\(CMR:\frac{\tan^2a-\sin^2a}{\cosh^2a-\cos^2a}=\tan^6a\)
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Lời giải:
a)
\(\frac{\sin ^2a+2\cos ^2a-1}{\cot ^2a}=\frac{(\sin ^2a+\cos ^2a)+\cos ^2a-1}{\cot ^2a}=\frac{1+\cos ^2a-1}{\cot ^2a}=\frac{\cos ^2a}{\cot ^2a}=\frac{\cos ^2a}{(\frac{\cos a}{\sin a})^2}=\sin ^2a\)
b)
\(\frac{1-\sin ^2a\cos ^2a}{\cos ^2a}-\cos ^2a=\frac{1}{\cos ^2a}-\sin ^2a-\cos ^2a\)
\(=\frac{\sin ^2a+\cos ^2a}{\cos ^2a}-(\sin ^2a+\cos ^2a)=\tan ^2a+1-1=\tan ^2a\)
c)
\(\frac{\sin ^2a-\tan ^2a}{\cos ^2a-\cot ^2a}=\frac{\sin ^2a-\frac{\sin ^2a}{\cos ^2a}}{\cos ^2a-\frac{\cos ^2a}{\sin ^2a}}=\frac{\sin ^4a(\cos ^2a-1)}{\cos ^4a(\sin ^2a-1)}\)
\(=\frac{\sin ^4a(-\sin ^2a)}{\cos ^4a(-\cos ^2a)}=\frac{\sin ^6a}{\cos ^6a}=\tan ^6a\)
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\(\frac{cosa}{1+sina}+\frac{sina}{cosa}=\frac{cos^2a+sina\left(1+sina\right)}{cosa\left(1+sina\right)}=\frac{1+sina}{cosa\left(1+sina\right)}=\frac{1}{cosa}\)
\(\frac{sin^2a+cos^2a+2sina.cosa}{\left(sina-cosa\right)\left(sina+cosa\right)}=\frac{\left(sina+cosa\right)^2}{\left(sina-cosa\right)\left(sina+cosa\right)}=\frac{sina+cosa}{sina-cosa}=\frac{\frac{sina}{cosa}+1}{\frac{sina}{cosa}-1}=\frac{tana+1}{tana-1}\)
\(\left(sin^2a\right)^3+\left(cos^2a\right)^3=\left(sin^2a+cos^2a\right)^3-3sin^2a.cos^2a\left(sin^2a+cos^2a\right)\)
\(=1-3sin^2a.cos^2a\)
\(sin^2a-tan^2a=tan^4a\left(\frac{sin^2a}{tan^4a}-\frac{1}{tan^2a}\right)=tan^4a\left(sin^2a.\frac{cos^2a}{sin^2a}-\frac{1}{tan^2a}\right)\)
\(=tan^4a\left(cos^2a-cot^2a\right)\) bạn ghi sai đề câu này
\(\frac{tan^3a}{sin^2a}-\frac{1}{sina.cosa}+\frac{cot^3a}{cos^2a}=tan^3a\left(1+cot^2a\right)-\frac{1}{sina.cosa}+cot^3a\left(1+tan^2a\right)\)
\(=tan^3a+tana-\frac{1}{sina.cosa}+cot^3a+cota\)
\(=tan^3a+cot^3a+\frac{sina}{cosa}+\frac{cosa}{sina}-\frac{1}{sina.cosa}\)
