Góc 2α =  A M H ^

a, Ta có:  sin 2 α = A H A M = 2 A H A M = 2 A B . A C B C 2 = 2 sin α . cos α

b,  1 + cos2α =  1 + H M A M = H C A M = 2 H C B C =  2 . A C 2 B C 2 = 2 cos 2 α

c, 1 – cos2α =  1 - H M A M = H B A M = 2 H B B C =  2 . A B 2 B C 2 = 2 sin 2 α

b) Ta có: \(\sin^2\alpha+\cos^2\alpha=1\)

\(\Leftrightarrow\cos^2\alpha=\dfrac{16}{25}\)

hay \(\cos\alpha=\dfrac{4}{5}\)

Ta có: \(A=5\cdot\sin^2\alpha+6\cdot\cos^2\alpha\)

\(=5\cdot\left(\dfrac{3}{5}\right)^2+6\cdot\left(\dfrac{4}{5}\right)^2\)

\(=5\cdot\dfrac{9}{25}+6\cdot\dfrac{16}{25}\)

\(=\dfrac{141}{25}\)

c) Ta có: \(\tan\alpha=\dfrac{1}{\cot\alpha}=\dfrac{1}{\dfrac{4}{3}}=\dfrac{3}{4}\)

\(D=\dfrac{\sin\alpha+\cos\alpha}{\sin\alpha-\cos\alpha}\)

\(=\dfrac{\dfrac{9}{16}+\dfrac{16}{9}}{\dfrac{9}{16}-\dfrac{16}{9}}=-\dfrac{337}{175}\)

Đáp án đúng : D

a: \(=\left(\sin^2\alpha+\cos^2\alpha\right)^2=1^2=1\)

\(\tan\alpha+\cot\alpha=3\)

=>\(\frac{\sin\alpha}{cos\alpha}+\frac{cos\alpha}{\sin\alpha}=3\)

=>\(\frac{\sin^2\alpha+cos^2\alpha}{\sin\alpha\cdot cos\alpha}=3\)

=>\(\frac{1}{\sin\alpha\cdot cos\alpha}=3\)

=>\(\sin\alpha\cdot cos\alpha=\frac13\)

\(\tan^2\alpha+\cot^2\alpha=\left(\tan\alpha+\cot\alpha\right)^2-2\cdot tan\alpha\cdot\cot a\)

\(=3^2-2\)

=9-2

=7

Ta có: \(tana+cota=3\Rightarrow\dfrac{sina}{cosa}+\dfrac{cosa}{sina}=3\)

\(\Rightarrow\dfrac{sin^2a+cos^2a}{sina\cdot cosa}=3\Rightarrow sina\cdot cosa=\dfrac{1}{3}\)

Ta có: \(\left(tana+cota\right)^2=9\)\(\Rightarrow tan^2a+cot^2a=9-2tana\cdot cota=9-2=7\)

a: \(A=cos^4a+2\cdot cos^2a\cdot\sin^2a+\sin^4a\)

\(=\left(cos^2a+\sin^2a\right)^2=1^2\)

=1

=>A không phụ thuộc vào biến

b: \(B=\sin^4a+cos^2a\cdot\sin^2a+cos^2a\)

\(=\sin^2a\left(\sin^2a+cos^2a\right)+cos^2a\)

\(=\sin^2a+cos^2a\)

=1

=>B không phụ thuộc vào biến

c: \(C=2\left(\sin a-cosa\right)^2-\left(\sin a+cosa\right)^2+6\cdot\sin a\cdot cosa\)

\(=2\left(1-2\cdot\sin a\cdot cosa\right)-\left(1+2\cdot\sin a\cdot cosa\right)+6\cdot\sin a\cdot cosa\)

\(=2-4\cdot\sin a\cdot cosa-1-2\cdot\sin a\cdot cosa+6\cdot\sin a\cdot cosa\)

