\(\sin^275^o+\sin^215^o-\cos^250^o-\cos^240^o+\cot45^o.\cot45^o\)
a,\(^{ }\cos^215^o+^{ }\cos^225^o+cos^235^o+cos^245^o+cos^255^o+cos^265^o+cos^275^o\)
b,\(\sin^210^o-sin^220^o+sin^230^o-sin^240^o-sin^250^o-sin^270^o+sin^280^o\)
c,\(sin15^o+sin75^o-cos15^o-cos75^o+sin30^o\)
d,\(sin35^o+sin67^o-cos23^o-cos55^o\)
e,\(cos^220^o+cos^240^o+cos^250^o+cos^270^o\)
f,\(sin^220^o-tan40^o+cot50^o-cos70^o\)
GIẢI GIÚP MIK VS MỌI NGƯỜI!!!!!!! MIK ĐANG CẦN GẤP LẮM
CÁC BN CHỈ CẦN LÀM CHO MIK CÂU D,E,F LÀ ĐC RỒI
d/ \(sin35+sin67-cos23-cos55\)
\(=sin35+sin67-sin67-sin35=0\)
e/ \(cos^220+cos^240+cos^250+cos^270\)
\(=cos^220+cos^240+sin^220+sin^240=1+1=2\)
f/ Đề sai.
bài 1: Tính giá trị của các biểu thức sau:
a, \(\cos^215^o+\cos^225^o+\cos^235^o+\cos^245^o+cos^255^o+cos^265^o+cos^275^o\)
b,\(\sin^210^o-sin^220^o-sin^230^o-sin^240^o-sin^250^o-sin^270^o+sin^280^o\)
c,\(\sin15^o+\sin75^o-cos15^o-cos75^o+\sin30^o\)
Giải giúp e vs m.n
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,\(\sin15+\sin75-\sin75-\cos15+\sin30=\sin30=\frac{1}{2}\)
Thực hiện phép tính
a) \(\tan40^o.\cot40^o+\frac{\sin50^o}{\cos40^o}\)
b) \(\cot44^o.\cot45^o.\cot46^o\)
c)\(\left(1+\tan^225^o\right).\sin^265^o\)
d) \(\tan35^o.\tan40^o.\tan45^o.\tan50^o.\tan55^o\)
e) \(\cos^220^o+\cos^240^o+\cos^250^o+\cos^270^o\)
f) \(\sin^227^o+\cos^227^o+\tan27^o-\cot73^o\)
a/ \(\tan40.\cot40+\frac{\sin50}{\cos40}\)
\(=1+\frac{\cos40}{\cos40}=1+1=2\)
bài 1: tính giá trị của các biểu thức sau
a) \(\cot^215^o+\cos^225^o+\cos^235^o+\cos^245^o+\cos^255^o+\cos^265^o+\cos^275^o\)
b) \(\sin^210^o-\sin^220^o-\sin^230^o-sin^240^o-\sin^250^o-\sin^270^o+\sin^280^o\)
c) \(\sin15^o+\sin75^o-\cos15^o-\cos75^o+\sin30^o\)
giải giúp mik vs mấy bạn~ mjk cần gấp lắm
câu a "cot" chuyển thành "cos" giùm mjk nha
https://hoc24.vn/hoi-dap/question/647714.html
a) ta có : cos215+cos225+cos235+cos245+cos255+cos265+cos275cos215+cos225+cos235+cos245+cos255+cos265+cos275
=cos215+cos275+cos225+cos265+cos235+cos255+cos245=cos215+cos275+cos225+cos265+cos235+cos255+cos245 =cos215+cos2(90−15)+cos225+cos2(90−25)+cos235+cos2(90−35)+cos245=cos215+cos2(90−15)+cos225+cos2(90−25)+cos235+cos2(90−35)+cos245 =cos215+sin215+cos225+sin225+cos235+sin235+cos245=cos215+sin215+cos225+sin225+cos235+sin235+cos245
tính nhanh
A=\(sin^242^o+sin^243^o+sin^244^o+sin^245^0+sin^246^o+sin^247^o+sin^248^o\)
B=\(\cos^215^o-cos^225^o+cos^235^o-cos^245^o+cos^255^o-cos^265^o+cos^275^o\)
\(ADCT:\sin^2\alpha+\cos^2\alpha=1\)
\(A=\left(\sin^242^0+\sin^248^0\right)+\left(\sin^243^0+\sin^247^0\right)+\left(\sin^244^0+\sin^246^0\right)+\sin45^0\)
