kẻ \(AH\perp BC\)
Ta có: \(VP=\)\(AC^2+BC^2-2AC.BC.\cos C\)
\(=AC^2+BC^2-2AC.BC.\dfrac{CH}{AC}\)
\(=AC^2+BC^2-2BC.CH\)
\(=AH^2+HC^2+BH^2+HC^2+2BH.CH-2BH.CH-2CH^2\)
\(=AH^2+BH^2\)
\(=AB^2=VT\)
Vậy đẳng thức được chứng minh.