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![](https://rs.olm.vn/images/avt/0.png?1311)
a)
^MAC = ^MCA = a ---> ^AMH = ^MAC + ^MCA = 2a
sin2a = sinAMH = AH/MA = 2AH/BC = 2(AH/AC).(AC/BC) = 2 sina.cosa
b)
1+cos2a = 1+cosAMH = 1+MH/MA = (MA+MH)/MA = CH/MA = 2CH/BC =
= 2 (CH/AC).(AC/BC) = 2 cosa.cosa = 2 cos^2 (a)
c)
1-cos2a = 1-cosAMH = 1-MH/MA = (MA-MH)/MA = BH/MA = 2BH/BC =
= 2 (BH/AB).(AB/BC) = 2 sinBAH.sinACB = 2 sin^2 (a)
(^BAH = ^ACB = a vì chúng cùng phụ với góc ABC)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: sin a=sin C=AB/BC
cos a=AC/BC=b/a
sin 2a=2sinacosa\(=2\cdot\dfrac{b}{a}\cdot\dfrac{AB}{BC}=\dfrac{2b\cdot AB}{a^2}\)
b: \(sin2a=sin\left(a+a\right)\)
\(=sina\cdot cosa+sina\cdot cosa\)
\(=2\cdot sina\cdot cosa\)
![](https://rs.olm.vn/images/avt/0.png?1311)
2/ \(\frac{sin^3a-cos^3a}{sin^3a+cos^3a}=\frac{tan^3a-1}{tan^3a+1}=\frac{3^3-1}{3^3+1}=\frac{13}{14}\) (chia tử mẫu cho cos3a)
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có :\(sin^2a+cos^2a=1\)=> \(1-cos^2a=sin^2a\)
ma \(1-cos^2a=2sin^2a\)
<=> \(sin^2a=2sin^2a\)
<=> 1/2 (vô lí)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{sin^2a-cos^2a+cos^4a}{cos^2a-sin^2a+sin^4a}=\frac{sin^2a-cos^2a\left(1-cos^2a\right)}{cos^2a-sin^2a\left(1-sin^2a\right)}=\frac{sin^2a-cos^2a.sin^2a}{cos^2a-sin^2a.cos^2a}\)
\(=\frac{sin^2a\left(1-cos^2a\right)}{cos^2a\left(1-sin^2a\right)}=\frac{sin^2a.sin^2a}{cos^2a.cos^2a}=tan^4a\)
\(sin^4a+cos^4a=\left(sin^2a+cos^2a\right)^2-sin^2a.cos^2a=1-2sin^2a.cos^2a\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, Sử dụng tích chéo:
Ta có:
+/ \(\cos\alpha.\cos\alpha=\cos^2\alpha\) (1)
+/ \(\left(1+\sin\alpha\right)\left(1-\sin\alpha\right)=1-\sin^2\alpha\)
Mà \(\sin^2\alpha+\cos^2\alpha=1\)
\(\Rightarrow1-\sin^2\alpha=\cos^2\alpha\)
hay \(\left(1+\sin\alpha\right)\left(1-\sin\alpha\right)=\cos^2\alpha\) (2)
Từ (1), (2)
\(\Rightarrow\)\(\cos\alpha.\cos\alpha=\)\(\left(1+\sin\alpha\right)\left(1-\sin\alpha\right)\)
\(\Rightarrow\)\(\dfrac{\cos\alpha}{1-\sin\alpha}=\dfrac{1+\sin\alpha}{\cos\alpha}\) (đpcm)
b/ xem lại đề
a: Xét ΔMAC có MA=MC
nên ΔMAC cân tại M
=>\(\widehat{AMC}=180^0-2\cdot\widehat{ACB}\)
mà \(\widehat{AMC}=180^0-\widehat{AMB}\)
nên \(\widehat{AMB}=2\cdot\widehat{ACB}\)
=>\(sin2\alpha=sin\left(\widehat{AMB}\right)=\dfrac{AH}{AM}=\dfrac{AH}{\dfrac{BC}{2}}=AH\cdot\dfrac{2}{BC}=\dfrac{2AH}{BC}\)
Xét ΔHAC vuông tại H có \(sinACH=\dfrac{AH}{AC}\)
Xét ΔABC vuông tại A có \(cosACB=\dfrac{AC}{CB}\)
=>\(sin\alpha\cdot cos\alpha=\dfrac{AH}{AC}\cdot\dfrac{AC}{CB}=\dfrac{AH}{CB}\)
=>\(2\cdot sin\alpha\cdot cos\alpha=2\cdot\dfrac{AH}{BC}\)
Do đó: \(sin2\alpha=2\cdot sin\alpha\cdot cos\alpha\)
b: Xét ΔABC vuông tại A có \(cos^2\alpha=cos^2C=\left(\dfrac{CA}{CB}\right)^2=\dfrac{CA^2}{CB^2}\)
=>\(2\cdot cos^2\alpha=\dfrac{2\cdot CA^2}{CB^2}=\dfrac{2\cdot CH\cdot CB}{CB^2}=\dfrac{2\cdot CH}{CB}\)
\(\Leftrightarrow2\cdot cos^2\alpha-1=\dfrac{2\cdot CH}{2\cdot CM}-1=\dfrac{CH}{CM}-1=\dfrac{CH-CM}{CM}=\dfrac{HM}{CM}\)
\(=\dfrac{HM}{AM}=cosAMB=cos2\alpha\)
=>\(1+cos2\alpha=2\cdot cos^2\alpha\)
c: \(1-2\cdot sin^2\alpha=1-2\cdot\left(1-cos^2\alpha\right)=2\cdot cos^2\alpha+1=cos2\alpha\)
=>\(1-cos2\alpha=2\cdot sin^2\alpha\)