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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) \(\dfrac{tan\alpha-tan\beta}{cot\beta-cot\alpha}=\dfrac{\dfrac{sin\alpha}{cos\alpha}-\dfrac{sin\beta}{cos\beta}}{\dfrac{cos\beta}{sin\beta}-\dfrac{cos\alpha}{sin\alpha}}\)
\(=\dfrac{\dfrac{sin\alpha cos\beta-cos\alpha sin\beta}{cos\alpha cos\beta}}{\dfrac{cos\beta sin\alpha-cos\alpha sin\beta}{sin\beta sin\alpha}}\)
\(=\dfrac{sin\beta sin\alpha}{cos\beta cos\alpha}=tan\alpha tan\beta\).
b) \(tan100^o+\dfrac{sin530^o}{1+sin640^o}=tan100^o+\dfrac{sin170^o}{1+sin280^o}\)
\(=-cot10^o+\dfrac{sin10^o}{1-sin80^o}\)\(=\dfrac{-cos10^o}{sin10^o}+\dfrac{sin10^o}{1-cos10^o}\)
\(=\dfrac{-cos10^o+cos^210^o+sin^210^o}{sin10^o\left(1-cos10^o\right)}\) \(=\dfrac{1-cos10^o}{sin10^o\left(1-cos10^o\right)}=\dfrac{1}{sin10^o}\) .
a)
\(\sin ^4a-\cos ^4a+1=(\sin ^2a-\cos ^2a)(\sin ^2a+\cos^2a)+1\)
\(=(\sin ^2a-\cos ^2a).1+1=\sin ^2a-\cos ^2a+\sin ^2a+\cos ^2a\)
\(=2\sin ^2a\)
b) \(\sin ^2a+2\cos ^2a-1=(\sin ^2a+\cos^2a)+\cos ^2a-1\)
\(=1+\cos ^2a-1=\cos ^2a\)
\(\Rightarrow \frac{\sin ^2a+2\cos ^2a-1}{\cot ^2a}=\frac{\cos ^2a}{\cot ^2a}=\frac{\cos ^2a}{\frac{\cos ^2a}{\sin ^2a}}=\sin ^2a\)
c)
\(\frac{1-\sin ^2a\cos ^2a}{\cos ^2a}-\cos ^2a=\frac{1}{\cos ^2a}-\sin ^2a-\cos ^2a\)
\(=\frac{1}{\cos ^2a}-(\sin ^2a+\cos ^2a)=\frac{1}{\cos ^2a}-1\)
\(=\frac{1-\cos ^2a}{\cos ^2a}=\frac{\sin ^2a}{\cos ^2a}=\tan ^2a\)
d)
\(\frac{\sin ^2a-\tan ^2a}{\cos ^2a-\cot ^2a}=\frac{\sin ^2a-\frac{\sin ^2a}{\cos ^2a}}{\cos ^2a-\frac{\cos ^2a}{\sin ^2a}}\) \(=\frac{\sin ^2a(1-\frac{1}{\cos ^2a})}{\cos ^2a(1-\frac{1}{\sin ^2a})}\)
\(=\frac{\sin ^2a.\frac{\cos ^2a-1}{\cos ^2a}}{\cos ^2a.\frac{\sin ^2a-1}{\sin ^2a}}\) \(=\frac{\sin ^2a.\frac{-\sin ^2a}{\cos ^2a}}{\cos ^2a.\frac{-\cos ^2a}{\sin ^2a}}=\frac{\sin ^6a}{\cos ^6a}=\tan ^6a\)
f)
\(\frac{(\sin a+\cos a)^2-1}{\cot a-\sin a\cos a}=\frac{\sin ^2a+\cos ^2a+2\sin a\cos a-1}{\frac{\cos a}{\sin a}-\sin a\cos a}\)
\(=\sin a.\frac{1+2\sin a\cos a-1}{\cos a-\cos a\sin ^2a}\)
\(=\sin a. \frac{2\sin a\cos a}{\cos a(1-\sin ^2a)}=\sin a. \frac{2\sin a\cos a}{\cos a. \cos^2 a}=\frac{2\sin ^2a}{\cos ^2a}=2\tan ^2a\)
\(\cot a=\dfrac{1}{3}\)
\(1+\tan^2a=\dfrac{1}{\cos^2a}=1+9=10\)
\(\Leftrightarrow\cos a=\dfrac{\sqrt{10}}{10}\)
\(\Leftrightarrow\sin a=\dfrac{3\sqrt{10}}{10}\)
\(A=5\cdot\dfrac{3\sqrt{10}}{10}-7\cdot\dfrac{1}{10}+9\cdot\dfrac{1}{9}=\dfrac{3\sqrt{10}}{2}-\dfrac{7}{10}+1=\dfrac{3+15\sqrt{10}}{10}\)
a) \(sin20^o+2sin40^o-sin100^o=sin20^o-sin100^o+2sin40^o\)
\(=2cos60^osin\left(-40^o\right)+2sin40^o\)\(=-2cos60^osin40^o+2sin40^o\)
\(=2sin40^o\left(-cos60^o+1\right)=2sin40^o.\left(-\dfrac{1}{2}+1\right)=sin40^o\)(đpcm).
b) \(\dfrac{sin\left(45^o+\alpha\right)-cos\left(45^o+\alpha\right)}{sin\left(45^o+\alpha\right)+cos\left(45^o+\alpha\right)}\)
\(=\dfrac{sin\left(45^o+\alpha\right)-sin\left(45^o-\alpha\right)}{sin\left(45^o+\alpha\right)+sin\left(45^o-\alpha\right)}=\dfrac{2cos45^o.sin\alpha}{2sin45^o.cos\alpha}\)
\(=tan\alpha\) (Đpcm).
a) \(0< \alpha< 90^o\)
b) \(90^o< \alpha< 180^o\)
c) \(0< \alpha< 90^o\)
d) \(90^o< \alpha< 180^o\)
a) \(sin\left(270^o-\alpha\right)=sin\left(-90^o-\alpha\right)=-sin\left(90^o+\alpha\right)\)\(=-cos\alpha\).
b) \(cos\left(270^o-\alpha\right)=cos\left(-90^o-\alpha\right)=cos\left(90^o+\alpha\right)\)\(=-sin\alpha\).
c) \(sin\left(270^o+\alpha\right)=sin\left(-90^o+\alpha\right)=-sin\left(90^o-\alpha\right)\)\(=-cos\alpha\).
d) \(cos\left(270^o+\alpha\right)=cos\left(-90^o+\alpha\right)=cos\left(90^o-\alpha\right)\)\(=sin\alpha\).
a) \(sin6\alpha cot3\alpha cos6\alpha=2.sin3\alpha.cos3\alpha\dfrac{cos3\alpha}{sin3\alpha}-cos6\alpha\)
\(=2cos^23\alpha-\left(2cos^23\alpha-1\right)=1\) (Không phụ thuộc vào x).
b) \(\left[tan\left(90^o-\alpha\right)-cot\left(90^o+\alpha\right)\right]^2\)\(-\left[cot\left(180^o+\alpha\right)+cot\left(270^o+\alpha\right)\right]^2\)
\(=\left[cot\alpha+cot\left(90^o-\alpha\right)\right]^2\)\(-\left[cot\alpha+cot\left(90^o+\alpha\right)\right]^2\)
\(=\left[cot\alpha+tan\alpha\right]^2-\left[cot\alpha-tan\alpha\right]^2\)
\(=4tan\alpha cot\alpha=4\). (Không phụ thuộc vào \(\alpha\)).
đáp án D đúng nhé bạn
Ok. Mình cảm ơn