Tính giá trị của biểu thức sau :
sin^2 10+sin^2 20+sin^2 45+sin^2 70+sin^2 80=?
SOS cíu mik vs
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A=sin^2 70°+sin^2 80°+sin^2 10°+sin^2 20°
\(=\sin^270^o+sin^280^o+sin^210^o+sin^220^o.\)
Nhập zô máy tính như sau:
\(=Sin\left(70\right)^2+Sin\left(80\right)^2+Sin\left(10\right)^2+Sin\left(20\right)^2\)
\(=2\)
Nếu bn ko đc dùng máy tính thì dùng bảng cx đc nha
\(M=\left(\sin^210^0+\sin^280^0\right)+\left(\sin^220^0+\sin^270^0\right)-3\tan39^0\cdot\cot39^0\\ M=\left(\sin^210^0+\cos^210^0\right)+\left(\sin^220^0+\cos^220^0\right)-3\cdot1=1+1-3=-1\)
A=(sin220°+sin270°)+(sin230°+sin260°)
+(sin240°+sin250°)-tan245°
=(sin220°+cos220°)+(sin230°+cos230°)+(sin240°+cos240°)-1
=1+1+1-1=2
\(A=sin^210^o+sin^220^o+sin^230^o+sin^240^o+sin^250^o+sin^260^o+sin^270^o+sin^280^o\)
\(A=\left(sin^210^o+sin^280^o\right)+\left(sin^220^o+sin^270^o\right)+\left(sin^230^o+sin^260^o\right)+\left(sin^240^o+sin^250^o\right)\)
\(A=\left(sin^210^o+cos^210^o\right)+\left(sin^220^o+cos^220^o\right)+\left(sin^230^o+cos^230^o\right)+\left(sin^240^o+cos^240^o\right)\)
\(A=1+1+1+1\)
\(A=4\)
a: \(A=sin^210^0+sin^280^0+cos^220^0+sin^270^0\)
\(=sin^210^0+cos^210^0+sin^270^0+sin^270^0\)
\(=2\cdot sin^270^0+1\)
b: \(=sin^215^0+sin^275^0+sin^235^0+sin^255^0\)
\(=sin^215^0+cos^215^0+sin^235^0+cos^235^0\)
=1+1
=2
\(A=sin^210^0+sin^280^0+cos^220^0+sin^270^0\)
\(=sin^210^0+cos^210^0+sin^270^0+sin^270^0\)
\(=2sin^270^0+1\)
\(B=sin^215^0+sin^275^0+sin^235^0+sin^255^0\)
\(=sin^215^0+cos^215^0+sin^235^0+cos^235^0\)
=1+1
=2
P=sin2200+sin2400+sin2450+sin2500+sin2700
đổi sin2500 thành cos2400,sin2700 thành cos2200 rồi thay vào ta được:
sin2200+cos2200+sin2400+cos2400+\(\left(\dfrac{\sqrt{2}}{2}\right)^2\)
=\(2+\dfrac{1}{2}=\dfrac{5}{2}=2,5\)
Ta có: \(\sin {70^o} = \cos {20^o};\;\cos {110^o} = - \cos {70^o} = - \sin {20^o}\)
\(\begin{array}{l} \Rightarrow A = {(\sin {20^o} + \cos {20^o})^2} + {(\cos {20^o} - \sin {20^o})^2}\\ = ({\sin ^2}{20^o} + {\cos ^2}{20^o} + 2\sin {20^o}\cos {20^o}) + ({\cos ^2}{20^o} + {\sin ^2}{20^o} - 2\sin {20^o}\cos {20^o})\\ = 2({\sin ^2}{20^o} + {\cos ^2}{20^o})\\ = 2\end{array}\)
Ta có: \(\tan {110^o} = - \tan {70^o} = - \cot {20^o};\;\cot {110^o} = - \cot {70^o} = - \tan {20^o}.\)
\( \Rightarrow B = \tan {20^o} + \cot {20^o} + ( - \cot {20^o}) + ( - \tan {20^o}) = 0\)
a) \(M = \sin {45^o}.\cos {45^o} + \sin {30^o}\)
Ta có: \(\left\{ \begin{array}{l}\sin {45^o} = \cos {45^o} = \frac{{\sqrt 2 }}{2};\;\\\sin {30^o} = \frac{1}{2}\end{array} \right.\)
Thay vào M, ta được: \(M = \frac{{\sqrt 2 }}{2}.\frac{{\sqrt 2 }}{2} + \frac{1}{2} = \frac{2}{4} + \frac{1}{2} = 1\)
b) \(N = \sin {60^o}.\cos {30^o} + \frac{1}{2}.\sin {45^o}.\cos {45^o}\)
Ta có: \(\sin {60^o} = \frac{{\sqrt 3 }}{2};\;\;\cos {30^o} = \frac{{\sqrt 3 }}{2};\;\sin {45^o} = \frac{{\sqrt 2 }}{2};\, \cos {45^o}= \frac{{\sqrt 2 }}{2}\)
Thay vào N, ta được: \(N = \frac{{\sqrt 3 }}{2}.\frac{{\sqrt 3 }}{2} + \frac{1}{2}.\frac{{\sqrt 2 }}{2}.\frac{{\sqrt 2 }}{2} = \frac{3}{4} + \frac{1}{4} = 1\)
c) \(P = 1 + {\tan ^2}{60^o}\)
Ta có: \(\tan {60^o} = \sqrt 3 \)
Thay vào P, ta được: \(Q = 1 + {\left( {\sqrt 3 } \right)^2} = 4.\)
d) \(Q = \frac{1}{{{{\sin }^2}{{120}^o}}} - {\cot ^2}{120^o}.\)
Ta có: \(\sin {120^o} = \frac{{\sqrt 3 }}{2};\;\;\cot {120^o} = \frac{{ - 1}}{{\sqrt 3 }}\)
Thay vào P, ta được: \(Q = \frac{1}{{{{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}}} - \;{\left( {\frac{{ - 1}}{{\sqrt 3 }}} \right)^2} = \frac{1}{{\frac{3}{4}}} - \;\frac{1}{3} = \;\frac{4}{3} - \;\frac{1}{3} = 1.\)
\(A=sin^210^o+cos^220^o+sin^280^o+cos^270^o\)
\(A=\left(sin^210^o+sin^280^o\right)+\left(cos^220^o+cos^270^o\right)\)
\(A=0+0\)
\(A=0\)
\(sin^210+sin^220+sin^245+sin^270+sin^280\)
\(=sin^210+sin^220+sin^245+cos^220+cos^210=1+1+sin^245=2+\dfrac{1}{2}=\dfrac{5}{2}\)