Question

Rút gọn các biểu thức (không dùng bảng số và máy tính)

a) \(\sin^2\left(180^0-\alpha\right)+\tan^2\left(180^0-\alpha\right).\tan^2\left(270^0+\alpha\right)+\sin\left(90^0+\alpha\right)\cos\left(\alpha-360^0\right)\)

b) \(\dfrac{\cos\left(\alpha-180^0\right)}{\sin\left(180^0-\alpha\right)}+\dfrac{\tan\left(\alpha-180^0\right)\cos\left(180^0+\alpha\right)\sin\left(270^0+\alpha\right)}{\tan\left(270^0+\alpha\right)}\)

c) \(\dfrac{\cos\left(-288^0\right)\cot72^0}{\tan\left(-162^0\right)\sin108^0}-\tan18^0\)

d) \(\dfrac{\sin20^0\sin30^0\sin40^0\sin50^0\sin60^0\sin70^0}{\cos10^0\cos50^0}\)

Answer 1

a)\(sin^2\left(180^o-\alpha\right)+tan^2\left(180-\alpha\right).tan^2\left(270^o+\alpha\right)\)\(+sin\left(90^o+\alpha\right)cos\left(\alpha-360^o\right)\)
\(=sin^2\alpha+tan^2\alpha.cot^2\alpha+cos\alpha cos\alpha\)
\(=sin^2\alpha+cos^2\alpha+\left(tan\alpha cot\alpha\right)^2=1+1=2\).

Answer 2

\(\dfrac{cos\left(\alpha-180^o\right)}{sin\left(180^o-\alpha\right)}+\dfrac{tan\left(\alpha-180^o\right)cos\left(180^o+\alpha\right)sin\left(270^o+\alpha\right)}{tan\left(270^o+\alpha\right)}\)
\(=\dfrac{cos\left(180^o-\alpha\right)}{sin\left(180^o-\alpha\right)}+\dfrac{-tan\left(180^o-\alpha\right).cos\alpha.sin\left(90^o+\alpha\right)}{-tan\left(90^o+\alpha\right)}\)
\(=tan\left(180^o-\alpha\right)+\dfrac{tan\alpha.cos\alpha.cos\alpha}{cot\alpha}\)
\(=-tan\alpha+tan^2\alpha cos^2\alpha\)
\(=tan\alpha\left(-1+tan\alpha cos^2\alpha\right)\)
\(=tan\alpha\left(sin\alpha cos\alpha-1\right)\).

Answer 3

c) \(\dfrac{cos\left(-288^o\right)cot72^o}{tan\left(-162^o\right)sin108^o}-tan18^o\)
\(=\dfrac{cos72^ocot72^o}{tan18^o.sin72^o}-tan18^o\)
\(=\dfrac{cos^272^o.cos18^o}{sin72^osin18^o.sin72^o}-tan18^o\)
\(=cot^272^ocot18^o-tan18^o\)
\(=tan^218^ocot18^o-tan18^o\)
\(=tan18^o-tan18^o=0\).

Answer 4

d) \(\dfrac{sin20^osin30^osin40^osin50^osin60^osin70^o}{cos10^ocos50^o}\)
\(=\dfrac{sin20^o.sin70^osin30^o.sin60^osin40^o.50^o}{cos10^ocos50^o}\)\(=\dfrac{\dfrac{1}{2}.\left(cos50^o-cos90^o\right)}{cos10^ocos50^o}\)\(.\dfrac{1}{2}\left(cos30^o-cos90^o\right).\dfrac{1}{2}\left(cos10^o-cos90^o\right)\)
\(=\dfrac{cos50^o.cos30^o.cos10^o}{8cos10^ocos50^o}=\dfrac{cos30^o}{8}=\dfrac{\sqrt{3}}{16}\).