Chương 6: CUNG VÀ GÓC LƯỢNG GIÁC. CÔNG THỨC LƯỢNG GIÁC

e: \(=\dfrac{1}{2}\cdot cos90+\dfrac{1}{2}\cdot cos60=\dfrac{1}{4}\)

g: \(=8\cdot cosx\cdot\dfrac{-1}{2}\cdot\left[cos5x-cosx\right]\)

\(=-4\cdot cosx\cdot cos5x+4cos^2x\)

\(=-4\cdot\dfrac{1}{2}\cdot\left[cos6x-cos4x\right]+4\cdot cos^2x\)

\(=-2\cdot cos6x+2\cdot cos4x+4\cdot cos^2x\)

\(D=\frac{\left(-sinx\right).\left(sinx\right)\left(tanx\right)}{\left(-cosx\right)\left(-cosx\right)\left(-tanx\right)}=\frac{sin^2x}{cos^2x}=tan^2x\)

\(sinx.sin\left(\frac{\pi}{3}-x\right)sin\left(\frac{\pi}{3}+x\right)=\frac{1}{2}sinx.\left[cos2x-cos\frac{2\pi}{3}\right]=\frac{1}{2}sinx\left(cos2x+\frac{1}{2}\right)\)

\(=\frac{1}{2}sinx.cos2x+\frac{1}{4}sinx=\frac{1}{4}\left(sin3x+sin\left(-x\right)\right)+\frac{1}{4}sinx\)

\(=\frac{1}{4}sin3x-\frac{1}{4}sinx+\frac{1}{4}sinx=\frac{1}{4}sin3x\)

Ta có :

\(VT=\sqrt{sin^4x+4cos^2x}+\sqrt{cos^4x+4sin^2x}\)

\(=\sqrt{sin^4x-4sin^2x+4}+\sqrt{cos^4x-4cos^2x+4}\)

\(=\sqrt{\left(sin^2x-2\right)^2}+\sqrt{\left(cos^2x-2\right)^2}\)

\(=\left|sin^2x-2\right|+\left|cos^2x-2\right|\)

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

\(=4-1=3\)

\(VT=\sqrt{sin^4x+4\left(1-sin^2x\right)}+\sqrt{cos^4x+4\left(1-cos^2x\right)}\)

\(=\sqrt{sin^4x-4sin^2x+4}+\sqrt{cos^4x-4cos^2x+4}\)

\(=\sqrt{\left(2-sin^2x\right)^2}+\sqrt{\left(2-cos^2x\right)^2}\)

\(=2-sin^2x+2-cos^2x=4-\left(sin^2x+cos^2x\right)=3\)

Sửa đề: DC=18

góc ADB=1/2*180=90 độ

=>BD vuông góc AC

BC^2=DC*DA

=18*50=900

=>BC=30

=>AB=40

=>R=40/2=20

2:

sin(11pi/3)=-căn 3/2

cos(11pi/3)=1/2

tan 11pi/3=-căn 3

1:

sin120=căn 3/2

cos120=-1/2

tan 120=-căn 3

Lời giải:

Ta thấy:

$S_{ABC}=\frac{a.h_a}{2}=\frac{b.h_b}{2}=\frac{c.h_c}{2}$

$\Rightarrow ah_a=bh_b=ch_c$

$\Rightarrow (ah_a)^2=bh_b.ch_c$

$\Leftrightarrow a^2.h_a^2=bc.(h_bh_c)$

Mà $a^2=bc\neq 0$ nên $h_a^2=h_bh_c$

Ta có đpcm.

Lời giải:

a) Sửa đề:

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

\(=1-\frac{\sin ^3a}{\sin a+\cos a}-\frac{\cos ^3a}{\sin a+\cos a}=1-\frac{\sin ^3a+\cos ^3a}{\sin a+\cos a}=1-\frac{(\sin a+\cos a)(\sin ^2a-\sin a\cos a+\cos ^2a)}{\sin a+\cos a}\)

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

\(=\sin a\cos a\) (đpcm)

b)

