1, a sin(B-C) + b sin(C-A) + c sin(A-B)=0 2, sin2 A + sin2 B + sin2 C = 2+2cosAcosBcosC 3, r= 4RsinA/2 sin B/2 sin C/2
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27 tháng 4 2022 lúc 1:49
1. cos 2a + cos 2b - 2 cos(a+b) cos( a-b)
2. cos2a + sin2b 1
3. cos a2 + sin b2 1
4. cos2 a + sin2 a 1
5. cos 2a cos2 a - 2 sin 2a
6. sin 2a - 2 sin a. cos a.
7. sin 2a cos2 a - sin2 a
8. sin 2a - sin 2b 2 sin ( a+b) cos ( a - b)
9. sin 2a - sin 2b 2 cos( a+b) sin ( a - b)
10. cos a2 + sin a2 1
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Đọc tiếp
1. cos 2a + cos 2b = - 2 cos(a+b) cos( a-b)
2. cos2a + sin2b = 1
3. cos a2 + sin b2= 1
4. cos2 a + sin2 a = 1
5. cos 2a = cos2 a - 2 sin 2a
6. sin 2a = - 2 sin a. cos a.
7. sin 2a = cos2 a - sin2 a
8. sin 2a - sin 2b= 2 sin ( a+b) cos ( a - b)
9. sin 2a - sin 2b= 2 cos( a+b) sin ( a - b)
10. cos a2 + sin a2 = 1
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Chung minh rang voi moi goc luong giac α lam cho bieu thuc xac dinh thi
a) \(\dfrac{1-sin2\alpha}{1+sin2\alpha}\)=cot\(^2\)(\(\dfrac{\pi}{4}\)+α) b) \(\dfrac{sin\alpha+sin\beta cos\left(\alpha+\beta\right)}{cos\alpha-sin\beta sin\left(\alpha+\beta\right)}\)=tan\(\left(\alpha+\beta\right)\).
a, \(\dfrac{1-sin2a}{1+sin2a}\)
\(=\dfrac{sin^2a+cos^2a-2sina.cosa}{sin^2a+cos^2a+2sina.cosa}\)
\(=\dfrac{\left(sina-cosa\right)^2}{\left(sina+cosa\right)^2}\)
\(=\dfrac{2sin^2\left(a-\dfrac{\pi}{4}\right)}{2sin^2\left(a+\dfrac{\pi}{4}\right)}\)
\(=\dfrac{sin^2\left(\dfrac{\pi}{4}-a\right)}{sin^2\left(a+\dfrac{\pi}{4}\right)}\)
\(=\dfrac{cos^2\left(\dfrac{\pi}{4}+a\right)}{sin^2\left(\dfrac{\pi}{4}+a\right)}=cot\left(\dfrac{\pi}{4}+a\right)\)
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b, \(\dfrac{sina+sinb.cos\left(a+b\right)}{cosa-sinb.sin\left(a+b\right)}\)
\(=\dfrac{sina+sinb.cosa.cosb-sinb.sina.sinb}{cosa-sinb.sina.cosb-sinb.cosa.sinb}\)
\(=\dfrac{sina.\left(1-sin^2b\right)+sinb.cosa.cosb}{cosa.\left(1-sin^2b\right)-sinb.sina.cosb}\)
\(=\dfrac{sina.cos^2b+sinb.cosa.cosb}{cosa.cos^2b-sinb.sina.cosb}\)
\(=\dfrac{\left(sina.cosb+sinb.cosa\right).cosb}{\left(cosa.cosb-sinb.sina\right).cosb}\)
\(=\dfrac{sin\left(a+b\right)}{cos\left(a+b\right)}=tan\left(a+b\right)\)
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giải các phương trình sau : a). sin 2x+sin2 x=1/2
b.2sin2 x +3 sin x cosx + cos2 x= 0
c.sin2 x/2 + sin x - 2 cos 2 x/2 = 1/2
Chứng minh các hệ thức sau :
a) sinalpha+sinleft(alpha+dfrac{14}{3}piright)+sinleft(alpha-dfrac{8}{3}piright)0
b) dfrac{sin4a}{1+cos4a}.dfrac{cos2a}{1+cos2a}cotleft(dfrac{3}{2}pi-aright)
c) left(cos a-cos bright)^2-left(sin a-sin bright)^2-4sin^2dfrac{a-b}{2}cosleft(a+bright)
d) sin^2left(45^0+alpharight)-sin^2left(30^0-alpharight)-sin15^0cosleft(15^0+2alpharight)sin2alpha
Đọc tiếp
Chứng minh các hệ thức sau :
a) \(\sin\alpha+\sin\left(\alpha+\dfrac{14}{3}\pi\right)+\sin\left(\alpha-\dfrac{8}{3}\pi\right)=0\)
b) \(\dfrac{\sin4a}{1+\cos4a}.\dfrac{\cos2a}{1+\cos2a}=\cot\left(\dfrac{3}{2}\pi-a\right)\)
