Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
f/
\(sin2A+sin2B+sin2C=2sin\left(A+B\right).cos\left(A-B\right)+2sinC.cosC\)
\(=2sinC.cos\left(A-B\right)+2sinC.cosC\)
\(=2sinC\left(cos\left(A-B\right)+cosC\right)\)
\(=2sinC\left[cos\left(A-B\right)-cos\left(A+B\right)\right]\)
\(=4sinC.sinA.sinB\)
g/
\(cos^2A+cos^2B+cos^2C=\frac{1}{2}+\frac{1}{2}cos2A+\frac{1}{2}+\frac{1}{2}cos2B+cos^2C\)
\(=1+\frac{1}{2}\left(cos2A+cos2B\right)+cos^2C\)
\(=1+cos\left(A+B\right).cos\left(A-B\right)+cos^2C\)
\(=1-cosC.cos\left(A-B\right)+cos^2C\)
\(=1-cosC\left(cos\left(A-B\right)-cosC\right)\)
\(=1-cosC\left[cos\left(A-B\right)+cos\left(A+B\right)\right]\)
\(=1-2cosC.cosA.cosB\)
d/ \(sinA+sinB+sinC=2sin\frac{A+B}{2}cos\frac{A-B}{2}+2sin\frac{C}{2}.cos\frac{C}{2}\)
\(=2cos\frac{C}{2}.cos\frac{A-B}{2}+2sin\frac{C}{2}.cos\frac{C}{2}\)
\(=2cos\frac{C}{2}\left(cos\frac{A-B}{2}+sin\frac{C}{2}\right)\)
\(=2cos\frac{C}{2}\left(cos\frac{A-B}{2}+cos\frac{A+B}{2}\right)\)
\(=4cos\frac{C}{2}.cos\frac{A}{2}.cos\frac{B}{2}\)
e/
\(cosA+cosB+cosC=2cos\frac{A+B}{2}cos\frac{A-B}{2}+1-2sin^2\frac{C}{2}\)
\(=1+2sin\frac{C}{2}.cos\frac{A-B}{2}-2sin^2\frac{C}{2}\)
\(=1+2sin\frac{C}{2}\left(cos\frac{A-B}{2}-sin\frac{C}{2}\right)\)
\(=1+2sin\frac{C}{2}\left(cos\frac{A-B}{2}-cos\frac{A+B}{2}\right)\)
\(=1+4sin\frac{C}{2}.sin\frac{A}{2}sin\frac{B}{2}\)
B=1-sin2a+cos2a
\(=\sin^2a+\cos^2a-\sin^2a+\cos^2a=2\cos^2a\)
C= 1-sina.cosa.tana
\(=1-\sin a.\cos a.\frac{\sin a}{\cos a}=1-\sin^2a=\cos^2a\)
biết có vậy thôi à
Theo đl sin có:
\(\dfrac{a}{sinA}=\dfrac{b}{sinB}=\dfrac{c}{sinC}\Rightarrow b=a\dfrac{sinB}{sinA};c=\dfrac{sinC}{sinA}.a\)
Mà `b+c=2a`
\(\Rightarrow a\dfrac{sinB}{sinA}+a\dfrac{sinC}{sinA}=2a\\ \Rightarrow\dfrac{sinB}{sinA}+\dfrac{sinC}{sinA}=2\\ \Leftrightarrow sinB+sinC=2sinA\)
Chọn B
a) Sin (B+C) = Sin (180-A) = Sin A
b) Cos (A+B) = Cos ( 180-A) = Cos A
c) Sin (\(\dfrac{B+C}{2}\)) = Sin \(\left(\dfrac{180-A}{2}\right)\)= Sin \(\left(90^0-\dfrac{A}{2}\right)\)= Cos \(\dfrac{A}{2}\)
d) Tan \(\left(\dfrac{A+C}{2}\right)\)= Tan\(\left(\dfrac{180-B}{2}\right)\)=Tan\(\left(90^0-\dfrac{B}{2}\right)\)= Cot \(\dfrac{B}{2}\)
\(V=cos^2A+cos^2B+cos^2C-1\)
\(V=\frac{1+cos2A}{2}+\frac{1+cos2B}{2}+cos^2C-1\)
\(V=\frac{1}{2}\left(cos2A+cos2B\right)+cos^2C\)
\(V=cos\left(A+B\right).cos\left(A-B\right)+cos^2C\)
\(V=-cosC.cos\left(A-B\right)+cos^2C\)
\(V=cosC\left[cos\left(A-B\right)-cosC\right]\)
\(V=-2cosC.sin\left(\frac{A-B+C}{2}\right).sin\left(\frac{A-B-C}{2}\right)\)
\(V=2cosC.cosB.cosA\)
\(V=0\Rightarrow\left[{}\begin{matrix}cosA=0\\cosB=0\\cosC=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}A=90^0\\B=90^0\\C=90^0\end{matrix}\right.\)
Chọn A.
Giả sử A = α; B + C = β.
Biểu thức trở thành P = sinα.cosβ - cosα.sinβ.
Trong tam giác ABC, có A + B + C = 1800 nên α + β = 1800.
Do hai góc α và β bù nhau nên sinα = sinβ và cosα = - cosβ.
Do đó, P = sinα.cosβ - cosα.sinβ = -sinα.cosα + cosα.cosβ = 0.