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

\(=\frac{\frac{\cos ^3a-\sin ^3a}{\sin a\cos a}}{\frac{\cos a-\sin a}{\sin a\cos a}}=\frac{\cos ^3a-\sin ^3a}{\cos a-\sin a}=\frac{(\cos a-\sin a)(\cos ^2a+\sin a\cos a+\sin ^2a)}{\cos a-\sin a}\)

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

(đpcm)

Lời giải:

a) Sửa đề:

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

\(=1-\frac{\sin ^3a}{\sin a+\cos a}-\frac{\cos ^3a}{\sin a+\cos a}=1-\frac{\sin ^3a+\cos ^3a}{\sin a+\cos a}=1-\frac{(\sin a+\cos a)(\sin ^2a-\sin a\cos a+\cos ^2a)}{\sin a+\cos a}\)

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

\(=\sin a\cos a\) (đpcm)

b)

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

\(=\frac{\frac{\cos ^3a-\sin ^3a}{\sin a\cos a}}{\frac{\cos a-\sin a}{\sin a\cos a}}=\frac{\cos ^3a-\sin ^3a}{\cos a-\sin a}=\frac{(\cos a-\sin a)(\cos ^2a+\sin a\cos a+\sin ^2a)}{\cos a-\sin a}\)

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

(đpcm)

Lời giải:

a) Sửa đề:

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

\(=1-\frac{\sin ^3a}{\sin a+\cos a}-\frac{\cos ^3a}{\sin a+\cos a}=1-\frac{\sin ^3a+\cos ^3a}{\sin a+\cos a}=1-\frac{(\sin a+\cos a)(\sin ^2a-\sin a\cos a+\cos ^2a)}{\sin a+\cos a}\)

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

\(=\sin a\cos a\) (đpcm)

b)

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

\(=\frac{\frac{\cos ^3a-\sin ^3a}{\sin a\cos a}}{\frac{\cos a-\sin a}{\sin a\cos a}}=\frac{\cos ^3a-\sin ^3a}{\cos a-\sin a}=\frac{(\cos a-\sin a)(\cos ^2a+\sin a\cos a+\sin ^2a)}{\cos a-\sin a}\)

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

(đpcm)

a: \(VT=1-\dfrac{sin^2a}{1+\dfrac{cosa}{sina}}-\dfrac{cos^2a}{1+\dfrac{sina}{cosa}}\)

\(=1-sin^2a:\dfrac{sina+cosa}{sina}-cos^2a:\dfrac{cosa+sina}{cosa}\)

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

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

b: \(=\dfrac{\left(sin^2a+cos^2a\right)^2-2\cdot sin^2a\cdot cos^2a-1}{\left(sin^2a+cos^2a\right)^3-3\cdot sin^2\cdot cos^2a-1}\)

\(=\dfrac{-2\cdot\left(sina\cdot cosa\right)^2}{-3\cdot\left(sina\cdot cosa\right)^2}=\dfrac{2}{3}\)

c: \(=\dfrac{1+2cosa+cos^2a-1+2cosa-cos^2a}{1-cos^2a}\)

\(=\dfrac{4cosa}{sin^2a}=\dfrac{4cota}{sina}\)

f: \(=\dfrac{sin^2a+cos^2a+1+2cosa}{sina\left(cosa+1\right)}=\dfrac{2cosa+2}{sina\left(cosa+1\right)}=\dfrac{2}{sina}\)

bạn chỉ cần nhớ rằng: sin²x+ cos²x= 1 và cotx*tanx= 1 rồi quy đồng lên và làm bình thường

\(=\dfrac{1+cotx-sin^2x}{1+\dfrac{cosx}{sinx}}-\dfrac{cos^2x}{1+\dfrac{sinx}{cosx}}\)

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

\(=\dfrac{sinx+cosx-sin^3x}{sinx}\cdot\dfrac{sinx}{sinx+cosx}-\dfrac{cos^3x}{cosx+sinx}\)

\(=\dfrac{sinx+cosx-sin^3x-cos^3x}{sinx+cosx}\)

\(=\dfrac{\left(sinx+cosx\right)-\left(sinx+cosx\right)\left(sin^2+cos^2x-sinx\cdot cosx\right)}{sinx+cosx}\)

\(=1-1+sinx\cdot cosx=\dfrac{1}{2}sin2x\)