Chương 6: CUNG VÀ GÓC LƯỢNG GIÁC. CÔNG THỨC LƯỢNG GIÁC

\(\alpha;\beta\in\left(0;\dfrac{\pi}{2}\right)\Rightarrow cos\alpha;sin\beta>0\)

Ta có : \(cos\alpha=\sqrt{1-sin^2\alpha}=\sqrt{1-\dfrac{1}{5}}=\dfrac{2}{\sqrt{5}}\)  

\(sin\beta=\sqrt{1-cos^2B}=\sqrt{1-\dfrac{1}{10}}=\dfrac{3}{\sqrt{10}}\)

\(cos\left(\alpha+\beta\right)=cos\alpha.cos\beta-sin\alpha.sin\beta\) = \(\dfrac{2}{\sqrt{5}}.\dfrac{1}{\sqrt{10}}-\dfrac{1}{\sqrt{5}}.\dfrac{3}{\sqrt{10}}=\dfrac{-\sqrt{2}}{10}\)

\(a;\sin\mu+cos\mu=\sqrt{2}.\left[\dfrac{1}{\sqrt{2}}sin\mu+\dfrac{1}{\sqrt{2}}cos\mu\right]=\sqrt{2}\left[sin\left(\dfrac{\pi}{4}\right).sin\mu+cos\left(\dfrac{\pi}{4}\right).cos\mu\right]=\sqrt{2}.sin\left(\mu+\dfrac{\pi}{4}\right)=\sqrt{2}.cos\left(\mu-\dfrac{\pi}{4}\right)\left(đpcm\right)\)

\(b;4sin\left(\alpha+\dfrac{\pi}{3}\right)sin\left(\alpha-\dfrac{\pi}{3}\right)=4\left[sin\alpha.cos\dfrac{\pi}{3}+cos\alpha.sin\dfrac{\pi}{3}\right].\left(sin\alpha.cos\left(\dfrac{\pi}{3}\right)-cos\alpha.sin\dfrac{\pi}{3}\right)=4\left[\dfrac{sin\alpha}{2}+\dfrac{\sqrt{3}cos\alpha}{2}\right]\left(\dfrac{sin\alpha}{2}-\dfrac{\sqrt{3}cos\alpha}{2}\right)=\left(sin\alpha+\sqrt{3}cos\alpha\right)\left(sin\alpha-\sqrt{3}cos\alpha\right)=sin^2\alpha-3cos^2\alpha=sin^2\alpha-3\left(1-sin^2\alpha\right)=4sin^2\alpha-3\)

\(c;cos\left(a+b\right).cos\left(a-b\right)=\left(cosa.cosb-sina.sinb\right)\left(cosa.cosb+sina.sinb\right)=cos^2a.cos^2b-sin^2a.sin^2b=cos^2a\left(1-sin^2b\right)-sin^2a\left(1-cos^2b\right)=cos^2a-cos^2a.sin^2b-sin^2b+sin^2b.cos^2b=cos^2a-sin^2a\)

nhường cho I don't know :))

Kiểm tra lại đề bài, \(cosa=\dfrac{1}{\sqrt{10}}\) hay \(cos\beta=\dfrac{1}{\sqrt{10}}\)?

\(sin^2A+sin^2B+sin^2C=2\)

\(\Leftrightarrow sin^2A+\dfrac{1-cos2B}{2}+\dfrac{1-cos2C}{2}=2\)

\(\Leftrightarrow sin^2A-\dfrac{1}{2}\left(cos2B+cos2C\right)=1\)

\(\Leftrightarrow1-cos^2A-cos\left(B+C\right)cos\left(B-C\right)=1\)

\(\Leftrightarrow cos^2A+cos\left(B+C\right)cos\left(B-C\right)=0\)

\(\Leftrightarrow cos^2A-cosA.cos\left(B-C\right)=0\)

\(\Leftrightarrow cosA\left[cosA-cos\left(B-C\right)\right]=0\)

