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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.
\(y=\frac{\cos^4a+\sin^2a-\cos^2a}{\sin^4a+\cos^2a-\sin^2a}\)
\(\Leftrightarrow y=\frac{\cos^4a+\left(1-\cos^2a\right)-\cos^2a}{\left(\sin^2a\right)^2+\cos^2a-\sin^2a}\)
\(\Leftrightarrow y=\frac{\cos^4a+1-2\cos^2a}{\left(1-\cos^2a\right)^2+\cos^2a-\left(1-\cos^2a\right)}\)
\(\Leftrightarrow y=\frac{\left(1-\cos^2a\right)^2}{1-2\cos^2a+\cos^4a+2\cos^2a-1}\)
\(\Leftrightarrow y=\frac{\left(\sin^2a\right)^2}{\cos^4a}\)
\(\Leftrightarrow y=\frac{\sin^4a}{\cos^4a}\)
\(\Leftrightarrow y=\tan^4a\)
Vậy \(y=\tan^4a\)
\(A=\frac{sina+sin3a+sin2a}{cosa+cos3a+cos2a}=\frac{2sin2a.cosa+sin2a}{2cos2a.cosa+cos2a}=\frac{sin2a\left(2cosa+1\right)}{cos2a\left(2cosa+1\right)}=\frac{sin2a}{cos2a}=tan2a\)
\(B=\frac{sin^2a\left(1+tan^2a\right)}{cos^2a\left(1+cot^2a\right)}=\frac{sin^2a.\frac{1}{cos^2a}}{cos^2a.\frac{1}{sin^2a}}=\frac{sin^4a}{cos^4a}=tan^4a\)
\(VT=\dfrac{sin^23a.cos^2a-sin^2a.cos^23a}{\left(sina.cosa\right)^2}\)
\(=\dfrac{\left(sin3a.cosa-sina.cos3a\right)\left(sin3a.cosa+sina.cos3a\right)}{\dfrac{1}{4}sin^22a}\)
\(=\dfrac{4sin2a.sin4a}{sin^22a}=\dfrac{4sin4a}{sin2a}=\dfrac{8.sin2a.cos2a}{sin2a}=8cos2a\)
\(\frac{cosa}{1+sina}+\frac{sina}{cosa}=\frac{cos^2a+sina\left(1+sina\right)}{cosa\left(1+sina\right)}=\frac{1+sina}{cosa\left(1+sina\right)}=\frac{1}{cosa}\)
\(\frac{sin^2a+cos^2a+2sina.cosa}{\left(sina-cosa\right)\left(sina+cosa\right)}=\frac{\left(sina+cosa\right)^2}{\left(sina-cosa\right)\left(sina+cosa\right)}=\frac{sina+cosa}{sina-cosa}=\frac{\frac{sina}{cosa}+1}{\frac{sina}{cosa}-1}=\frac{tana+1}{tana-1}\)
\(\left(sin^2a\right)^3+\left(cos^2a\right)^3=\left(sin^2a+cos^2a\right)^3-3sin^2a.cos^2a\left(sin^2a+cos^2a\right)\)
\(=1-3sin^2a.cos^2a\)
\(sin^2a-tan^2a=tan^4a\left(\frac{sin^2a}{tan^4a}-\frac{1}{tan^2a}\right)=tan^4a\left(sin^2a.\frac{cos^2a}{sin^2a}-\frac{1}{tan^2a}\right)\)
\(=tan^4a\left(cos^2a-cot^2a\right)\) bạn ghi sai đề câu này
\(\frac{tan^3a}{sin^2a}-\frac{1}{sina.cosa}+\frac{cot^3a}{cos^2a}=tan^3a\left(1+cot^2a\right)-\frac{1}{sina.cosa}+cot^3a\left(1+tan^2a\right)\)
\(=tan^3a+tana-\frac{1}{sina.cosa}+cot^3a+cota\)
\(=tan^3a+cot^3a+\frac{sina}{cosa}+\frac{cosa}{sina}-\frac{1}{sina.cosa}\)
\(=tan^3a+cot^3a+\frac{sin^2a+cos^2a-1}{sina.cosa}=tan^3a+cot^3a\)
Đề bài không chính xác, biểu thức này vẫn phụ thuộc a
Đề bài đúng phải là: \(\sqrt{sin^4a+4cos^2a}+\sqrt{cos^4a+4sin^2a}\)
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)
\(\frac{\sin ^2a-\tan ^2a}{\cos ^2a-\cot ^2a}=\frac{\sin ^2a-\frac{\sin ^2a}{\cos ^2a}}{\cos ^2a-\frac{\cos ^2a}{\sin ^2a}}=\frac{\sin ^4a(\cos ^2a-1)}{\cos ^4a(\sin ^2a-1)}\)
\(=\frac{\sin ^4a(-\sin ^2a)}{\cos ^4a(-\cos ^2a)}=\frac{\sin ^6a}{\cos ^6a}=\tan ^6a\)
\(\frac{2+\sin^2\cos^2}{1+\cos^2}=\frac{2+\left(1-\cos^2\right)\cos^2}{1+\cos^2}=\frac{\left(1-\cos^4\right)+\left(1+\cos^2\right)}{1+\cos^2}\)
\(=\frac{\left(1+\cos^2\right)\left(1-\cos^2+1\right)}{1+\cos^2}=1+\sin^2\)
a/ \(VT=\frac{\sin^4x+2\sin x.\cos x-\left(1-\sin^2x\right)^2}{\frac{\sin2x}{\cos2x}-1}\)
\(=\frac{\sin^4x+2\sin x.\cos x-1+2\sin^2x-\sin^4x}{\frac{\sin2x-\cos2x}{\cos2x}}\) \(=\frac{1-2\sin^2x-\sin2x}{\frac{\cos2x-\sin2x}{\cos2x}}=\frac{\cos2x-\sin2x}{\frac{\cos2x-\sin2x}{\cos2x}}=\cos2x=VP\)