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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).
Chú ý rằng: sin450 = cos450, sin400 = cos500, sin500 = cos400
Ta được:
\(\dfrac{\cos50^0-\cos45^0+\cos50^0}{\cos40^0-\cos45^0+\cos50^0}-\dfrac{6\times3\left(\dfrac{\sqrt{3}}{3}+\tan15^0\right)}{3\left(1-\dfrac{\sqrt{3}}{3}\tan15^0\right)}\)
\(=1-6\left(\dfrac{\tan30^0+\tan15^0}{1-\tan30^0\times\tan15^0}\right)\)
\(=1-6\tan45^0=-5\)
Bấm MODE nhập 5 nhập 3
a, bấm 5 = -3 = -7 = ta được \(x_1=\dfrac{3+\sqrt{149}}{10};x_2=\dfrac{3-\sqrt{149}}{10}\)
Tương tự cho các câu còn lại
a)\(sin^2\left(180^o-\alpha\right)+tan^2\left(180-\alpha\right).tan^2\left(270^o+\alpha\right)\)\(+sin\left(90^o+\alpha\right)cos\left(\alpha-360^o\right)\)
\(=sin^2\alpha+tan^2\alpha.cot^2\alpha+cos\alpha cos\alpha\)
\(=sin^2\alpha+cos^2\alpha+\left(tan\alpha cot\alpha\right)^2=1+1=2\).
\(\dfrac{cos\left(\alpha-180^o\right)}{sin\left(180^o-\alpha\right)}+\dfrac{tan\left(\alpha-180^o\right)cos\left(180^o+\alpha\right)sin\left(270^o+\alpha\right)}{tan\left(270^o+\alpha\right)}\)
\(=\dfrac{cos\left(180^o-\alpha\right)}{sin\left(180^o-\alpha\right)}+\dfrac{-tan\left(180^o-\alpha\right).cos\alpha.sin\left(90^o+\alpha\right)}{-tan\left(90^o+\alpha\right)}\)
\(=tan\left(180^o-\alpha\right)+\dfrac{tan\alpha.cos\alpha.cos\alpha}{cot\alpha}\)
\(=-tan\alpha+tan^2\alpha cos^2\alpha\)
\(=tan\alpha\left(-1+tan\alpha cos^2\alpha\right)\)
\(=tan\alpha\left(sin\alpha cos\alpha-1\right)\).
a)\(sin^4\dfrac{\pi}{16}+sin^4\dfrac{3\pi}{16}+sin^4\dfrac{5\pi}{16}+sin^4\dfrac{7\pi}{16}\)
\(=\left(sin^4\dfrac{\pi}{16}+sin^4\dfrac{7\pi}{16}\right)+\left(sin^4\dfrac{3\pi}{16}+sin^4\dfrac{5\pi}{16}\right)\)
\(=\left(sin^4\dfrac{\pi}{16}+cos^4\dfrac{\pi}{16}\right)+\left(sin^4\dfrac{3\pi}{16}+cos^4\dfrac{3\pi}{16}\right)\)
\(=1-2sin^2\dfrac{\pi}{16}cos^2\dfrac{\pi}{16}+1-2sin^2\dfrac{3\pi}{16}cos^2\dfrac{3\pi}{16}\)
\(=2-\dfrac{1}{2}sin^2\dfrac{\pi}{8}-\dfrac{1}{2}sin^2\dfrac{3\pi}{8}\)
\(=2-\dfrac{1}{2}\left(sin^2\dfrac{\pi}{8}+sin^2\dfrac{3\pi}{8}\right)\)
\(=2-\dfrac{1}{2}\left(sin^2\dfrac{\pi}{8}+cos^2\dfrac{\pi}{8}\right)\)
\(=2-\dfrac{1}{2}=\dfrac{3}{2}\).
Có: \(cotx-tanx=\dfrac{cosx}{sinx}-\dfrac{sinx}{cosx}=\dfrac{cos^2x-sin^2x}{sinxcosx}=\dfrac{2cos2x}{sin2x}\)
Vì vậy:
\(cot7,5^o+tan67,5^o-tan7,5^o-cot67,5^o\)
\(=\left(cot7,5^o-tan7,5^o\right)-\left(cot67,5^o-tan67,5^o\right)\)
\(=\dfrac{2cos15^o}{sin15^o}-\dfrac{2cos135^o}{sin135^o}\)
\(=2\left(\dfrac{cos15^osin135^o-sin15^ocos135^o}{sin15^osin135^o}\right)\)
\(=2.\dfrac{sin120^o}{\dfrac{1}{2}\left(cos120^o-cos150^o\right)}\)
\(=\dfrac{4.\dfrac{\sqrt{3}}{2}}{\dfrac{-1}{2}+\dfrac{\sqrt{3}}{2}}=\dfrac{4\sqrt{3}}{\sqrt{3}-1}\)
a) Do \(90^o< \alpha< 180^o\) nên \(sin\alpha>0;cos\alpha< 0\).
b) Do \(180^o< \alpha< 270^o\) nên \(sin\alpha< 0;cos\alpha< 0\).
c) Do \(270^o< \alpha< 360^o\) nên \(sin\alpha< 0;cos\alpha>0\).
d) \(\alpha=1280^o=3.360^o+200^o\)
\(sin1280^o=sin\left(3.360^o+200^o\right)=sin200^o< 0\).
e)
\(sin\left(-235^o\right)=sin\left(-235^o+360^o\right)=sin125^o>0\).
\(cos\left(-235^o\right)=cos\left(-235^o+360^o\right)=cos125^o< 0\).
d) \(sin\left(-1876\right)=sin\left(-1876^o+1800^o\right)=sin\left(-76^o\right)\)\(=-sin76^o< 0\).
\(cos\left(-1876^o\right)=cos\left(-76^o\right)=cos76^o>0\).
cau a : ĐỀ SAI.