c) \(\left(\cos a-\cos b\right)^2-\left(\sin a-\sin b\right)^2=-4\sin^2\dfrac{a-b}{2}\cos\left(a+b\right)\)
d) \(\sin^2\left(45^0+\alpha\right)-\sin^2\left(30^0-\alpha\right)-\sin15^0\cos\left(15^0+2\alpha\right)=\sin2\alpha\)
Bài 1: Rút gọn:
A= \(\dfrac{sin2\alpha+sin\alpha}{1+cos2\alpha+cos2\alpha}\)
B= \(\dfrac{4sin^2\alpha}{1-cos^2\dfrac{\alpha}{2}}\)
C= \(\dfrac{1+cos\alpha-sin\alpha}{1-cos\alpha-sin\alpha}\)
trang 42 SGK Toán 11 tập 1 - Chân trời sáng tạo
16 tháng 7 2023 lúc 20:38
Cho \(cos\alpha = \frac{1}{3}\) và \( - \frac{\pi }{2} < \alpha < 0\). Tính
\(\begin{array}{l}a)\;sin\alpha \\b)\;sin2\alpha \\c)\;cos\left( {\alpha + \frac{\pi }{3}} \right)\end{array}\)
a, Ta có: \({\sin ^2}x + co{s^2}x = 1\)
\(\begin{array}{l} \Leftrightarrow {\sin ^2}\alpha + {\left( {\frac{1}{3}} \right)^2} = 1\\ \Leftrightarrow \sin \alpha = \pm \sqrt {1 - {{\left( {\frac{1}{3}} \right)}^2}} = \pm \frac{{2\sqrt 2 }}{3}\end{array}\)
Vì \( - \frac{\pi }{2} < \alpha < 0\) nên \(sin\alpha < 0 \Rightarrow \sin \alpha = - \frac{{2\sqrt 2 }}{3}\).
\(b)\;\,sin2\alpha = 2sin\alpha .cos\alpha = 2.\left( { - \frac{{2\sqrt 2 }}{3}} \right).\frac{1}{3} = - \frac{{4\sqrt 2 }}{9}\)
\(c)\;cos(\alpha + \frac{\pi }{3}) = cos\alpha .cos\frac{\pi }{3} - sin\alpha .sin\frac{\pi }{3}\)\( = \frac{1}{3}.\frac{1}{2} - \left( { - \frac{{2\sqrt 2 }}{3}} \right).\frac{{\sqrt 3 }}{2} = \frac{{2\sqrt 6 + 1}}{6}\).
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Rút gọn các biểu thức :
a) \(\dfrac{\sin2\alpha+\sin\alpha}{1+\cos2\alpha+\cos\alpha}\)
b) \(\dfrac{4\sin^2\alpha}{1-\cos^2\dfrac{\alpha}{2}}\)
c) \(\dfrac{1+\cos\alpha-\sin\alpha}{1-\cos\alpha-\sin\alpha}\)
d) \(\dfrac{1+\sin\alpha-2\sin^2\left(45^0-\dfrac{\alpha}{2}\right)}{4\cos\dfrac{\alpha}{2}}\)
a) \(\dfrac{\sin2\text{a}+\cos a}{1+\cos2\text{a}+\cos a}=2\tan a\)
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a) \(\dfrac{sin2\alpha+sin\alpha}{1+cos2\alpha+cos\alpha}=\dfrac{2sin\alpha cos\alpha+sin\alpha}{2cos^2\alpha+cos\alpha}\)\(=\dfrac{sin\alpha\left(2cos\alpha+1\right)}{cos\alpha\left(2cos\alpha+1\right)}=\dfrac{sin\alpha}{cos\alpha}=tan\alpha\).
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b) \(\dfrac{4sin^2\alpha}{1-cos^2\dfrac{\alpha}{2}}=\dfrac{4sin^2\alpha}{sin^2\dfrac{\alpha}{2}}=\dfrac{4.sin^2\dfrac{\alpha}{2}.cos^2\dfrac{\alpha}{2}}{sin^2\dfrac{\alpha}{2}}=4sin^2\dfrac{\alpha}{2}\).
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Chứng minh rằng khi góc \(\alpha\) nhọn thì :
a) \(\sin2\alpha=2\sin\alpha\cos\alpha\)
b) \(\cos2\alpha=1-2\sin^2\alpha\)
a: \(\sin2a=\sin\left(a+a\right)\)
\(=\sin a\cdot\cos a+\cos a\cdot\sin a\)
\(=2\sin a\cdot\cos a\)
b: \(\cos2a=\cos^2a-\sin^2a\)
\(=1-\sin^2a-\sin^2a\)
\(=1-2\sin^2a\)
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Bài 11: Cho tam giác ABC vuông tại A có BC = a; CA = b; AB = c, đường cao AH. a. Chứng minh: 1 + tan2 B = 1 cos2 B ; tan C 2 = c a+b . b. Chứng minh: AH = a. sin B . cos B , BH = a. cos2 B , CH = a. sin2 B.