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a)Do \(0^o< \alpha< 90^o\) nên \(0< sin\alpha< 1;0< cos\alpha< 1\).
Giả sử: \(tan\alpha< sin\alpha\Leftrightarrow\dfrac{sin\alpha}{cos\alpha}< sin\alpha\)
\(\Leftrightarrow sin\alpha< sin\alpha cos\alpha\)
\(\Leftrightarrow sin\alpha\left(1-cos\alpha\right)< 0\)
\(\Leftrightarrow1-cos\alpha< 0\)
\(\Leftrightarrow cos\alpha>1\) (vô lý).
b) \(sin\alpha+cos\alpha=sin\alpha+sin\left(\dfrac{\pi}{2}-\alpha\right)\)
\(=2.sin\dfrac{\pi}{4}cos\left(\dfrac{\pi}{4}-\alpha\right)=\sqrt{2}cos\left(\dfrac{\pi}{4}-\alpha\right)\)
\(=\sqrt{2}sin\left(\dfrac{\pi}{4}+\alpha\right)=\sqrt{2}sin\left(45^o+\alpha\right)\).
Do \(0^o< \alpha< 90^o\) nên \(45^o< \alpha+45^o< 135^o\).
Vì vậy \(\dfrac{\sqrt{2}}{2}< sin\left(\alpha+45^o\right)< 1\).
Từ đó suy ra \(\sqrt{2}.sin\left(45^o+\alpha\right)>\sqrt{2}.\dfrac{\sqrt{2}}{2}=1\) (Đpcm).
a) \(0< \alpha< 90^o\)
b) \(90^o< \alpha< 180^o\)
c) \(0< \alpha< 90^o\)
d) \(90^o< \alpha< 180^o\)
a)
\(A=\left(sin\alpha+cos\alpha\right)^2+\left(sin\alpha-cos\alpha\right)^2\)
\(=1+2sin\alpha cos\alpha+1-2sin\alpha cos\alpha=2\) (không phụ thuộc vào \(\alpha\)).
b)
\(B=sin^4\alpha-cos^4\alpha-2sin^2\alpha+1\)
\(=\left(sin^2\alpha+cos^2\alpha\right)\left(sin^2\alpha-cos^2\alpha\right)-2sin^2\alpha+1\)
\(=sin^2\alpha-cos^2\alpha-2sin^2\alpha+1\)
\(=-sin^2\alpha-cos^2\alpha+1\)
\(=-\left(sin^2\alpha+cos^2\alpha\right)+1=-1+1=0\).
a)\(\sin^2\alpha+\cos^2\alpha=1\Rightarrow\sin^2\alpha=1-\cos^2\alpha\)
\(\Rightarrow1-2^2=-3\) \(\Rightarrow\cos=-\sqrt{3}\left(0< \alpha< \dfrac{\pi}{2}\right)\)
b) \(\tan\alpha\times\cot\alpha=1\Rightarrow\tan\alpha=\dfrac{1}{\cot\alpha}\Rightarrow\tan=\dfrac{1}{4}\)
a)Do \(0< \alpha< \dfrac{\pi}{2}\) nên các giá trị lượng giác của \(\alpha\) đều dương.
\(cos\alpha=2sin\alpha\)(1)
Nếu \(sin\alpha=0\Rightarrow cos\alpha\) (vô lý).
Vì vậy \(sin\alpha\ne0\) . Từ (1) \(\Rightarrow\dfrac{cos\alpha}{sin\alpha}=2\)\(\Leftrightarrow cot\alpha=2\).
Suy ra: \(tan\alpha=\dfrac{1}{2}\).
\(sin\alpha=\sqrt{\dfrac{1}{1+cot^2\alpha}}=\dfrac{1}{\sqrt{3}}\).
\(cos\alpha=\sqrt{1-sin^2\alpha}=\sqrt{\dfrac{2}{3}}\).
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
\(A=\dfrac{3sin\alpha-cos\alpha}{sin\alpha+cos\alpha}=\dfrac{\dfrac{3sin\alpha}{cos\alpha}-1}{\dfrac{sin\alpha}{cos\alpha}-1}=\dfrac{3tan\alpha-1}{tan\alpha-1}\)\(=\dfrac{3\sqrt{2}-1}{\sqrt{2}-1}=5+2\sqrt{2}\).
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) \(B=4\left(sin^4\alpha+cos^4\alpha\right)-cos4\alpha\)
\(=4\left(sin^4\alpha+cos^4\alpha+2sin^2\alpha cos^2\alpha\right)-8sin^2\alpha cos^2\alpha\)\(-\left(1-2sin^22\alpha\right)\)
\(=4.\left(sin^2\alpha+cos^2\alpha\right)^2-2sin^22\alpha-1+2sin^22\alpha\)
\(=4-1=3\).