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a.
\(\dfrac{sina+sin5a+sin3a}{cosa+cos5a+cos3a}=\dfrac{2sin3a.cosa+sin3a}{2cos3a.cosa+cos3a}=\dfrac{sin3a\left(2cosa+1\right)}{cos3a\left(2cosa+1\right)}=\dfrac{sin3a}{cos3a}=tan3a\)
b.
\(\dfrac{1+cosa}{1-cosa}.\dfrac{sin^2\dfrac{a}{2}}{cos^2\dfrac{a}{1}}-cos^2a=\dfrac{1+cosa}{1-cosa}.\dfrac{\dfrac{1-cosa}{2}}{\dfrac{1+cosa}{2}}-cos^2a\)
\(=\dfrac{1+cosa}{1-cosa}.\dfrac{1-cosa}{1+cosa}-cos^2a=1-cos^2a=sin^2a\)
\(\frac{sin2a-2sina}{sin2a+2sina}=\frac{2sina.cosa-2sina}{2sina.cosa+2sina}=\frac{2sina\left(cosa-1\right)}{2sina\left(cosa+1\right)}=\frac{cosa-1}{cosa+1}\)
\(=\frac{1-2sin^2\frac{a}{2}-1}{2cos^2\frac{a}{2}-1+1}=\frac{-sin^2\frac{a}{2}}{cos^2\frac{a}{2}}=-tan^2\frac{a}{2}\)
\(\frac{sin^4x-sin^2x+cos^2x}{cos^4x-cos^2x+sin^2x}=\frac{sin^2x\left(sin^2x-1\right)+cos^2x}{cos^2x\left(cos^2x-1\right)+sin^2x}=\frac{-sin^2x.cos^2x+cos^2x}{-cos^2x.sin^2x+sin^2x}\)
\(=\frac{cos^2x\left(1-sin^2x\right)}{sin^2x\left(1-cos^2x\right)}=\frac{cos^4x}{sin^4x}=cot^4x\)
\(\frac{sin^3a-cos^3a}{sina-cosa}=\frac{\left(sina-cosa\right)\left[sin^2a+cos^2a+sina.cosa\right]}{sina-cosa}=1+sina.cosa=1+\frac{1}{2}sin2a\)
\(\dfrac{cosa+cos5a+cos3a}{sina+sin5a+sin3a}=\dfrac{2cos3a.cos2a+cos3a}{2sin3a.cos2a+sin3a}\)
\(=\dfrac{cos3a\left(2cos2a+1\right)}{sin3a\left(2cos2a+1\right)}=\dfrac{cos3a}{sin3a}=cot3a\)
\(\left(\dfrac{cosa}{sinb}+\dfrac{sina}{cosb}\right)\left(\dfrac{1-cos4b}{cos\left(a-b\right)}\right)=\dfrac{\left(cosa.cosb+sina.sinb\right)}{sinb.cosb}.\dfrac{2sin^22b}{cos\left(a-b\right)}\)
\(=\dfrac{cos\left(a-b\right)}{\dfrac{1}{2}sin2b}.\dfrac{2sin^22b}{cos\left(a-b\right)}=4sin2b\)
\(\frac{1+sin^2x}{1-sin^2x}=\frac{1+sin^2x}{cos^2x}=\frac{1}{cos^2x}+\frac{sin^2x}{cos^2x}=1+tan^2x+tan^2x=1+2tan^2x\)
\(\frac{sin^3a-cos^3a}{sina-cosa}-sina.cosa=\frac{\left(sina-cosa\right)\left(sin^2a+cos^2a+sina.cosa\right)}{sina-cosa}-sina.cosa\)
\(=sin^2a+cos^2a+sina.cosa-sina.cosa=1\)
\(\frac{1+cos2x+cosx+cos3x}{2cos^2x-1+cosx}=\frac{1+2cos^2x-1+2cosx.cos2x}{cos2x+cosx}=\frac{2cosx\left(cosx+cos2x\right)}{cos2x+cosx}=2cosx\)
\(\frac{1-2sin^2a}{cosa+sina}+\frac{2cos^2a-1}{cosa-sina}=\frac{cos^2a-sin^2a}{cosa+sina}+\frac{cos^2a-sin^2a}{cosa-sina}\)
\(=\frac{\left(cosa+sina\right)\left(cosa-sina\right)}{cosa+sina}+\frac{\left(cosa+sina\right)\left(cosa-sina\right)}{cosa-sina}=cosa-sina+cosa+sina=2cosa\)
\(\frac{1-cosx+cos2x}{sin2x-sinx}=\frac{1-cosx+2cos^2x-1}{2sinx.cosx-sinx}=\frac{cosx\left(2cosx-1\right)}{sinx\left(2cosx-1\right)}=\frac{cosx}{sinx}=cotx\)
\(sina+cosa=\sqrt{2}\Leftrightarrow\left(sina+cosa\right)^2=2\\ \)
\(\Leftrightarrow\sin^2a+2\sin a.cosa+cos^2a=2\)
\(\Leftrightarrow1+2.sina.cosa=2\)
\(\Leftrightarrow2.sina.cosa=2-1=1\)
\(\Leftrightarrow\sin a.cosa=\frac{1}{2}\)
Vậy P=sina.cosa=\(\frac{1}{2}\)
\(Q=\sin^4a+cos^4a\)
\(\Leftrightarrow\left(sin^2a\right)^2+\left(cos^2a\right)^2\)
\(\Leftrightarrow\left(sin^2a+cos^2a\right)^2-2.sin^2a.cos^2a\)
\(\Leftrightarrow1^2-2.sin^2a.cos^2a\) tách tiếp rồi thế vào là được .tương tự phàn P ý
còn R thì tách sin^3a=sin^2a+sina tương tự cos mũ 3 a cụng vậy
theo tớ là như thế còn có sai thì đừng có ném đá ném gạch na
\(\dfrac{sina}{sina-cosa}-\dfrac{cosa}{cosa-sina}=\dfrac{sina+cosa}{sina-cosa}=\dfrac{1+cota}{1-cota}=\dfrac{\left(1+cota\right)^2}{1-cot^2a}\)
Đề bài ko đúng
\(sina.cos^5a-cosa.sin^5a=sina.cosa\left(cos^4a-sin^4a\right)\)
\(=\frac{1}{2}sin2a\left(cos^2a-sin^2a\right)\left(cos^2a+sin^2a\right)\)
\(=\frac{1}{2}sin2a.cos2a=\frac{1}{4}sin4a\)