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Nhân cả tử và mẫu của phân số chứa tan với \(sina.cosa\)
\(A=\frac{sin^2x-cos^2x}{sin^2x+cos^2x}+cos2x=sin^2x-cos^2x+cos2x=-cos2x+cos2x=0\)
\(B=\frac{1+sin4a-cos4a}{1+sin4a+cos4a}=\frac{1+2sin2a.cos2a-\left(1-2sin^22a\right)}{1+2sin4a.cos4a+2cos^22a-1}\)
\(B=\frac{2sin2a\left(sin2a+cos2a\right)}{2cos2a\left(sin2a+cos2a\right)}=\frac{sin2a}{cos2a}=tan2a\)
\(C=\frac{3-4cos2a+2cos^22a-1}{3+4cos2a+2cos^22a-1}=\frac{2\left(cos^22a-2cos2a-1\right)}{2\left(cos^22a+2cos2a+1\right)}\)
\(C=\frac{\left(cos2a-1\right)^2}{\left(cos2a+1\right)^2}=\frac{\left(1-2sin^2a-1\right)^2}{\left(2cos^2a-1+1\right)^2}=\frac{sin^4a}{cos^4a}=tan^4a\)
\(D=\frac{sin^22a+4sin^4a-\left(2sina.cosa\right)^2}{4-4sin^2a-sin^22a}=\frac{sin^22a+4sin^4a-sin^22a}{4\left(1-sin^2a\right)-\left(2sina.cosa\right)^2}=\frac{4sin^4a}{4cos^2a-4sin^2a.cos^2a}\)
\(=\frac{sin^4a}{cos^2a\left(1-sin^2a\right)}=\frac{sin^4a}{cos^2a.cos^2a}=\frac{sin^4a}{cos^4a}=tan^4a\)
\(\dfrac{1+cos2a-sin2a}{1+cos2a+sin2a}=\dfrac{2cos^2a-2sina.cosa}{2cos^2a+2sinacosa}\)
\(=\dfrac{2cosa\left(cosa-sina\right)}{2cosa\left(cosa+sina\right)}=\dfrac{cosa-sina}{cosa+sina}=\dfrac{\sqrt{2}sin\left(\dfrac{\pi}{4}-a\right)}{\sqrt{2}cos\left(\dfrac{\pi}{4}-a\right)}=tan\left(\dfrac{\pi}{4}-a\right)\)
\(\dfrac{1+cos2a-cosa}{sin2a-sina}=\dfrac{2cos^2a-cosa}{2sina.cosa-sina}=\dfrac{cosa\left(2cosa-1\right)}{sina\left(2cosa-1\right)}=\dfrac{cosa}{sina}=cota\)
\(sin^6a+cos^6a=\left(sin^2x\right)^3+\left(cos^2x\right)^3\)
\(=\left(sin^2x+cos^2x\right)\left(sin^4x+cos^4x-sin^2x.cos^2x\right)\)
\(=sin^4x+2sin^2xcos^2x+cos^4x-3sin^2x.cos^2x\)
\(=\left(sin^2x+cos^2x\right)^2-\frac{3}{4}.\left(2sinx.cosx\right)^2\)
\(=1-\frac{3}{4}sin^22x=1-\frac{3}{4}\left(\frac{1}{2}-\frac{1}{2}cos4x\right)=\frac{5}{8}+\frac{3}{8}cos4x\)
2/
\(\frac{1+sin2a-cos2a}{1+cos2a}=\frac{1+2sina.cosa-\left(1-2sin^2a\right)}{1+2cos^2a-1}=\frac{2sina.cosa+2sin^2a}{2cos^2a}\)
\(=\frac{2sina.cosa}{2cos^2a}+\frac{2sin^2a}{2cos^2a}=tana+tan^2a\)
\(\dfrac{sina+sin5a+sin3a}{cosa+cos5a+cos3a}=\dfrac{2sin3a.cos2a+sin3a}{2cos3a.cos2a+cos3a}=\dfrac{sin3a\left(2cos2a+1\right)}{cos3a\left(2cos2a+1\right)}=\dfrac{sin3a}{cos3a}=tan3a\)
