Chọn C.

Ta có: 

A = sin( a-16⁰) .cos( a + 14⁰) – sin( a + 14⁰) .cos(a - 16⁰) = sin[ ( a - 17⁰) – (a + 13⁰) ] = sin( -30⁰) = -0,5.

=cos0+cos180+cos20+cos160+cos40+cos140+cos60+cos120+cos80+cos100

=0+0+...+0

=0

`cos 0^o +cos 20^o +cos 40^o +...+cos 160^o +cos 180^o`

`=(cos 0^o +cos 180^o)+(cos 20^o +cos 160^o)+....+(cos 80^o +cos 100^o)`

`=(cos 0^o -cos 0^o)+(cos 20^o -cos 20^o)+....+(cos 80^o -cos 80^o)`

`=0`

Áp dụng: `cos \alpha = -cos(180^o -\alpha)=-cos(\pi - \alpha)`.

a) \(A = \cos {0^o} + \cos {40^o} + \cos {120^o} + \cos {140^o}\)

Tra bảng giá trị lượng giác của một số góc đặc biệt, ta có:

 \(\cos {0^o} = 1;\;\cos {120^o} =  - \frac{1}{2}\)

Lại có: \(\cos {140^o} =  - \cos \left( {{{180}^o} - {{40}^o}} \right) =  - \cos {40^o}\)  

\(\begin{array}{l} \Rightarrow A = 1 + \cos {40^o} + \left( { - \frac{1}{2}} \right) - \cos {40^o}\\ \Leftrightarrow A = \frac{1}{2}.\end{array}\)

b) \(B = \sin {5^o} + \sin {150^o} - \sin {175^o} + \sin {180^o}\)

Tra bảng giá trị lượng giác của một số góc đặc biệt, ta có:

 \(\sin {150^o} = \frac{1}{2};\;\sin {180^o} = 0\)

Lại có: \(\sin {175^o} = \sin \left( {{{180}^o} - {{175}^o}} \right) = \sin {5^o}\)  

\(\begin{array}{l} \Rightarrow B = \sin {5^o} + \frac{1}{2} - \sin {5^o} + 0\\ \Leftrightarrow B = \frac{1}{2}.\end{array}\)

c) \(C = \cos {15^o} + \cos {35^o} - \sin {75^o} - \sin {55^o}\)

Ta có: \(\sin {75^o} = \cos\left( {{{90}^o} - {{75}^o}} \right) = \cos {15^o}\); \(\sin {55^o} = \cos\left( {{{90}^o} - {{55}^o}} \right) = \cos {35^o}\)

\(\begin{array}{l} \Rightarrow C = \cos {15^o} + \cos {35^o} - \cos {15^o} - \cos {35^o}\\ \Leftrightarrow C = 0.\end{array}\)

d) \(D = \tan {25^o}.\tan {45^o}.\tan {115^o}\)

Ta có: \(\tan {115^o} =  - \tan \left( {{{180}^o} - {{115}^o}} \right) =  - \tan {65^o}\)

Mà: \(\tan {65^o} = \cot \left( {{{90}^o} - {{65}^o}} \right) = \cot {25^o}\)

\(\begin{array}{l} \Rightarrow D = \tan {25^o}.\tan {45^o}.(-\cot {25^o})\\ \Leftrightarrow D =- \tan {45^o} = -1\end{array}\)

e) \(E = \cot {10^o}.\cot {30^o}.\cot {100^o}\)

Ta có: \(\cot {100^o} =  - \cot \left( {{{180}^o} - {{100}^o}} \right) =  - \cot {80^o}\)

Mà: \(\cot {80^o} = \tan \left( {{{90}^o} - {{80}^o}} \right) = \tan {10^o}\Rightarrow \cot {100^o}  =- \tan {10^o}\)

\(\begin{array}{l} \Rightarrow E = \cot {10^o}.\cot {30^o}.(-\tan {10^o})\\ \Leftrightarrow E = -\cot {30^o} =- \sqrt 3 .\end{array}\)

a) sin 40 - cos 50 =0

b) sin²30 + sin²40 + sin²50 + sin²60 = 2

c) cos²10 - cos²20 + cos²30 - cos²40 - cos²50 - cos²70 + cos²80 = - sin²30

\(a.sin40^o-cos50^o=sin40^o-sin40^o=0\)
\(b.sin^230^o+sin^240^o+sin^250^o+sin^260^o=\left(sin^230^0+sin^260^o\right)+\left(sin^240^0+sin^250^o\right)=\left(sin^230^0+cos^230^o\right)+\left(sin^240+cos^240^o\right)=1+1=2\)
\(c.\left(cos^210^o+cos^280^o\right)-\left(cos^220^o+cos^270^0\right)-\left(cos^240^o-cos^250^o\right)+cos^230^o=\left(cos^210^o+sin^210^o\right)-\left(cos^220^o+sin^220^o\right)-\left(cos^240^o+sin^240^0\right)+cos^230^0=1-1-1+\dfrac{3}{4}=-\dfrac{1}{4}\)

a) 0

b)2

c)3/4-1

a) Ta có: \(sin^2x+sin^2\left(90-x\right)=sin^2x+cos^2x=1.\)

áp dụng: A = 2

b)Ta có: \(cos\left(x\right)=-cos\left(180-x\right)\)

áp dụng: B = 0

c) Ta có: \(tan\left(x\right)\cdot tan\left(90-x\right)=\frac{sinx}{cosx}\cdot\frac{sin\left(90-x\right)}{cos\left(90-x\right)}=\frac{sinx}{cosx}\cdot\frac{cosx}{sinx}=1\)

áp dụng: C = 1

quá sai

trả dép em về

Lời giải:
$D=\frac{1+\cos a+2\cos ^2a-1+4\cos ^3a-3\cos a}{\cos a+2\cos ^2a-1}$

$=\frac{4\cos ^3a+2\cos ^2a-2\cos a}{\cos a+2\cos ^2a-1}$

$=\frac{2\cos a(\cos a+2\cos ^2a-1)}{\cos a+2\cos ^2a-1}$

$=2\cos a$

ta có :

\(\left(1-cosa\right)\left(1+cosa\right)=1-cos^2a=sin^2a\)

\(P=cos20+cos160+cos40+cos140+...+cos80+cos100+cos180\)

\(=2cos90.cos70+2cos90.cos50+...+2cos90.cos10+cos180\)

\(=cos90\left(2cos70+2cos50+...+2cos10\right)+cos180\)

\(=cos180=-1\) (do \(cos90=0\))

\(\begin{array}{l}\cos \left( {a + b} \right) + \cos \left( {a - b} \right) = \cos a.\cos b - \sin a.\sin b + \sin a.\sin b + \cos a.\cos b = 2\cos a.\cos b\\\cos \left( {a + b} \right) - \cos \left( {a - b} \right) = \cos a.\cos b - \sin a.\sin b - \sin a.\sin b - \cos a.\cos b =  - 2\sin a.\sin b\\\sin \left( {a + b} \right) + \sin \left( {a - b} \right) = \sin a.\cos b + \cos a.\sin b + \sin a.\cos b - \cos a.\sin b = 2\sin a.\cos b\end{array}\)