\(=tan^3a+cot^3a+\frac{sin^2a+cos^2a-1}{sina.cosa}=tan^3a+cot^3a\)
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cái câu 1 kia lạ thật, phần phía trc có ngoặc thì phải nhân vs hạng tử nào đó chứ nhỉ? Và mk tính ra kq là \(-\cos^22\alpha\)
\(VT=\cos^4\alpha+\sin^4\alpha-2\cos^6\alpha-2\sin^6\alpha\)
\(=\sin^4\alpha\left(1-2\sin^2\alpha\right)-\cos^4\alpha\left(2\cos^2\alpha-1\right)\)
\(=\sin^4\alpha.\cos2\alpha-\cos^4\alpha.\cos2\alpha\)
\(=\cos2\alpha\left(\sin^2\alpha.\sin^2\alpha-\cos^4\alpha\right)\)
\(=\cos2\alpha.\left[\left(1-\cos^2\alpha\right)^2-\cos^4\alpha\right]\)
\(=\cos2\alpha.\left(1-2\cos^2\alpha\right)\)
\(=-\cos^22\alpha\)
2/ \(VT=\frac{1-\cos^2\alpha+\cos^2\alpha}{1+\sin2\alpha}=\frac{1}{1+\sin2\alpha}\)
\(VP=\frac{\frac{\sin\alpha}{\cos\alpha}-1}{\frac{\sin\alpha}{\cos\alpha}+1}=\frac{\frac{\sin\alpha-\cos\alpha}{\cos\alpha}}{\frac{\sin\alpha+\cos\alpha}{\cos\alpha}}=\frac{\sin\alpha-\cos\alpha}{\sin\alpha+\cos\alpha}\)
hmm, câu 2 có vẻ vô lí, bn thử nhân chéo lên mà xem, nó ko ra KQ = nhau đâu
1)
\((\cos^4a+\sin ^4a)-2(\cos^6a+\sin ^6a)=(\cos ^4a+\sin ^4a)-2(\cos ^2a+\sin ^2a)(\cos ^4a-\cos ^2a\sin ^2a+\sin ^4a)\)
\(=(\cos ^4a+\sin ^4a)-2(\cos ^4a-\cos ^2a\sin ^2a+\sin ^4a)\)
\(=-(\cos ^4a-2\sin ^2a\cos ^2a+\sin ^4a)=-(\cos ^2a-\sin ^2a)^2=-\cos ^22a\)
(bạn xem lại đề. Nếu thay $(\cos ^4a+\sin ^4a)$ thành $3(\cos ^4a+\sin ^4a)$ thì kết quả thu được là $(\cos ^2a+\sin ^2a)^2=1$ như yêu cầu)
2) Sửa đề:
\(\frac{\sin ^2a-\cos ^2a}{1+2\sin a\cos a}=\frac{(\sin a-\cos a)(\sin a+\cos a)}{\sin ^2a+\cos ^2a+2\sin a\cos a}=\frac{(\sin a-\cos a)(\sin a+\cos a)}{(\sin a+\cos a)^2}\)
\(=\frac{\sin a-\cos a}{\sin a+\cos a}=\frac{\frac{\sin a}{\cos a}-1}{\frac{\sin a}{\cos a}+1}=\frac{\tan a-1}{\tan a+1}\)
Bạn lưu ý viết đề bài chuẩn hơn.
