ta thấy cos12^o = sin78^o nên cos²12^o = sin²78^o

Sau khi biến đổi ta được: S= sin²78^o + cos²78^o + sin²89^o + cos²89^o=2

đáp án C

`sin^2x+cos^2x=1`

`<=>sin^2x+(1/2)^2=1`

`<=> sinx=\pm \sqrt3/2`

• `sinx=\sqrt3/2 => P=3. (\sqrt3/2)^2 +1=13/4`

• `sinx=-\sqrt3/2 => P = 3.(-\sqrt3/2) +1=13/4`

`=>` A.

\(P=3sin^2x+1=3\left(1-cos^2x\right)+1=3\left(1-\dfrac{1}{4}\right)+1=\dfrac{13}{4}\)

Chọn B

\(\frac{1}{cos^2a}=1+tan^2a\Rightarrow cos^2a=\frac{1}{1+tan^2a}=\frac{1}{10}\)

a/ \(\frac{sina-cosa}{sina+cosa}=\frac{\frac{sina}{cosa}-\frac{cosa}{cosa}}{\frac{sina}{cosa}+\frac{cosa}{cosa}}=\frac{tana-1}{tana+1}=\frac{3-1}{3+1}\)

b/ \(\frac{2sina+3cosa}{3sina-5cosa}=\frac{3tana+3}{3tana-5}=\frac{3.3+3}{3.3-5}\)

c/ \(\frac{1+2cos^2a}{1-cos^2a-cos^2a}=\frac{1+2cos^2a}{1-2cos^2a}=\frac{1+2.\frac{1}{10}}{1-2.\frac{1}{10}}\)

d/ \(\frac{\left(1-cos^2a\right)^2+\left(cos^2a\right)^2}{1+1-cos^2a}=\frac{\left(1-\frac{1}{10}\right)^2+\left(\frac{1}{10}\right)^2}{2-\frac{1}{10}}\)

\(P=sin^2x+3cos^2x=1-cos^2x+3cos^2x=1+2cos^2x=1+2.\left(\dfrac{1}{4}\right)^2=\dfrac{9}{8}\)

a)

\(\sin ^4a-\cos ^4a+1=(\sin ^2a-\cos ^2a)(\sin ^2a+\cos^2a)+1\)

\(=(\sin ^2a-\cos ^2a).1+1=\sin ^2a-\cos ^2a+\sin ^2a+\cos ^2a\)

\(=2\sin ^2a\)

b) \(\sin ^2a+2\cos ^2a-1=(\sin ^2a+\cos^2a)+\cos ^2a-1\)

\(=1+\cos ^2a-1=\cos ^2a\)

\(\Rightarrow \frac{\sin ^2a+2\cos ^2a-1}{\cot ^2a}=\frac{\cos ^2a}{\cot ^2a}=\frac{\cos ^2a}{\frac{\cos ^2a}{\sin ^2a}}=\sin ^2a\)

c)

\(\frac{1-\sin ^2a\cos ^2a}{\cos ^2a}-\cos ^2a=\frac{1}{\cos ^2a}-\sin ^2a-\cos ^2a\)

\(=\frac{1}{\cos ^2a}-(\sin ^2a+\cos ^2a)=\frac{1}{\cos ^2a}-1\)

\(=\frac{1-\cos ^2a}{\cos ^2a}=\frac{\sin ^2a}{\cos ^2a}=\tan ^2a\)

d)

\(\frac{\sin ^2a-\tan ^2a}{\cos ^2a-\cot ^2a}=\frac{\sin ^2a-\frac{\sin ^2a}{\cos ^2a}}{\cos ^2a-\frac{\cos ^2a}{\sin ^2a}}\) \(=\frac{\sin ^2a(1-\frac{1}{\cos ^2a})}{\cos ^2a(1-\frac{1}{\sin ^2a})}\)

\(=\frac{\sin ^2a.\frac{\cos ^2a-1}{\cos ^2a}}{\cos ^2a.\frac{\sin ^2a-1}{\sin ^2a}}\) \(=\frac{\sin ^2a.\frac{-\sin ^2a}{\cos ^2a}}{\cos ^2a.\frac{-\cos ^2a}{\sin ^2a}}=\frac{\sin ^6a}{\cos ^6a}=\tan ^6a\)

f)

\(\frac{(\sin a+\cos a)^2-1}{\cot a-\sin a\cos a}=\frac{\sin ^2a+\cos ^2a+2\sin a\cos a-1}{\frac{\cos a}{\sin a}-\sin a\cos a}\)

\(=\sin a.\frac{1+2\sin a\cos a-1}{\cos a-\cos a\sin ^2a}\)

\(=\sin a. \frac{2\sin a\cos a}{\cos a(1-\sin ^2a)}=\sin a. \frac{2\sin a\cos a}{\cos a. \cos^2 a}=\frac{2\sin ^2a}{\cos ^2a}=2\tan ^2a\)

\(\dfrac{1+sin^4x-cos^4x}{1-sin^6x-cos^6x}=\dfrac{1+sin^2x-cos^2x}{1-\left(sin^4x+cos^4x-sin^2xcos^2x\right)}=\dfrac{1+sin^2x-\left(1-sin^2x\right)}{1-\left(1-3sin^2xcos^2x\right)}=\dfrac{2sin^2x}{3sin^2xcos^2x}=\dfrac{2}{3cos^2x}\)