\(\Leftrightarrow cosA.sin\left(\dfrac{A+B-C}{2}\right)sin\left(\dfrac{A+C-B}{2}\right)=0\)

\(\Leftrightarrow cosA.sin\left(90^0-C\right)sin\left(90^0-B\right)=0\)

\(\Leftrightarrow cosA.cosB.cosC=0\)

\(\Leftrightarrow\left[{}\begin{matrix}A=90^0\\B=90^0\\C=90^0\end{matrix}\right.\) hay tam giác ABC vuông

Theo tính chất của tam giác, ta có:

\(A+B+C=180^0\)

\(\Rightarrow\dfrac{A+B+C}{2}=90^0\)

\(\Rightarrow\dfrac{B+C}{2}=90^0-\dfrac{A}{2}\)

\(\Rightarrow tan\left(\dfrac{B+C}{2}\right)=tan\left(90^0-\dfrac{A}{2}\right)\)

\(\Rightarrow tan\left(\dfrac{B+C}{2}\right)=cot\left(\dfrac{A}{2}\right)\)

Tính:F=Cos(π/4+α) x cos(π/4-α)

G=Sin(π/3+α) x cos(π/3-α)

H=cos(π/2-α) x sin(π/2+α)

I=sin(π/4+α) - cos(π/4-α)

K=cos(π/6-x) - sin(π/3+x)

vì\(\dfrac{\pi}{2}< x< 0\Rightarrow sin< 0\)

Ta có: sin²x + cos²x=1\(\Rightarrow\)sin²x=1-cos²x=1-\(\left(\dfrac{3}{5}\right)^2=\dfrac{16}{25}\Rightarrow\sin x=\dfrac{-4}{5}\)

\(\sin\left(\dfrac{\pi}{2}-x\right)=\sin\dfrac{\pi}{2}\times\cos x-\cos\dfrac{\pi}{2}\times\sin x\)

\(=1\times\dfrac{3}{5}-0\times-\dfrac{4}{5}\)

=\(\dfrac{3}{5}\)

\(sin\left(\dfrac{\pi}{2}-x\right)=cosx=\dfrac{3}{5}\)

6:

b: \(=\dfrac{sina\left(2cosa+1\right)}{1+2cos^2a-1+cosa}\)

\(=\dfrac{sina\left(2cosa+1\right)}{cosa\left(2cosa+1\right)}=\dfrac{sina}{cosa}=tana\)

c: \(=\dfrac{\dfrac{sin2a}{cos2a}\cdot\dfrac{sina}{cosa}}{\dfrac{sin2a}{cos2a}-\dfrac{sina}{cosa}}\)

\(=\dfrac{sin2a\cdot sina}{sin2a\cdot cosa-sina\cdot cos2a}\)

\(=\dfrac{sina^2\cdot2\cdot cosa}{sina\cdot cos^2a\cdot2-sina\cdot\left(2cos^2a-1\right)}\)

\(=\dfrac{sin^2a\cdot2\cdot cosa}{sina\left(2cos^2a-2cos^2a+1\right)}=2\cdot sina\cdot cosa=sin2a\)

\(P=cos2a=1-2sin^2a=1-2.\left(\dfrac{4}{5}\right)^2=-\dfrac{7}{25}\)

Đặt \(\alpha=12^o\)

Ta có : \(B=\dfrac{cos\left(\dfrac{9}{2}\pi+\alpha\right).cot\left(-3\pi+\alpha\right)}{cos\left(-5\pi+\alpha\right)}\)   \(=\dfrac{cos\left(\dfrac{\pi}{2}+\alpha\right).cot\left(\alpha-\pi\right)}{cos\left(\alpha-\pi\right)}\)

\(=\dfrac{-sin\alpha.-cot\left(\pi-\alpha\right)}{-cos\alpha}\)  \(=\dfrac{-sin\alpha.cot\alpha}{-cos\alpha}=tan\alpha.cot\alpha=1\)