§3. Các hệ thức lượng giác trong tam giác và giải tam giác

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31 tháng 3 2022 lúc 9:17

\(a;\left(\cos a-\sin a\right)\left(cosa+sina\right)=cos^2a-sin^2a=1-sin^2a-sin^2a=1-2sin^2a\)

\(b;VP=\left(2cosa-1\right)\left(2cosa+1\right)=4cos^2a-1=4\left(1-sin^2a\right)-1=3-4sin^2a=VT\)

e;\(\dfrac{1}{1+tana}+\dfrac{1}{1+cota}=1\Leftrightarrow cota+tana+2=\left(cota+1\right)\left(tana+1\right)\Leftrightarrow cota+tana+2=cota.tana+cota+tana+1\Leftrightarrow cota+tana+2=1+cota+tana+1\Leftrightarrow0=0\left(đúng\right)\Rightarrow VT=VP\)

\(d;sin^3a+cos^3a=\left(sina+cosa\right)\left(sin^2a-sina.cosa+cos^2a\right)=\left(sina+cosa\right)\left(1-sina.cosa\right)\left(đpcm\right)\left(hđt:a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\right)\)

\(c;sin^2a.cosa+sina.cos^2a=\left(sina.cosa\right)\left(sin^2+cos^2\right)=sina.cosa\)

\(f;;tana+\dfrac{cosa}{1+sina}=\dfrac{sina}{cosa}+\dfrac{cosa}{1+sina}=\dfrac{sina+sin^2a+cos^2a}{cosa\left(1+sina\right)}=\dfrac{1+sina}{cosa\left(1+sina\right)}=\dfrac{1}{cosa}\)

\(g;1+cot^2a=\dfrac{1}{sin^2a}=\dfrac{1}{1-cos^2a}=\dfrac{1}{\left(1-cosa\right)\left(1+cosa\right)}\left(đpcm\right)\)

\(h;\dfrac{1+cosa}{1-cosa}-\dfrac{1-cosa}{1+cosa}=\dfrac{\left(cosa+1\right)^2-\left(cosa-1\right)^2}{1-cosa^2}=\dfrac{\left(cosa+1-cosa+1\right)\left(cosa+1+cosa-1\right)}{1-cos^2a}=\dfrac{4cosa}{sin^2a}\left(đpcm\right)\)

\(k;\dfrac{1+cosa}{sina}-\dfrac{sina}{1+cosa}=\dfrac{\left(cosa+1\right)^2-sin^2a}{sina\left(1+cosa\right)}=\dfrac{cos^2a+2cosa+1-sin^2a}{sina\left(1+cosa\right)}=\dfrac{2cos^2a+2cosa}{sina\left(1+cosa\right)}=\dfrac{2cosa\left(1+cosa\right)}{sina\left(1+cosa\right)}=\dfrac{2cosa}{sina}=2cota\left(đpcm\right)\)

\(m;;;\Leftrightarrow sin^3a=cosa\left(1+cosa\right)\left(tana-sina\right)=\left(cosa+cos^2a\right)\left(tana-sina\right)\Leftrightarrow sin^3a=\left(cosa+cos^2a\right)\left(\dfrac{sina}{cosa}-sina\right)=sina-sina.cosa+cosa.sina-cos^2a.sina\Leftrightarrow sin^3a=sina-cos^2a.sina\Leftrightarrow sin^3a-sina\left(1-cos^2a\right)=0\Leftrightarrow sin^3a-sina.sin^2a=0\Leftrightarrow0=0\left(đúng\right)\Rightarrowđpcm\)