\(sin15^o+sin75^o-cos15^o-cos75^o+sin30^o\)

\(=\left(sin15+sin75^o\right)-\left(cos15^o+cos75^o\right)+sin30^o\)

\(=\dfrac{\sqrt{6}}{2}-\dfrac{\sqrt{6}}{2}+\dfrac{1}{2}\)

\(=0+\dfrac{1}{2}\)

\(=\dfrac{1}{2}\)

\(sin15^o+sin75^o-cos15^0-cos75^o+sin30^o\)

\(=cos75^o+cos15^0-cos15^0-cos75^o+sin30^o\)

\(=sin30^o=\dfrac{1}{2}\)

Chọn A.

Ta có:

a, \(\cos^215+\cos^225+\cos^235+\cos^245+\sin^235+\sin^225+\sin^215\)

=\(\left(\cos^215+\sin^215\right)+\left(\cos^225+\sin^225\right)+\left(\cos^235+\sin^235\right)+\cos^245\)

=\(1+1+1+\frac{1}{2}=\frac{7}{2}\)

b.\(\sin^210-\sin^220-\sin^230-\sin^240-\cos^240-\cos^220+\cos^210\)

=\(\left(\sin^210+\cos^210\right)-\left(\sin^220+\cos^220\right)-\left(\sin^240+\cos^240\right)-\sin^230\)

=\(1-1-1-\frac{1}{4}=-\frac{5}{4}\)

c,\(\sin15+\sin75-\sin75-\cos15+\sin30=\sin30=\frac{1}{2}\)

a)
\(2sin30+3sin45^o-sin60^o=2.\dfrac{1}{2}+3.\dfrac{\sqrt{2}}{2}-\dfrac{\sqrt{3}}{2}\)\(=\dfrac{2+3\sqrt{2}-\sqrt{3}}{2}\).
b)\(2cos30^o+3sin45^o-cos60^o=2.\dfrac{\sqrt{3}}{2}+3.\dfrac{\sqrt{2}}{2}-\dfrac{1}{2}\)\(=\dfrac{2\sqrt{3}+3\sqrt{2}-1}{2}\).

Hai góc 30⁰ và 150⁰ bù nhau nên  sin 30 ° = sin 150 °

Hai góc 15⁰ và 165⁰ bù nhau nên cos 15 ° = − cos 165 ° .

Do đó P = sin 30 ° cos 15 ° + sin 150 ° cos 165 ° = sin 30 ° . c os15 0 + sin 30 ° . ( − cos 15 ° ) = 0 .

Chọn B.

câu a "cot" chuyển thành "cos" giùm mjk nha

https://hoc24.vn/hoi-dap/question/647714.html

a) ta có : 

Ta có : \(cos30^0=sin60^0\)

\(cos15^0=sin75^0\)

Sắp xếp : \(sin30^0,sin40^0,sin60^0,sin75^0,sin89^0.\)

Ta có: \(\cos30^o=\sin60^0\), \(\cos15^0=\sin75^0\)

mà \(\sin30^0< \sin40^0< \sin60^0< \sin75^0< \sin89^0\)

\(\Leftrightarrow\sin30^0< \sin40^0< \cos60^0< \cos75^0< \sin89^0\)