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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
\(\sin39^013'=0,6322\)
\(\cos52^018'=0,6115\)
\(\tan13^020'=0,2370\)
\(\cot10^017'=5,5118\)
\(\sin54^0=0,8090\)
\(\cos45^0=0,7071\)
a) ta có : \(A=\dfrac{sin33}{cos57}+\dfrac{tan32}{cot58}-2\left(sin20.cos70+cos20.sin70\right)\)
\(\Leftrightarrow A=\dfrac{sin33}{cos\left(90-33\right)}+\dfrac{tan32}{cot\left(90-32\right)}-2\left(sin20.cos\left(90-20\right)+cos20.sin\left(90-20\right)\right)\)
\(\Leftrightarrow A=\dfrac{sin33}{sin33}+\dfrac{tan32}{tan32}-2\left(sin20.sin20+cos20.cos20\right)\)\(\Leftrightarrow A=1+1-2\left(sin^220+cos^220\right)=1+1-2=0\)
b) sữa đề chút nha
ta có : \(B=\dfrac{sin^215+sin^275-sin^212-sin^278}{cos^213+cos^277+cos^21+cos^289}+\dfrac{2tan55}{cot35}\)
\(\Leftrightarrow B=\dfrac{sin^215+sin^2\left(90-15\right)-sin^212-sin^2\left(90-12\right)}{cos^213+cos^2\left(90-13\right)+cos^21+cos^2\left(90-1\right)}+\dfrac{2tan\left(90-35\right)}{cot35}\)
\(\Leftrightarrow B=\dfrac{sin^215+cos^215-sin^212-cos^212}{cos^213+sin^213+cos^21+sin^21}+\dfrac{2cot35}{cot35}\) \(\Leftrightarrow B=\dfrac{sin^215+cos^215-\left(sin^212+cos^212\right)}{cos^213+sin^213+cos^21+sin^21}+\dfrac{2cot35}{cot35}\)\(\Leftrightarrow B=\dfrac{1-1}{cos^213+sin^213+cos^21+sin^21}+2=0+2=2\)
câu 1 : ta có : \(A=\left(sin^4x+cos^4x+sin^2x.cos^2x\right)^2-\left(sin^8x+cos^8x\right)\)
\(=\left(1-sin^2x.cos^2x\right)^2-\left(1-3sin^2x.cos^2x\right)\)
\(=\left(1-sin^2x.cos^2x\right)^2-\left(1-sin^2x.cos^2x\right)+2sin^2xcos^2x\)
\(=-sin^2x.cos^2x\left(1-sin^2x.cos^2x\right)+2sin^2x.cos^2x\)
\(=sin^2x.cos^2x\left(1+sin^2x.cos^2x\right)\)
tới đây mk xin sử dụng kiến thức lớp 10 một chút
\(=\dfrac{sin^22x}{4}\left(1+\dfrac{sin^22x}{4}\right)=\dfrac{sin^22x}{4}+\dfrac{sin^42x}{16}\)
vẩn phụ thuộc vào x \(\Rightarrow\) đề sai .
câu 1 : câu này bn có thể tìm trong trang của mk , mk nhớ đã làm nó rồi nhưng tìm hoài không đc . nếu đc bn có thể chờ mk đi hok về mk sẽ kiếm cho bn hoắc có thể là lm lại cho bn nha :)
câu 2 : https://hoc24.vn/hoi-dap/question/657072.html
câu 3 : https://hoc24.vn/hoi-dap/question/657069.html
câu 4 : https://hoc24.vn/hoi-dap/question/656635.html
câu 5 : https://hoc24.vn/hoi-dap/question/657071.html
Câu 1:
Ta có: \(\cos\left(90^0-\alpha\right)=\sin\alpha\)
\(\Leftrightarrow\sin\alpha=1:\sqrt{\dfrac{1^2+2^2}{1}}=1:\sqrt{5}=\dfrac{\sqrt{5}}{5}\)
Câu 2:
a) \(\cos\alpha=\sqrt{1-\sin^2\alpha}=\sqrt{1-\dfrac{16}{25}}=\dfrac{3}{5}\)
\(\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}=\dfrac{4}{5}:\dfrac{3}{5}=\dfrac{4}{3}\)
ĐKXĐ: x>=0; x<>1
\(B=\dfrac{\left(\sqrt{x}+1\right)^2+\sqrt{x}\left(\sqrt{x}-1\right)+\sqrt{x}}{x-1}:\dfrac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2}{x-1}\)
\(=\dfrac{x+2\sqrt{x}+1+x-\sqrt{x}+\sqrt{x}}{x-1}\cdot\dfrac{x-1}{x+2\sqrt{x}+1-x+2\sqrt{x}-1}\)
\(=\dfrac{2x+2\sqrt{x}+1}{4\sqrt{x}}\)
Khi \(x=\dfrac{2-\sqrt{3}}{2}=\dfrac{4-2\sqrt{3}}{4}=\left(\dfrac{\sqrt{3}-1}{2}\right)^2\) thì:
\(B=\dfrac{2\cdot\dfrac{2-\sqrt{3}}{2}+2\cdot\dfrac{\sqrt{3}-1}{2}+1}{4\cdot\dfrac{\sqrt{3}-1}{2}}\)
\(=\dfrac{2-\sqrt{3}+\sqrt{3}-1+1}{2\left(\sqrt{3}-1\right)}=\dfrac{2}{2\left(\sqrt{3}-1\right)}=\dfrac{1}{\sqrt{3}-1}=\dfrac{\sqrt{3}+1}{2}\)
ĐKXĐ: \(abc\ne0\)
\(a^3+b^3+3ab\left(a+b\right)+c^3-3ab\left(a+b\right)-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2+2ab-ac-bc\right)-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)
TH1: \(a+b+c=0\)
\(P=\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\dfrac{\left(-c\right)\left(-a\right)\left(-b\right)}{abc}=-1\)
TH2: \(a=b=c\Rightarrow P=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)
Bạn kiểm tra lại đề, \(f\left(x\right)=\dfrac{x^3}{1-3x-3x^2}\) hay \(f\left(x\right)=\dfrac{x^3}{1-3x+3x^2}\)
Đề sai em
Đề đúng: \(\dfrac{\left(sina+cosa\right)^2-\left(sina-cosa\right)^2}{sina.cosa}=4\)
a: \(A\simeq0.1787\)
b: \(B\simeq0.2582\)