\(A=\left(\sin^242^0+\cos^242^0\right)+\left(\sin^243^0+\cos^243^0\right)+\left(\sin^244^0+\cos^244^0\right)+\frac{\sqrt{2}}{2}\)
\(A=1+1+1+\frac{\sqrt{2}}{2}=\frac{6+\sqrt{2}}{2}\)
Câu b lm tương tự
\(2sin^275^o+2sin^215^o-cos^250^o-cos^240^o+cot40^o.cot50^o\)
\(=2sin^275+2cos^2\left(90-15\right)-sin^2\left(90-50\right)-cos^240+cot40.tan\left(90-50\right)\)
\(=2\left(sin^275+cos^275\right)-\left(sin^240+cos^240\right)+tan40.cot40\)
\(=2.1-1+1=2\)
Tính giá trị các biểu thức sau:
a) Cos\(^2\)15\(^o\) +cos\(^2\)25\(^o\) + cos\(^2\)35\(^o\) +cos\(^2\) 45\(^o\)+cos\(^2\) 55\(^o\) +cos\(^2\) 65\(^o\) +\(cos^275^o\)
b)\(Sin^210^o-sin^220^o+sin^230^o-sin^240^o-sin^250^o-sin^270^o+sin^280^o\)
~Giúp với ạ~
Thank
mk bỏ dấu độ nha . trong toán người ta cho phép
a) ta có : \(cos^215+cos^225+cos^235+cos^245+cos^255+cos^265+cos^275\)
\(=cos^215+cos^275+cos^225+cos^265+cos^235+cos^255+cos^245\) \(=cos^215+cos^2\left(90-15\right)+cos^225+cos^2\left(90-25\right)+cos^235+cos^2\left(90-35\right)+cos^245\) \(=cos^215+sin^215+cos^225+sin^225+cos^235+sin^235+cos^245\)\(=1+1+1+\dfrac{1}{2}=\dfrac{7}{2}\)
b) ta có : \(sin^210-sin^220+sin^230-sin^240-sin^250-sin^270+sin^280\)
\(=sin^210+sin^280-sin^220-sin^270-sin^240-sin^250+sin^230\) \(=sin^210+sin^2\left(90-10\right)-sin^220-sin^2\left(90-20\right)-sin^240-sin^2\left(90-40\right)+sin^230\) \(=sin^210+cos^210-sin^220-cos^220-sin^240-cos^240+sin^230\) \(=1-1-1+\dfrac{1}{4}=\dfrac{-3}{4}\)
bài 1 : không dùng bảng số hoặc máy tính, hãy tính :
a, A = \(\cos^220^o+\cos^230^o+\cos^240^o+.....+\cos^270^o\)
b, B = \(\sin^25^o+\sin^225^o+\sin^245^o+\sin^265^o+\sin^285^o\)
c, C = \(\sin^210^o-\sin^220^o+\sin^230^o-\sin^240^o-\sin^250^o-\sin^270^o+\sin^280^o\)
bài 2 : cho tam giác ABC vuông tại A, biết sin B = \(\frac{1}{4}\) C. Tính C ?
B1... Cho \(\Delta ABC\) nhọn AB=c , AC= b , CB=a
CMR : \(\dfrac{a}{\sin a}=\dfrac{b}{\sin b}=\dfrac{c}{\sin c}\)
B2... Không dùng bảng số và m.tính . Hãy tính
a) \(\sin^212^o+\sin^222^o+\sin^232^o+\sin^258^o+\sin^268^o+\sin^278^o\)
b)\(\cos^215^o+\cos^225^o+\cos^235^o+\cos^255^o+\cos^265^o+\cos^275^o-3\)
c) \(4\cos^2\alpha-6\sin^2\alpha,\) biết \(\sin\alpha=\dfrac{1}{5}\)
d) \(\sin\alpha.\cos\alpha\) biết \(\tan\alpha+\cot\alpha=3\)
a) Ta có : sin\(^2\)12o=cos278o=> sin212o+sin278o=1.
tương tự => A=3
b) tương tự câu (a) ta có: cos215o=sin275o ( do 15+75=90 nha bạn ) => cos215o+cos275o=1. Tương tự => B=0
c) Ta có: C= 4cos2a -6sin2a=4(cos2a+sin2a)-10sin2a=4-10.\(\dfrac{1}{5^2}\)=4-0,4=3,6