\(\sin A=\sin\left(\Pi-B-C\right)=\sin\left(B+C\right)\)

\(=\sin B\cos C+\cos B\sin C\)

Vô lý thế nhỉ: đề có cho giằng buộc A,B,C đâu?

A+B+C=(pi) ???

TL:

sinA+sinB+sinC=1-cosA+cosB+cosC => Tam giác ABC Vuông tại A

Vế trái = sinA + sinB + sinC

= 2sin(A + B)/2.cos(A - B)/2 + 2sinC/2.cosC/2

= 2cosC/2.cos(A - B)/2 + 2sinC/2.cosC/2

= 2cosC/2[cos(A - B)/2 + sinC/2]

=2.cosC/2.[cos(A - B)/2 + cos(A + B)/2]

= 4.cosC/2.cosB/2.cosA/2

Vế phải = 1 - cosA + cosB + cosC

= 2sin²A/2 + 2cos(B + C)/2.cos(B - C)/2

= 2.sinA/2[sinA/2 + cos(B - C)/2] (vì cos(B + C)/2 = sinA/2)

= 2.sinA/2[cos(B + C)/2 + cos(B - C)/2

= 4.sinA/2.cosB/2.cosC/2

Vậy sinA + sinB + sinC = 1 - cosA + cosB + cosC

<=> cosA/2.cosB/2.cosC/2 = sinA/2.cosB/2.cosC/2

<=> cosB/2.cosC/2(sinA/2 - cosA/2) = 0

mà cosB/2 ≠ 0 và cosC/2 ≠ 0

=> sinA/2 = cosA/2

<=> A/2 = 45^o

<=> A = 90^o

tam giác ABC vuông tại A

\(sin^4x+cos^4x=sin^4x+cos^4x+2sin^2x.cos^2x-2sin^2x.cos^2x\)

\(=\left(sin^2x+cos^2x\right)^2-\frac{1}{2}\left(2sinx.cosx\right)^2\)

\(=1-\frac{1}{2}sin^22x\Rightarrow\left\{{}\begin{matrix}a=1\\b=1\\c=2\end{matrix}\right.\) \(\Rightarrow a+3b+c=?\)

\(\frac{sin\left(A-B\right)}{sinC}=\frac{sin\left(A-B\right).sinC}{sin^2C}=\frac{sin\left(A-B\right).sin\left(A+B\right)}{sin^2C}=\frac{-\frac{1}{2}\left(cos2A-cos2B\right)}{sin^2C}\)

\(=\frac{-\frac{1}{2}\left(1-2sin^2A-1+2sin^2B\right)}{sin^2C}=\frac{sin^2A-sin^2B}{sin^2C}=\frac{\left(\frac{a}{2R}\right)^2-\left(\frac{b}{2R}\right)^2}{\left(\frac{c}{2R}\right)^2}=\frac{a^2-b^2}{c^2}\)

Câu 3:

a/ Đề dị dị, là \(\frac{cosA+cosB}{sinB+sinC}\) hay \(\frac{cosB+cosC}{sinB+sinC}\) bạn?

b/ \(cos\left(B-C\right)-cos\left(B+C\right)=1+cosA\)

\(\Leftrightarrow cos\left(B-C\right)+cosA=1+cosA\)

\(\Leftrightarrow cos\left(B-C\right)=1\)

\(\Rightarrow B=C\Rightarrow\Delta ABC\) cân tại A

\(2sinB.sinC=1+cosA\Leftrightarrow cos\left(B-C\right)-cos\left(B+C\right)=1+cosA\)

\(\Leftrightarrow cos\left(B-C\right)+cosA=1+cosA\)

\(\Leftrightarrow cos\left(B-C\right)=1\)

\(\Rightarrow B-C=0\Rightarrow B=C\)

\(sinA=\frac{cosA+cosB}{sinB+sinC}=\frac{cosA+cosB}{2sinB}\) (do \(B=C\))

\(\Leftrightarrow2sinA.sinB=cosA+cosB\)

\(\Leftrightarrow cos\left(A-B\right)-cos\left(A+B\right)=cosA+cosB\)

\(\Leftrightarrow cos\left(A-B\right)+cosC=cosA+cosB\)

\(\Leftrightarrow cos\left(A-B\right)+cosB=cosA+cosB\)

\(\Leftrightarrow cos\left(A-B\right)=cosB\)

\(\Rightarrow A-B=B\Rightarrow A=2B=B+C\)

Mà \(A+B+C=180^0\Rightarrow2A=180^0\Rightarrow A=90^0\)

\(\Rightarrow\Delta ABC\) vuông cân tại A

1.

\(sinA+sinB-sinC=2sin\dfrac{A+B}{2}.cos\dfrac{A-B}{2}-sin\left(A+B\right)\)

\(=2sin\dfrac{A+B}{2}.cos\dfrac{A-B}{2}-2sin\dfrac{A+B}{2}.cos\dfrac{A+B}{2}\)

\(=2sin\dfrac{A+B}{2}.\left(cos\dfrac{A-B}{2}-cos\dfrac{A+B}{2}\right)\)

\(=2sin\dfrac{A+B}{2}.2sin\dfrac{A}{2}.sin\dfrac{B}{2}\)

\(=4sin\dfrac{A}{2}.sin\dfrac{B}{2}.cos\dfrac{C}{2}\)

Sao t lại đc như này v, ai check hộ phát

\(\dfrac{cosa+cos5a+cos3a}{sina+sin5a+sin3a}=\dfrac{2cos3a.cos2a+cos3a}{2sin3a.cos2a+sin3a}\)

\(=\dfrac{cos3a\left(2cos2a+1\right)}{sin3a\left(2cos2a+1\right)}=\dfrac{cos3a}{sin3a}=cot3a\)

\(\left(\dfrac{cosa}{sinb}+\dfrac{sina}{cosb}\right)\left(\dfrac{1-cos4b}{cos\left(a-b\right)}\right)=\dfrac{\left(cosa.cosb+sina.sinb\right)}{sinb.cosb}.\dfrac{2sin^22b}{cos\left(a-b\right)}\)

\(=\dfrac{cos\left(a-b\right)}{\dfrac{1}{2}sin2b}.\dfrac{2sin^22b}{cos\left(a-b\right)}=4sin2b\)

Bạn ơi mình khong hiểu sao đoạn ( 1/cos4b)/ cos(a-b) lại tách thành 2sin^22b / cos(a-b)

Yêu cầu đề bài là gì vậy bạn?