Tính P = \(\cos^20^0+\cos^21^0+\cos^22^0+...+\cos^2180^0\)
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NH
1
TG
14 tháng 8 2019
Ta có \(\cos1^o=\sin89^o\)
\(\cos2^o=sin88^o\)
................
\(\cos44^o=\sin46^o\)
\(\cos45^o=\frac{\sqrt{2}}{2}\)
\(\Rightarrow\cos^21^o=\sin^289^o\)
\(\cos^22^o=\sin^288^o\)
....................................
\(\cos^244^o=\sin^246^o\)
\(\cos^245^o=\frac{2}{4}=\frac{1}{2}\)
Khi đó \(B=\sin^289^o+\sin^288^o+...+\sin^246^o+\cos^245^o+\cos^246^o+...+\cos^289^o\)
\(=\left(\sin^289^o+\cos^289^o\right)+\left(\sin^288^o+\cos^288^o\right)+...+\left(\sin^246^o+\cos^246^o\right)+\cos^245^o\)
\(=1+1+...+1+\frac{1}{2}\)(44 số 1)
\(=44+\frac{1}{2}=\frac{89}{2}=44,5\)
5 tháng 6 2019
\(\cos^21^o+\cos^289^o=\cos^21^o+\cos^2\left(90^o-1^o\right)=\cos^21^o+\sin^21^o=1\)
\(\cos^22^o+\cos^288^o=\cos^22^o+\cos^2\left(90^o-2^o\right)=\cos^22^o+\sin^22^o=1\)
.......
\(\cos^244^o+\cos^246^o=\cos^244^o+\cos^2\left(90^o-44^o\right)=\cos^244^o+\sin^244^o=1\)
\(\cos^245^o=\left(\frac{\sqrt{2}}{2}\right)^2=\frac{1}{2}\)
=> \(A=1.44+\frac{1}{2}-\frac{1}{2}=44\)
NH
0
BT
18 tháng 5 2017
a)
\(4a^2cos^260^o+2ab.cos^2180^o+\dfrac{4}{3}cos^230^o\)
\(=4a^2.\left(\dfrac{1}{2}\right)^2+2ab.\left(-1\right)^2+\dfrac{4}{3}.\left(\dfrac{\sqrt{3}}{2}\right)^2\)
\(=4a^2.\dfrac{1}{4}+2ab+\dfrac{4}{3}.\dfrac{3}{4}\)
\(=a^2+2ab+1\).
b)
\(\left(asin90^o+btan45^o\right)\left(acos0^o+bcos180^o\right)\)
\(=\left(a+b\right)\left(a-b\right)=a^2-b^2\).
NT
1
26 tháng 2 2020
\(A=sin^21^o+c\text{os}^22^o+sin^23^o+c\text{os}^24^o+...+sin^2179^o+c\text{os}^2180^o\)
\(=sin^21^o+c\text{os}^22^o+sin^23^o+c\text{os}^24^o+...+c\text{os}^290^o-sin^289^o-c\text{os}^288^o-...-sin^21^o-c\text{os}^20^o\)
\(=c\text{os}^290^o-c\text{os}^20^o\)
\(=-1\)
Chúc bn học tốt
NT
1
NV
Nguyễn Việt Lâm
Giáo viên
10 tháng 6 2020
\(P=cos20+cos160+cos40+cos140+...+cos80+cos100+cos180\)
\(=2cos90.cos70+2cos90.cos50+...+2cos90.cos10+cos180\)
\(=cos90\left(2cos70+2cos50+...+2cos10\right)+cos180\)
\(=cos180=-1\) (do \(cos90=0\))
13 tháng 9 2023
=cos0+cos180+cos20+cos160+cos40+cos140+cos60+cos120+cos80+cos100
=0+0+...+0
=0
31 tháng 10 2021
\(P=4\left[\left(cos^21^0+cos^289^0\right)+\left(cos^22^0+cos^288^0\right)+...+\left(cos^244^0+cos^246^0\right)+cos^245^0\right]\)
\(=4\left[\left(cos^21^0+sin^21^0\right)+\left(cos^22^0+sin^22^0\right)+...+\left(cos^244^0+sin^244^0\right)+cos^245^0\right]\)
\(=4\left(1+1+...+1+\frac{\sqrt{2}}{2}\right)\)