a.

\(tana=\dfrac{sina}{cosa}=\dfrac{1}{15}\Rightarrow sina=\dfrac{cosa}{15}\)

\(\Rightarrow sin2a=2sina.cosa=\dfrac{2cosa}{15}.cosa=\dfrac{2}{15}cos^2a=\dfrac{2}{15}.\dfrac{1}{1+tan^2a}=\dfrac{2}{15}.\dfrac{1}{1+\dfrac{1}{15^2}}=\dfrac{15}{113}\)

b.

\(5^2=\left(3sina+4cosa\right)^2\le\left(3^2+4^2\right)\left(sin^2+cos^2a\right)=25\)

Đẳng thức xảy ra khi và chỉ khi: \(\left\{{}\begin{matrix}\dfrac{sina}{3}=\dfrac{cosa}{4}\\3sina+4cosa=5\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}sina=\dfrac{3}{5}\\cosa=\dfrac{4}{5}\end{matrix}\right.\)

c.

\(\dfrac{1}{tan^2a}+\dfrac{1}{cot^2a}+\dfrac{1}{sin^2a}+\dfrac{1}{cos^2a}=7\)

\(\Leftrightarrow\dfrac{cos^2a}{sin^2a}+\dfrac{sin^2a}{cos^2a}+\dfrac{1}{sin^2a}+\dfrac{1}{cos^2a}=7\)

\(\)\(\Leftrightarrow\dfrac{sin^4a+cos^4a}{sin^2a.cos^2a}+\dfrac{sin^2a+cos^2a}{sin^2a.cos^2a}=7\)

\(\Leftrightarrow\dfrac{\left(sin^2a+cos^2a\right)^2-2sin^2a.cos^2a}{sin^2a.cos^2a}+\dfrac{1}{sin^2a.cos^2a}=7\)

\(\Leftrightarrow\dfrac{2}{sin^2a.cos^2a}=9\)

\(\Leftrightarrow\dfrac{8}{\left(2sina.cosa\right)^2}=9\)

\(\Leftrightarrow\dfrac{8}{sin^22a}=9\)

\(\Leftrightarrow sin^22a=\dfrac{8}{9}\)

\(\frac{sin2a-2sina}{sin2a+2sina}=\frac{2sina.cosa-2sina}{2sina.cosa+2sina}=\frac{2sina\left(cosa-1\right)}{2sina\left(cosa+1\right)}=\frac{cosa-1}{cosa+1}\)

\(=\frac{1-2sin^2\frac{a}{2}-1}{2cos^2\frac{a}{2}-1+1}=\frac{-sin^2\frac{a}{2}}{cos^2\frac{a}{2}}=-tan^2\frac{a}{2}\)

\(\frac{sin^4x-sin^2x+cos^2x}{cos^4x-cos^2x+sin^2x}=\frac{sin^2x\left(sin^2x-1\right)+cos^2x}{cos^2x\left(cos^2x-1\right)+sin^2x}=\frac{-sin^2x.cos^2x+cos^2x}{-cos^2x.sin^2x+sin^2x}\)

\(=\frac{cos^2x\left(1-sin^2x\right)}{sin^2x\left(1-cos^2x\right)}=\frac{cos^4x}{sin^4x}=cot^4x\)

\(\frac{sin^3a-cos^3a}{sina-cosa}=\frac{\left(sina-cosa\right)\left[sin^2a+cos^2a+sina.cosa\right]}{sina-cosa}=1+sina.cosa=1+\frac{1}{2}sin2a\)

MN K BT?

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

\(\Rightarrow sin\left(A+B\right)=sin\left(180^0-C\right)=sinC\)

\(cos\left(A+B\right)=cos\left(180^0-C\right)=-cosC\)

\(tan\left(A+B\right)=tan\left(180^0-C\right)=-tanC\)

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

\(\Rightarrow sin\frac{A+B}{2}=sin\left(90^0-\frac{C}{2}\right)=cos\frac{C}{2}\)

\(cos\frac{A+B}{2}=cos\left(90^0-\frac{C}{2}\right)=sin\frac{C}{2}\)

\(tan\frac{A+B}{2}=tan\left(90-\frac{C}{2}\right)=cot\frac{C}{2}\)

c/ \(A+B=180^0-C\Rightarrow tan\left(A+B\right)=-tanC\)

\(\Leftrightarrow\frac{tanA+tanB}{1-tanA.tanB}=-tanC\)

\(\Leftrightarrow tanA+tanB=-tanC+tanA.tanB.tanC\)

\(\Leftrightarrow tanA+tanB+tanC=tanA.tanB.tanC\)

d/ \(sinA+sinB+sinC=2sin\frac{A+B}{2}cos\frac{A-B}{2}+2sin\frac{C}{2}.cos\frac{C}{2}\)

\(=2cos\frac{C}{2}.cos\frac{A-B}{2}+2sin\frac{C}{2}.cos\frac{C}{2}\)

\(=2cos\frac{C}{2}\left(cos\frac{A-B}{2}+sin\frac{C}{2}\right)\)

\(=2cos\frac{C}{2}\left(cos\frac{A-B}{2}+cos\frac{A+B}{2}\right)\)

\(=4cos\frac{C}{2}.cos\frac{A}{2}.cos\frac{B}{2}\)

e/

\(cosA+cosB+cosC=2cos\frac{A+B}{2}cos\frac{A-B}{2}+1-2sin^2\frac{C}{2}\)

\(=1+2sin\frac{C}{2}.cos\frac{A-B}{2}-2sin^2\frac{C}{2}\)

\(=1+2sin\frac{C}{2}\left(cos\frac{A-B}{2}-sin\frac{C}{2}\right)\)

\(=1+2sin\frac{C}{2}\left(cos\frac{A-B}{2}-cos\frac{A+B}{2}\right)\)

\(=1+4sin\frac{C}{2}.sin\frac{A}{2}sin\frac{B}{2}\)

f/

\(sin2A+sin2B+sin2C=2sin\left(A+B\right).cos\left(A-B\right)+2sinC.cosC\)

\(=2sinC.cos\left(A-B\right)+2sinC.cosC\)

\(=2sinC\left(cos\left(A-B\right)+cosC\right)\)

\(=2sinC\left[cos\left(A-B\right)-cos\left(A+B\right)\right]\)

\(=4sinC.sinA.sinB\)

g/

\(cos^2A+cos^2B+cos^2C=\frac{1}{2}+\frac{1}{2}cos2A+\frac{1}{2}+\frac{1}{2}cos2B+cos^2C\)

\(=1+\frac{1}{2}\left(cos2A+cos2B\right)+cos^2C\)

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

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

\(=1-cosC\left(cos\left(A-B\right)-cosC\right)\)

\(=1-cosC\left[cos\left(A-B\right)+cos\left(A+B\right)\right]\)

\(=1-2cosC.cosA.cosB\)

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