\(\dfrac{1+sin4a-cos4a}{1+sin4a+cos4a}=\dfrac{1+2sin2a.cos2a-\left(1-2sin^22a\right)}{1+2sin2a.cos2a+2cos^22a-1}=\dfrac{2sin2a\left(sin2a+cos2a\right)}{2cos2a\left(sin2a+cos2a\right)}=\dfrac{sin2a}{cos2a}=tan2a\)
\(96\sqrt{3}sin\left(\dfrac{\pi}{48}\right)cos\left(\dfrac{\pi}{48}\right)cos\left(\dfrac{\pi}{24}\right)cos\left(\dfrac{\pi}{12}\right)cos\left(\dfrac{\pi}{6}\right)=48\sqrt{3}sin\left(\dfrac{\pi}{24}\right)cos\left(\dfrac{\pi}{24}\right)cos\left(\dfrac{\pi}{12}\right)cos\left(\dfrac{\pi}{6}\right)\)
\(=24\sqrt{3}sin\left(\dfrac{\pi}{12}\right)cos\left(\dfrac{\pi}{12}\right)cos\left(\dfrac{\pi}{6}\right)=12\sqrt{3}sin\left(\dfrac{\pi}{6}\right)cos\left(\dfrac{\pi}{6}\right)\)
\(=6\sqrt{3}sin\left(\dfrac{\pi}{3}\right)=6\sqrt{3}.\dfrac{\sqrt{3}}{2}=9\)
\(A+B+C=\pi\Rightarrow A+B=\pi-C\Rightarrow tan\left(A+B\right)=tan\left(\pi-C\right)\)
\(\Rightarrow\dfrac{tanA+tanB}{1-tanA.tanB}=-tanC\Rightarrow tanA+tanB=-tanC+tanA.tanB.tanC\)
\(\Rightarrow tanA+tanB+tanC=tanA.tanB.tanC\)
a) \(A=2\left(sin^6\alpha+cos^6\alpha\right)-3\left(sin^4\alpha+cos^4\alpha\right)\)
\(=2\left(sin^2\alpha+cos^2\alpha\right)\left(sin^4\alpha-sin^2\alpha cos^2\alpha+cos^4\alpha\right)\)\(-3\left(sin^4\alpha+cos^4\alpha\right)\)
\(=2\left(sin^4\alpha+cos^4\alpha-sin^2\alpha cos^2\alpha\right)-3\left(sin^4\alpha+cos^4\alpha\right)\)
\(=-\left(sin^4\alpha+cos^4\alpha+2sin^2\alpha cos^2\alpha\right)\)
\(=-\left(sin^2\alpha+cos^2\alpha\right)^2=-1\) (Không phụ thuộc vào \(\alpha\)).
b) \(B=4\left(sin^4\alpha+cos^4\alpha\right)-cos4\alpha\)
\(=4\left(sin^4\alpha+cos^4\alpha+2sin^2\alpha cos^2\alpha\right)-8sin^2\alpha cos^2\alpha\)\(-\left(1-2sin^22\alpha\right)\)
\(=4.\left(sin^2\alpha+cos^2\alpha\right)^2-2sin^22\alpha-1+2sin^22\alpha\)
\(=4-1=3\).
a) P = sin2α + sin2α.\(\frac{cos\text{α}}{sin\text{α}}\) + cos2α - cos2α.\(\frac{sin\text{α}}{cos\text{α}}\)
=sin2α + sinα.cosα + cos2α - cosα.sinα
=sin2α + cos2α
=1
Vậy P không phụ thuộc vào α
b) Q= -cos4α(2cos2α -1 -2) +sin4α(1 -2sin2α+2)
= -cos4α(cos2α -2) +sin4α(cos2α +2)
=-cos4α.cos2α +2cos4α +sin4α.cos2α +2sin4α
=cos2α(sin4α -cos4α) +2(sin4α +cos4α)
=cos2α [\(\left(\frac{1-cos^22\text{α}}{2}\right)^2-\left(\frac{1+cos^22\text{α}}{2}\right)^2\)]+2.[\(\left(\frac{1-cos^22\text{α}}{2}\right)^2+ \left(\frac{1+cos^22\text{α}}{2}\right)^2\)]
= -cos2α.cos2α +1+cos22α
= -cos22α +1+cos22α
=1
Vậy Q không phụ thuộc vào α