Question

Bài 3: Tính:

a. 2sin30^o  - 2cos60^o + tan45^o

b. cos² 20^o  + cos² 40^o + cos² 50^o  + cos² 70^o 

Answer 1

Answer 2

a, \(2sin30^o-2cos60^o+tan45^o\)

\(=2cos60^o-2cos60^o+tan45^o\)

\(=tan45^o\)

\(=1\)

Answer 3

b, \(cos^220^o+cos^240^o+cos^250^o+cos^270^o\)

\(=cos^220^o+cos^240^o+sin^240^o+sin^220^o\)

\(=cos^220^o+sin^220^o+cos^240^o+sin^240^o\)

\(=1+1=2\)

Answer 4

a: Ta có: \(2\cdot\sin30^0-2\cdot\cos60^0+\tan45^0\)

\(=2\cdot\dfrac{1}{2}-2\cdot\dfrac{1}{2}+1\)

=1