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/

\(cosA+cosB+cosC=2cos\frac{A+B}{2}cos\frac{A-B}{2}+1-2sin^2\frac{C}{2}\)

\(=1+2sin\frac{C}{2}.cos\frac{A-B}{2}-2sin^2\frac{C}{2}\)

\(=1+2sin\frac{C}{2}\left(cos\frac{A-B}{2}-sin\frac{C}{2}\right)\)

\(=1+2sin\frac{C}{2}\left(cos\frac{A-B}{2}-cos\frac{A+B}{2}\right)\)

\(=1+4sin\frac{C}{2}.sin\frac{A}{2}sin\frac{B}{2}\)

(1): Gọi G là trọng tâm của ΔABC, M là trung điểm của BC, N là trung điểm của AC

Theo đề, ta có: AM=3; BN=3/2; \(S_{ABC}=3\)

Xét ΔABC có

AM,BN là các đường trung tuyến

G là trọng tâm

Do đó: G là giao điểm của AM và BN

Xét ΔABC có

AM là đường trung tuyến

G là trọng tâm

Do đó: \(AG=\frac23AM=\frac23\cdot3=2\)

Xét ΔABC có

BN là đường trung tuyến

G là trọng tâm

Do đó: \(BG=\frac23BN=\frac23\cdot\frac32=1\)

G là trọng tâm của ΔABC

=>\(S_{BGA}=\frac13\cdot S_{ABC}=\frac13\cdot3=1\)

=>\(\frac12\cdot GA\cdot GB\cdot\sin AGB=1\)

=>\(\frac12\cdot2\cdot1\cdot\sin AGB=1\)

=>sinAGB=1

=>\(\hat{AGB}=90^0\)

=>AM⊥BN tại G

ΔGAB vuông tại G

=>\(GA^2+GB^2=AB^2\)

=>\(AB^2=1^2+2^2=5\)

=>\(c=AB=\sqrt5\)

Gọi K là trung điểm của AB

=>\(GK=\frac13\cdot m_{C}\)

ΔGAB vuông tại G

mà GK là đường trung tuyến

nên \(GK=\frac{AB}{2}=\frac{\sqrt5}{2}\)

=>\(m_{C}=\frac{3\sqrt5}{2}\)

Xét ΔABC có AM,BN là các đường trung tuyến

nên \(MA^2=m_{a}^2=\frac{2(b^2 + c^2) - a^2}{4}\) và \(BN^2=m_{b}^2=\frac{2(a^2 + c^2) - b^2}{4}\)

=>\(m_a^2 + m_b^2 = \frac{a^2 + b^2 + 4c^2}{4}\)

Xét ΔABC có CK là đường trung tuyến

nên \(CK^2=m_{c}^2=\frac{2(a^2 + b^2) - c^2}{4}\implies2(a^2+b^2)=4m_{c}^2+c^2\)

=>\(4m_c^2 + c^2 = 4 \cdot \frac{45}{4} + 5 = 50 \implies a^2 + b^2 = 25\)

\(m_{a}^2+m_{b}^2=\frac{a^2 + b^2 + 4c^2}{4}\)

\(m_{a}^2+m_{b}^2=\) \(3^2+\left(\frac{3}{2}\right)^2=\frac{25 + 4(5)}{4}=\frac{45}{4}\quad\)

\(m_{a}^2-m_{b}^2=\frac{3(b^2 - a^2)}{4}\)

\(\implies9-\frac{9}{4}=\frac{27}{4}=\frac{3(b^2 - a^2)}{4}\)

\(\implies b^2-a^2=9\)

Do đó, ta có hệ phương trình:

\(\begin{cases}a^2+b^2=25\\ b^2-a^2=9\end{cases}\)

=>\(2a^2=16\)

\(\implies a^2=8\)

\(\implies a=2\sqrt{2}\)

(2): \(a^2 = \frac{8}{9}(m_b^2 + m_c^2) - \frac{4}{9} m_a^2\)

=>\(a^2=\frac{8}{9}(2+3)-\frac{4}{9}(1)=\frac{36}{9}=4\)

=>a=2

\(b^2 = \frac{8}{9}(m_a^2 + m_c^2) - \frac{4}{9} m_b^2\)

=>\(b^2 = \frac{8}{9}(1 + 3) - \frac{4}{9}(2) = \frac{24}{9} = \frac{8}{3}\)

\(c^2 = \frac{8}{9}(m_a^2 + m_b^2) - \frac{4}{9} m_c^2\)

=>\(c^2 = \frac{8}{9}(1 + 2) - \frac{4}{9}(3) = \frac{12}{9} = \frac{4}{3}\)

\(\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{\frac{8}{3} + \frac{4}{3} - 4}{2 \cdot \sqrt{\frac{8}{3}} \cdot \sqrt{\frac{4}{3}}} = \frac{0}{2bc} = 0\)

\(\cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{4 + \frac{4}{3} - \frac{8}{3}}{2 \cdot 2 \cdot \frac{2}{\sqrt{3}}} = \frac{\frac{8}{3}}{\frac{8}{\sqrt{3}}} = \frac{\sqrt{3}}{3}\)