=2-1

=1

=>C không phụ thuộc vào biến

d: \(D=\left(\tan a-\cot a\right)^2-\left(\tan a+\cot a\right)^2\)

\(=\tan^2a-2\cdot\tan a\cdot\cot a+\cot^2a-\left(\tan^2a+2\cdot\tan a\cdot\cot a+\cot^2a\right)\)

\(=-4\cdot\tan a\cdot\cot a=-4\)

=>D không phụ thuộc vào biến

e: \(E=4\cdot cos^2a+\left(\sin a-cosa\right)^2+\left(\sin a+cosa\right)^2+2\left(\sin^2a-cos^2a\right)\)

\(=4\cdot cos^2a+\sin^2a+cos^2a-2\cdot\sin a\cdot cosa+\sin^2a+cos^2a+2\cdot\sin a\cdot cosa+2\left(\sin^2a-cos^2a\right)\)

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

\(=2\cdot\sin^2a+2\cdot cos^2a+2=2+2=4\)

=>E không phụ thuộc vào biến

f: \(F=\frac{1}{1+\sin a}+\frac{1}{1-\sin a}-2\cdot\tan^2a\)

\(=\frac{1-\sin a+1+\sin a}{\left(1+\sin a\right)\left(1-\sin a\right)}-2\cdot\tan^2a\)

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

=>F không phụ thuộc vào biến

a: MA=MB

=>ΔMAB cân tại M

MA=MC

=>ΔMAC cân tại M

Xét ΔMAC có \(\hat{AMB}\) là góc ngoài tại đỉnh M

nên \(\hat{AMB}=\hat{MAC}+\hat{MCA}=2\cdot\hat{ACB}\)

=>\(\sin2a=\sin\left(2\cdot\hat{ACB}\right)=\sin AMB\)

Xét ΔAMH vuông tại H có sin AMH=\(\frac{AH}{AM}\)

=>\(\sin AMB=\frac{AH}{AM}\)

Xét ΔABC vuông tại A có

\(\sin\alpha=\sin C=\frac{AB}{BC};cos\alpha=cosC=\frac{AC}{BC}\)

\(2\cdot\sin a\cdot cosa=2\cdot\frac{AB}{BC}\cdot\frac{AC}{BC}=\frac{2\cdot AH\cdot BC}{BC\cdot BC}=2\cdot\frac{AH}{BC}=\frac{2\cdot AH}{2\cdot AM}=\frac{AH}{AM}\)

=>\(\sin2\alpha=2\cdot\sin\alpha\cdot cos\alpha\)

b: Xét ΔAMH vuông tại M có

cos AMH=\(\frac{HM}{AM}\)

=>\(cos2\alpha=\frac{HM}{AM}\)

=>\(1+cos2a=1+\frac{HM}{AM}=\frac{HM+AM}{AM}=\frac{HM+MC}{AM}=\frac{HC}{AM}\)

\(2\cdot cos^2a=2\cdot cos^2C=2\cdot\left(\frac{CA}{CB}\right)^2\)

\(=\frac{2\cdot CA^2}{CB^2}=\frac{2\cdot CH\cdot BC}{CB^2}=\frac{2\cdot CH}{CB}=\frac{2\cdot CH}{2\cdot AM}=\frac{CH}{AM}\)

Do đó: \(1+cos2a=2\cdot cos^2a\)

c: \(1-cos2a=1-\frac{HM}{AM}=\frac{AM-HM}{AM}=\frac{BM-HM}{AM}=\frac{BH}{AM}\)

\(2\cdot\sin^2a=2\cdot\sin^2C=2\cdot\left(\frac{AB}{BC}\right)^2\)

\(=\frac{2\cdot AB^2}{BC^2}=\frac{2\cdot BH\cdot BC}{BC^2}=\frac{2\cdot BH}{BC}=\frac{2\cdot BH}{2\cdot AM}=\frac{BH}{AM}\)

Do đó: \(1-cos2a=2\cdot\sin^2a\)