\(A=sin42^0-cos48^0=cos\left(90^0-42^0\right)-cos48^0=cos48^0-cos48^0=0\)

\(B=cot56^0-tan34^0=tan\left(90^0-56^0\right)-tan34^0=tan34^0-tan34^0=0\)

\(C=sin30^0-cot50^0-cos60^0+tan40^0\)

\(=cos\left(90^0-30^0\right)-tan\left(90^0-50^0\right)-cos60^0+tan40^0\)

\(=cos60^0-tan40^0-cos60^0+tan40^0=0\)

\(A=\sin42^0-\cos48^0=\sin42^0-\sin42^0=0\)

\(B=\cot56^0-\tan34^0=\tan34^0-\tan34^0=0\)

\(A=2cos120^0.cos10^0+sin\left(90^0+10^0\right)\)

\(=2.\left(-\dfrac{1}{2}\right).cos10^0+cos10^0\)

\(=-cos10^0+cos10^0=0\)

\(B=\dfrac{sin9}{cos9}+\dfrac{sin81}{cos81}-\dfrac{sin27}{cos27}-\dfrac{sin63}{cos63}\)

\(=\dfrac{sin9.cos81+sin81.cos9}{cos9.cos81}-\dfrac{sin27.cos63+sin63.cos27}{cos27.cos63}\)

\(=\dfrac{sin90}{\dfrac{1}{2}cos90+\dfrac{1}{2}cos72}-\dfrac{sin90}{\dfrac{1}{2}cos90+\dfrac{1}{2}cos36}\)

\(=\dfrac{2}{cos72}-\dfrac{2}{cos36}=\dfrac{2\left(cos36-cos72\right)}{cos36.cos72}=\dfrac{4sin54.sin18}{cos36.cos72}\)

\(=\dfrac{4sin\left(90-36\right).sin\left(90-72\right)}{cos36.cos72}=\dfrac{4cos36.cos72}{cos36.cos72}=4\)

b: \(\cos30^0=\dfrac{\sqrt{3}}{2}\)

a) =-36

b)=-107

c)=-70

d) =-90

a) =-36

b)=-107

c)=-70

d) =-90

a, -36
b, -107
c,  -70
d, -90

a,

Đổi `tan 12^o = cot 78^o ; tan 28^o = cot 62^o ; tan 58^o = cot 32^o`

Vì `32^o<61^o<62^o<78^o<79^15'`

`->cot 32^o>cot 61^o>cot 62^o > cot 78^o > cot 79^o15'`

`->tan 58^o>cot 61^o > tan 28^o > tan 12^o > cot 79^o15'`

b,

Đổi `sin 56^o = cos 34^o ; sin 74^o=cos 16^o`

Vì `16^o<24^o<63^o41'<67^o<85 ^o`

`->cos 16^o>cos 34^o>cos 63^o41'>cos 67^o>cos 85 ^o`

`->sin 74^o>sin 56^o>cos 63^o41'>cos 67^o>cos 85 ^o`

a) Vì \(\frac{\pi }{2} < a < \pi \) nên \(\cos a < 0\). Do đó \(\cos a = \sqrt {1 - {{\sin }^2}a}  = \sqrt {1 - \frac{1}{3}}  =  - \frac{{\sqrt 6 }}{3}\)

Ta có: \(\cos \left( {a + \frac{\pi }{6}} \right) = \cos a\cos \frac{\pi }{6} - \sin a\sin \frac{\pi }{6} =  - \frac{{\sqrt 6 }}{3}.\frac{{\sqrt 3 }}{2} - \frac{1}{{\sqrt 3 }}.\frac{1}{2} =  - \frac{{\sqrt 3  + 3\sqrt 2 }}{6}\)

b) Vì \(\pi  < a < \frac{{3\pi }}{2}\) nên \(\sin a < 0\). Do đó \(\sin a = \sqrt {1 - {{\cos }^2}a}  = \sqrt {1 - \frac{1}{9}}  =  - \frac{{2\sqrt 2 }}{3}\)

Suy ra \(\tan a\; = \frac{{\sin a}}{{\cos a}} = \frac{{ - \frac{{2\sqrt 2 }}{3}}}{{ - \frac{1}{3}}} = 2\sqrt 2 \)

Ta có: \(\tan \left( {a - \frac{\pi }{4}} \right) = \frac{{\tan a - \tan \frac{\pi }{4}}}{{1 + \tan a\tan \frac{\pi }{4}}} = \frac{{\frac{{\sin a}}{{\cos a}} - 1}}{{1 + \frac{{\sin a}}{{\cos a}}}} = \frac{{2\sqrt 2  - 1}}{{1 + 2\sqrt 2 }} = \frac{{9 - 4\sqrt 2 }}{7}\)

Sin 15 độ , Sin 30 độ , Cos 58 độ , Sin 48 độ , Cos 36 độ , Có 29 độ

Có phải giải thích ko bạn .

@Nguyễn Thành Trương @Nguyễn Ngọc Lộc giúp cj ý giải bài này với ạ. ( Cj ý nhờ e tag tên giùm; xl vì đã tag tên tự tiện ạ)