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:
\(cot7,5^o+tan67,5^o-tan7,5^o-cot67,5^o\)
\(=\left(cot7,5^o-tan7,5^o\right)-\left(cot67,5^o-tan67,5^o\right)\)
\(=\dfrac{2cos15^o}{sin15^o}-\dfrac{2cos135^o}{sin135^o}\)
\(=2\left(\dfrac{cos15^osin135^o-sin15^ocos135^o}{sin15^osin135^o}\right)\)
\(=2.\dfrac{sin120^o}{\dfrac{1}{2}\left(cos120^o-cos150^o\right)}\)
\(=\dfrac{4.\dfrac{\sqrt{3}}{2}}{\dfrac{-1}{2}+\dfrac{\sqrt{3}}{2}}=\dfrac{4\sqrt{3}}{\sqrt{3}-1}\)

\(\cos^2a+sin^2a=1 ;\frac{\sin a}{\cos a}=\tan a\)

Thanks

cong là j vậy?

\(A=tan18^otan288+sin32^osin148^o-sin302^osin122^o\)
\(=tan18^o.tan\left(-72^o\right)+sin32^o.sin32^o+sin58^o.sin58^o\)
\(=-tan18^o.cot18^o+sin^232^o+sin^258^o\)
\(=-1+sin^232^o+cos^232^2=-1+1=0\).

b) \(B=\dfrac{1+sin^4\alpha-cos^4\alpha}{1-sin^6\alpha-cos^6\alpha}\)
\(=\dfrac{1+\left(sin^2\alpha+cos^2\alpha\right)\left(sin^2\alpha-cos^2\alpha\right)}{1-\left(sin^6\alpha+cos^6\alpha\right)}\)
\(=\dfrac{1+sin^2\alpha-cos^2\alpha}{1-\left(sin^2\alpha+cos^2\alpha\right)\left(sin^2\alpha-sin\alpha cos\alpha+cos^2\alpha\right)}\)
\(=\dfrac{sin^2\alpha+1-cos^2\alpha}{1-\left(1-sin\alpha.cos\alpha\right)}\)
\(=\dfrac{sin^2\alpha+sin^2\alpha}{sin\alpha cos\alpha}\)
\(=\dfrac{2sin^2\alpha}{sin\alpha cos\alpha}=\dfrac{2sin\alpha}{cos\alpha}=2tan\alpha\).

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).

d) \(sin200^osin310^o+cos340^ocos50^o\)
\(=sin20^o.sin50^o+cos20^ocos50^o\)
\(=cos\left(50^o-20^o\right)=cos30^o\).

\(\cos\alpha=\sqrt{1-\sin^2\alpha}=\sqrt{1-\frac{4}{9}}=\frac{\sqrt{5}}{3}\)

\(\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{\frac{2}{3}}{\frac{\sqrt{5}}{3}}=\frac{2\sqrt{5}}{5}\)

\(\cot=\frac{1}{\tan}=\frac{1}{\frac{2\sqrt{5}}{5}}=\frac{\sqrt{5}}{2}\)

a) Ta có: \(\sin^2\alpha+\cos^2\alpha=1\Rightarrow\cos^2a=1-\sin^2\alpha=1-\left(\frac{\sqrt{3}}{2}\right)^2=\frac{1}{4}\)

\(\Rightarrow\cos\alpha=\frac{1}{2}\)(do \(\cos\alpha>0\))

b) \(Q=\sin^2\alpha+\cot^2\alpha.\sin^2\alpha=\sin^2\alpha\left(1+\cot^2\alpha\right)\)\(=\sin^2\alpha\left(1+\frac{\cos^2\alpha}{\sin^2\alpha}\right)=\sin^2\alpha.\frac{\sin^2\alpha+\cos^2\alpha}{\sin^2\alpha}=1\)

a) \(\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3}\)

Bài 2:

Sửa đề: \(\sin\alpha=\dfrac{3}{5}\)

Ta có: \(\sin^2\alpha+\cos^2\alpha=1\)

\(\Leftrightarrow\cos^2\alpha=1-\left(\dfrac{3}{5}\right)^2=1-\dfrac{9}{25}=\dfrac{16}{25}\)

\(\Leftrightarrow\cos\alpha=\dfrac{4}{5}\)

Ta có: \(\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}\)

\(=\dfrac{3}{5}:\dfrac{4}{5}=\dfrac{3}{5}\cdot\dfrac{5}{4}=\dfrac{3}{4}\)

Ta có: \(\cot\alpha=\dfrac{1}{\tan\alpha}\)

\(=\dfrac{1}{\dfrac{3}{4}}=\dfrac{4}{3}\)