=1+1-1=1

thanks ạ

a) sin²30 độ - sin²40 độ - sin²50 độ + sin² 60 độ

= cos²60^o - cos²50^o - sin²50^o + sin²60^o

= (cos²60^o + sin²60^o) - (cos²50^o + sin²50^o)

= 1 - 1 = 0

b) cos²25 độ - cos²35độ + cos²45 độ -cos² 55 độ + cos² 65 độ

= sin²65^o - sin²55^o + cos²45^o - cos²55^o + cos²65^o

= (sin²65^o + cos²65^o) - (sin²55^o + cos²55^o) + cos²45^o

=  1 - 1 +1/2

= 1/2

Chọn B.

Ta có: A = sin²2⁰ + sin²4⁰ + sin² 6⁰ +…+ sin²84⁰ + sin²86⁰ + sin²88⁰

A = ( sin²2⁰ + sin² 88⁰) + ( sin² 4⁰ + sin²86⁰) +...+ (sin²44⁰ + sin²46⁰)

A = ( sin²2⁰ + cos²2⁰) + ( sin² 4⁰ +…+ cos²4⁰) +...+ ( sin²44⁰ + cos²44⁰) ( do 2 góc phụ nhau sin góc này bằng cos góc kia).

A = 1 + 1 + 1 + ... + 1 = 22

Ta có:

Nên A = ( sin 2 1 0 + sin 2 89 0 ) + ( sin 2 2 0  +  sin 2 88 0 ) +… + ( sin 2 44 0  +  sin 2 46 0 ) +  sin 2 45 0  +  sin 2 90 0

= ( sin 2 1 0  +  c o s 2 1 0 ) + ( sin 2 2 0  +  c o s 2 2 0 ) + … + ( sin 2 44 0  +  c o s 2 44 0 ) +  sin 2 45 0  +  sin 2 45 0

Vậy A =  91 2

Đáp án cần chọn là: C

Chọn B.

Ta có: A= ( sin²3⁰ + sin²87⁰) + ( sin²75⁰ + sin²15⁰)

A= (sin²3⁰ + cos²3⁰)  + ( sin²15⁰ + cos²15⁰)

= 1 + 1 = 2

\(=\dfrac{1}{cos\left(180+110\right)}+\dfrac{1}{\sqrt{3}sin\left(360-110\right)}\)

\(=-\dfrac{1}{cos110}-\dfrac{1}{\sqrt{3}sin110}\)

\(=\dfrac{-\sqrt{3}sin110-cos110}{\sqrt{3}\cdot sin110\cdot cos110}\)

\(=\dfrac{-2\left(\dfrac{\sqrt{3}}{2}\cdot sin110+\dfrac{1}{2}\cdot cos110\right)}{\dfrac{\sqrt{3}}{2}\cdot sin220}\)

\(=\dfrac{-2\cdot sin\left(110+30\right)}{\dfrac{\sqrt{3}}{2}\cdot sin220}=\dfrac{-2\cdot sin140}{\dfrac{\sqrt{3}}{2}\cdot sin220}\)

HS tự làm

a) \(\sin220^0< \sin10^0< \sin40^0< \sin90^0\)

b) \(\cos138^0< \cos90^0< \cos15^0< \cos0^0\)

a)

\(\cos225^0=\cos\left(180^0+45^0\right)=-\cos45^0=-\dfrac{\sqrt{2}}{2}\)

\(\sin240^0=\sin\left(180^0+60^0\right)=-\sin60^0=-\dfrac{\sqrt{3}}{2}\)

\(\cos\left(-15^0\right)=-\cot15^0=-\tan75^0=-\tan\left(30^0+45^0\right)\)

\(=\dfrac{-\tan30^0-\tan45^0}{1-\tan30^0\tan45^0}=\dfrac{-\dfrac{1}{\sqrt{3}}-1}{1-\dfrac{1}{\sqrt{3}}}=-\dfrac{\sqrt{3}+1}{\sqrt{3}-1}\)

\(=-\dfrac{\left(\sqrt{3}+1\right)^2}{2}=-2-\sqrt{3}\)

\(\tan75^0=\cot15^0=2+\sqrt{3}\)

b)

\(\sin\dfrac{7\pi}{12}=\sin\left(\dfrac{\pi}{3}+\dfrac{\pi}{4}\right)=\sin\dfrac{\pi}{3}\cos\dfrac{\pi}{4}+\cos\dfrac{\pi}{3}\sin\dfrac{\pi}{4}\)

\(=\dfrac{\sqrt{2}}{2}\left(\dfrac{\sqrt{3}}{2}+\dfrac{1}{2}\right)=\dfrac{\sqrt{6}+\sqrt{2}}{4}\)

\(\cos\left(-\dfrac{\pi}{12}\right)=\cos\left(\dfrac{\pi}{4}-\dfrac{\pi}{3}\right)=\cos\dfrac{\pi}{4}\cos\dfrac{\pi}{3}+\sin\dfrac{\pi}{3}\sin\dfrac{\pi}{4}\)

\(=\dfrac{\sqrt{2}}{2}\left(\dfrac{\sqrt{3}}{2}+\dfrac{1}{2}\right)=0,9659\dfrac{\sqrt{2}}{2}\left(\dfrac{\sqrt{3}}{2}+\dfrac{1}{2}\right)=0,9659\)

\(\tan\dfrac{13\pi}{12}=\tan\left(\pi+\dfrac{\pi}{12}\right)=\tan\dfrac{\pi}{12}=\tan\left(\dfrac{\pi}{3}-\dfrac{\pi}{4}\right)\)

\(=\dfrac{\tan\dfrac{\pi}{3}-\tan\dfrac{\pi}{4}}{1+\tan\dfrac{\pi}{3}\tan\dfrac{\pi}{4}}=\dfrac{\sqrt{3}-1}{1+\sqrt{3}}=2-\sqrt{3}\)