\(=\left(\sin100^0+\sin80^0\right)+\left(\cos16^0+\cos164^0\right)=1\)

\(B=\sin^230^0+\sin^240^0+\sin^250^0+\sin^260^0\)

\(B=\sin^230^0+\sin^240^0+\cos^2\left(90^0-50^0\right)+\cos^2\left(90^0-60^0\right)\)

\(B=\sin^230^0+\sin^240^0+\cos^240^0+\cos^230^0\)

\(B=\left(\sin^230^0+\cos^230^0\right)\left(\sin^240^0+\cos^240^0\right)\)

\(B=1+1\)

\(B=2\)

Chúc bạn hok tốt!!! vvvvvvvv

Ta có :\(\sin\left(60\right)=\cos\left(30\right)\)(phụ nhau)

\(\Leftrightarrow sin^2\left(60\right)=\cos^2\left(30\right)\)

và :\(sin^2\left(50\right)=\cos^2\left(40\right)\)(tương tự như trên nha bạn)

Thay vào biểu thức B ta có :

\(B=\sin^2\left(30\right)+sin^2\left(40\right)+\cos^2\left(30\right)+\cos^2\left(40\right)\)

\(B=1+1\)

\(B=2\)

chúc bạn học tốt :)

\(A=sin\alpha-sin\alpha\cdot cos^2\alpha\)

\(A=sin\alpha\left(1-cos^2\alpha\right)\)

\(A=sin\alpha\cdot sin^2\alpha\)

\(A=sin^3\alpha\)

https://hoc24.vn/hoi-dap/question/829326.html

=>CÂU NÀY CÓ BN VỪA HỎI XONG,LẦN SAU XEM KĨ R MS HỎI NHA.Mai Linh

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:  \(\left\{ \begin{array}{l}\sin {100^o} = \sin \left( {{{180}^o} - {{80}^o}} \right) = \sin {80^o}\\\cos {164^o} = \cos \left( {{{180}^o} - {{16}^o}} \right) =  - \cos {16^o}\end{array} \right.\)

\( \Rightarrow \sin {100^o} + \sin {80^o} + \cos {16^o} + \cos {164^o}\)\( = \sin {80^o} + \sin {80^o} + \cos {16^o}-\cos {16^o}\)\( = 2\sin {80^o}.\)

b) 

Ta có:

\(\left\{ \begin{array}{l}\sin \left( {{{180}^o} - \alpha } \right) = \sin \alpha \\\cos \left( {{{180}^o} - \alpha } \right) =  - \cos \alpha \\\tan \left( {{{180}^o} - \alpha } \right) =  - \tan \alpha \\\cot \left( {{{180}^o} - \alpha } \right) =  - \cot \alpha \end{array} \right.\quad ({0^o} < \alpha  < {90^o})\)\( \Rightarrow 2\sin \left( {{{180}^o} - \alpha } \right).\cot \alpha  - \cos \left( {{{180}^o} - \alpha } \right).\tan \alpha .\cot \left( {{{180}^o} - \alpha } \right)\) \( = 2\sin \alpha .\cot \alpha  - \left( { - \cos \alpha } \right).\tan \alpha .\left( { - \cot \alpha } \right)\)\( = 2\sin \alpha .\cot \alpha  - \cos \alpha .\tan \alpha .\cot \alpha \)

\( = 2\sin \alpha .\frac{{\cos \alpha }}{{\sin \alpha }} - \cos \alpha .\left( {\tan \alpha .\cot \alpha } \right)\)\( = 2\cos \alpha  - \cos \alpha .1 = \cos \alpha .\)

\(\sin^4\alpha+\cos^4\alpha+2\sin^2\alpha.\cos^2\alpha=\left(\sin^2\alpha+\cos^2\alpha\right)^2=1\)

\(\tan^2\alpha\left(2.\cos^2\alpha+\sin^2\alpha-1\right)=\tan^2\alpha\left(\cos^2\alpha+\left(\sin^2\alpha+\cos^2\alpha\right)-1\right)\)\(=\tan^2\alpha.\cos^2\alpha=\left(\frac{1}{\cos^2\alpha}-1\right)\cos^2\alpha=1-\cos^2\alpha=\sin^2\alpha\)

a, = \(\sin^2\alpha+2\sin\alpha.\cos\alpha+\cos^2\alpha\)+ \(\sin^2\alpha-2\sin\alpha\cos\alpha+\cos^2\alpha\)

= \(2\sin^2\alpha+2\cos^2\alpha\)= 4

b,=\(\sin\alpha\cos\alpha\)(\(\frac{\sin\alpha}{\cos\alpha}+\frac{\cos\alpha}{\sin\alpha}\))

= \(\sin\alpha\cos\alpha.\frac{\sin^2\alpha+\cos^2\alpha}{\sin\alpha\cos\alpha}\)

=1

#mã mã#

b, =1 

mk viết thiếu

#mã mã#

Mình chỉ biết \(\text{ rút gọn biểu thức lượng giác}\) thôi :

\(\left(sina+cosa\right)^2+\left(sina-cosa\right)^2\)

\(\text{ rút gọn biểu thức lượng giác}\)

\(2\)