a, \(\sin25^0\)< \(\sin70^0\)

b, \(\cos40^0\)> \(\cos75^0\)

c, \(\sin35^0\)= \(\cos55^0\)

\(\cos55^0\)< \(\cos35^0\)

\(\Rightarrow\)\(\sin35^0\)< \(\cos35^0\)

#mã mã#

a) $tg{28}^{\circ }=\frac{\sin {28}^{\circ }}{\cos {28}^{\circ }}=\sin {28}^{\circ }.\frac{1}{\cos {28}^{\circ }}$ (1)

Vì 0 < cos28° < 1 nên $\frac{1}{\cos {28}^{\circ }}>1\Rightarrow \sin {28}^{\circ }.\frac{1}{\cos {28}^{\circ }}>\sin {28}^{\circ }$ (2)

Từ (1) và (2) suy ra: tg28° > sin28°

b) Ta có: $\cot g{42}^{\circ }=\frac{\cos {42}^{\circ }}{\sin {42}^{\circ }}=c{os42}^{\circ }.\frac{1}{\sin {42}^{\circ }}$ (1)

Vì 0 < sin42° < 1 nên $\frac{1}{\sin {42}^{\circ }}>1\Rightarrow \cos {42}^{\circ }.\frac{1}{\sin {42}^{\circ }}>\cos {42}^{\circ }$ (2)

Từ (1) và (2) suy ra: cotg42° > cos42°

c) Ta có: 17° +73° =90° (1)

$\cot g{73}^{\circ }=\frac{\cos {73}^{\circ }}{\sin {73}^{\circ }}=\cos {73}^{\circ }.\frac{1}{\sin {73}^{\circ }}$ (2)

Vì 0 <sin73° <1 nên $\frac{1}{\sin {73}^{\circ }}>1\Rightarrow c{os73}^{\circ }.\frac{1}{\sin {73}^{\circ }}>c{os73}^{\circ }$ (3)

Từ (1), (2) và (3) suy ra: cotg73° > sin17°

d) Ta có: 32° +58° = 90° (1)

$tg{32}^{\circ }=\frac{\sin {32}^{\circ }}{\cos {32}^{\circ }}=\sin {32}^{\circ }.\frac{1}{\cos {32}^{\circ }}$ (2)

Vì 0 < cos32° < 1 nên $\frac{1}{{cos32}^{\circ }}>1\Rightarrow \sin {32}^{\circ }.\frac{1}{{cos32}^{\circ }}>\sin {32}^{\circ }$ (3)

Từ (1), (2) và (3) suy ra: tg32° > cos58°

a: \(\tan50^028'< \tan63^0\)

b: \(\cot14^0>\cot35^012'\)

c: \(\tan27^0=\cot63^0< \cot27^0\)

d: \(\tan65^0=\cot25^0>\cot65^0\)

\(a,=\)

\(b,=\)

\(c,\)ko biết

a/ Có 25⁰ < 70⁰

=> sin 25⁰< sin 70⁰

b/ tương tự

c/ Có sin 35⁰ = cos 55⁰

Có 55⁰ > 35⁰

=> sin 35⁰ > cos 35⁰

CUNT

Chú ý rằng: sin45⁰ = cos45⁰, sin40⁰ = cos50⁰, sin50⁰ = cos40⁰

Ta được:

\(\dfrac{\cos50^0-\cos45^0+\cos50^0}{\cos40^0-\cos45^0+\cos50^0}-\dfrac{6\times3\left(\dfrac{\sqrt{3}}{3}+\tan15^0\right)}{3\left(1-\dfrac{\sqrt{3}}{3}\tan15^0\right)}\)

\(=1-6\left(\dfrac{\tan30^0+\tan15^0}{1-\tan30^0\times\tan15^0}\right)\)

\(=1-6\tan45^0=-5\)

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

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

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

dấu nào là sao z b oi