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a) Vì \(\frac{\pi }{2} < a < \pi \) nên \(\cos a < 0\)
Ta có: \({\sin ^2}a + {\cos ^2}a = 1\)
\(\Leftrightarrow \frac{1}{9} + {\cos ^2}a = 1\)
\(\Leftrightarrow {\cos ^2}a = 1 - \frac{1}{9}= \frac{8}{9}\)
\(\Leftrightarrow \cos a =\pm\sqrt { \frac{8}{9}} = \pm \frac{{2\sqrt 2 }}{3}\)
Vì \(\cos a < 0\) nên \(cos a =-\frac{{2\sqrt 2 }}{3}\)
Suy ra \(\tan a = \frac{{\sin a}}{{\cos a}} = \frac{{\frac{1}{3}}}{{ - \frac{{2\sqrt 2 }}{3}}} = - \frac{{\sqrt 2 }}{4}\)
Ta có: \(\sin 2a = 2\sin a\cos a = 2.\frac{1}{3}.\left( { - \frac{{2\sqrt 2 }}{3}} \right) = - \frac{{4\sqrt 2 }}{9}\)
\(\cos 2a = 1 - 2{\sin ^2}a = 1 - \frac{2}{9} = \frac{7}{9}\)
\(\tan 2a = \frac{{2\tan a}}{{1 - {{\tan }^2}a}} = \frac{{2.\left( { - \frac{{\sqrt 2 }}{4}} \right)}}{{1 - {{\left( { - \frac{{\sqrt 2 }}{4}} \right)}^2}}} = - \frac{{4\sqrt 2 }}{7}\)
b) Vì \(\frac{\pi }{2} < a < \frac{{3\pi }}{4}\) nên \(\sin a > 0,\cos a < 0\)
\({\left( {\sin a + \cos a} \right)^2} = {\sin ^2}a + {\cos ^2}a + 2\sin a\cos a = 1 + 2\sin a\cos a = \frac{1}{4}\)
Suy ra \(\sin 2a = 2\sin a\cos a = \frac{1}{4} - 1 = - \frac{3}{4}\)
Ta có: \({\sin ^2}a + {\cos ^2}a = 1\;\)
\( \Leftrightarrow \left( {\frac{1}{2} - {\cos }a} \right)^2 + {\cos ^2}a - 1 = 0\)
\( \Leftrightarrow \frac{1}{4} - \cos a + {\cos ^2}a + {\cos ^2}a - 1 = 0\)
\( \Leftrightarrow 2{\cos ^2}a - \cos a - \frac{3}{4} = 0\)
\( \Rightarrow \cos a = \frac{{1 - \sqrt 7 }}{4}\) (Vì \(\cos a < 0)\)
\(\cos 2a = 2{\cos ^2}a - 1 = 2.{\left( {\frac{{1 - \sqrt 7 }}{4}} \right)^2} - 1 = - \frac{{\sqrt 7 }}{4}\)
\(\tan 2a = \frac{{\sin 2a}}{{\cos 2a}} = \frac{{ - \frac{3}{4}}}{{ - \frac{{\sqrt 7 }}{4}}} = \frac{{3\sqrt 7 }}{7}\)
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Đặt \(\sin^2\alpha=a;\cos^2\alpha=1\)
Theo đề, ta có: \(a^3+b^3=1-3ab\) và \(a+b=1\)
\(a^3+b^3+3ab=\left(a+b\right)^3-3ab\left(a+b\right)+3ab\)
\(=1^3-3ab+3ab=1\)
Do đó: \(a^3+b^3=1-3ab\)(đpcm)
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sin3x=sin(2x+x)=sin2xcoxx+cox2xsinx
=2sinxcox^2 x+(1-2sin^2 x)sinx
=2sinxcox^2 x+ sinx-2sin^3 x
=sinx(2cos^2 x +1) - 2sin^3 x
=sinx(2-2sin^2 x +1) - 2sin^3 x
=3sinx - 4 sin^3 x.
cos3x=cox(2x+x)=cos2xcosx-sin2xsinx
=(2cos^2 x-1)cosx-2sin^2 xcosx
=2cos^3 x-cosx-(2-cos^2 x)cosx
=2cos^3 x -cosx-2coxx+2cos^3 x
=4cos^3 x - 3cosx.
=> tan 3a= sin3a/cos3a rồi ra
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\(A=\frac{sin^2a-tan^2a}{cos^2a-cot^2a}=\frac{sin^2a-\frac{sin^2a}{cos^2a}}{cos^2a-\frac{cos^2a}{sin^2a}}=\frac{\frac{sin^2a\left(cos^2a-1\right)}{cos^2a}}{\frac{cos^2a\left(sin^2a-1\right)}{sin^2a}}=\frac{sin^4a.\left(-sin^2a\right)}{cos^4a.\left(-cos^2a\right)}=\frac{sin^6a}{cos^6a}=tan^6a\)