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a) Cần chứng minh \(\dfrac{1-cos\alpha}{sin\alpha}=\dfrac{sin\alpha}{1+cos\alpha}\)
\(\Rightarrow sin^2\alpha=\left(1-cos\alpha\right)\left(1+cos\alpha\right)\Rightarrow sin^2\alpha=1-cos^2\alpha\)
\(\Rightarrow sin^2\alpha+cos^2\alpha=1\)
Giả sử tam giác ABC vuông tại A
Ta có: \(\left\{{}\begin{matrix}sin^2B=\dfrac{AC^2}{BC^2}\\cos^2B=\dfrac{AB^2}{BC^2}\end{matrix}\right.\Rightarrow sin^2B+cos^2B=\dfrac{AC^2+AB^2}{BC^2}=\dfrac{BC^2}{BC^2}=1\)
a)\(\dfrac{1-cosa}{sina}=\dfrac{sina}{1+cosa}\)
<=>\(\left(1-cosa\right)\left(1+cosa\right)=sin^2a\)
<=>\(1-cos^2a=sin^2a\) (lđ)
b)Ta có VT=\(\dfrac{cosa}{1+sina}+tga=\dfrac{cosa}{1+sina}+\dfrac{sina}{cosa}=\dfrac{cos^2a+sin^2a+sina}{\left(1+sina\right)cosa}=\dfrac{1+sina}{\left(1+sina\right)cosa}=\dfrac{1}{cosa}=vp\left(dpcm\right)\)
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
a) Áp dụng tính chất của tỉ số lượng giác ta có:
+) Sin2α + Cos2α=1
hay \(\left(\dfrac{1}{3}\right)^2\)+Cos2α=1
\(\dfrac{1}{9}\)+Cos2α=1
Cos2α=\(\dfrac{8}{9}\)
⇒Cos α=\(\sqrt{\dfrac{8}{9}}\)=\(\dfrac{2\sqrt{2}}{3}\)
+) \(\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}=\dfrac{\dfrac{1}{3}}{\dfrac{2\sqrt{2}}{3}}=\dfrac{\sqrt{2}}{4}\)
+)\(\cot\alpha=\dfrac{\cos\alpha}{\sin\alpha}=\dfrac{\dfrac{2\sqrt{2}}{3}}{\dfrac{1}{3}}\)=\(2\sqrt{2}\)
\(\dfrac{\left(sina+cosa\right)^2-\left(sina-cosa\right)^2}{sina.cosa}=4\\ VT=\dfrac{sin^2a+2sinacosa+cos^2a-sin^2a+2sinacosa-cos^2a}{sinacosa}\\ =\dfrac{4sinacosa}{sinacosa}=4=VP\)
a: \(S=cos^2a\left(1+tan^2a\right)=cos^2a\cdot\dfrac{1}{cos^2a}=1\)
b: \(VP=\dfrac{1+sin2a-1+sin2a}{\dfrac{1}{2}\cdot sin2a}=\dfrac{2\cdot sin2a}{\dfrac{1}{2}\cdot sin2a}=4=VT\)
Câu 1:
\(1+\cot^2a=\dfrac{1}{\sin^2a}\)
nên \(\dfrac{1}{\sin^2a}=1+5^2=26\)
\(\Leftrightarrow\sin^2a=\dfrac{1}{26}\)
\(\Leftrightarrow\sin a=\dfrac{\sqrt{26}}{26}\)
\(\cos a=\sqrt{1-\dfrac{1}{26}}=\dfrac{5\sqrt{26}}{26}\)
\(A=\dfrac{\sin a+\cos a}{\sin a-\cos a}=\left(\dfrac{\sqrt{26}+5\sqrt{26}}{26}\right):\left(\dfrac{\sqrt{26}-5\sqrt{26}}{26}\right)\)
\(=\dfrac{6\sqrt{26}}{-4\sqrt{26}}=\dfrac{-3}{2}\)
\(\frac{sin^2a-cos^2a+cos^4a}{cos^2a-sin^2a+sin^4a}=\frac{sin^2a-cos^2a\left(1-cos^2a\right)}{cos^2a-sin^2a\left(1-sin^2a\right)}=\frac{sin^2a-cos^2a.sin^2a}{cos^2a-sin^2a.cos^2a}\)
\(=\frac{sin^2a\left(1-cos^2a\right)}{cos^2a\left(1-sin^2a\right)}=\frac{sin^2a.sin^2a}{cos^2a.cos^2a}=tan^4a\)
\(sin^4a+cos^4a=\left(sin^2a+cos^2a\right)^2-sin^2a.cos^2a=1-2sin^2a.cos^2a\)
a) ta có : \(sin^2\alpha+cos^2\alpha=1\Leftrightarrow sin^2\alpha=1-cos^2\alpha\)
\(\Leftrightarrow sin^2\alpha=\left(1-cos\alpha\right)\left(1+cos\alpha\right)\Leftrightarrow\dfrac{sin\alpha}{1+cos\alpha}=\dfrac{1-cos\alpha}{sin\alpha}\left(đpcm\right)\)
b) ta có : \(tan^2\alpha-sin^2\alpha=sin^2\alpha\left(\dfrac{1}{cos^2\alpha}-1\right)=sin^2\alpha\left(\dfrac{1-cos^2\alpha}{cos^2\alpha}\right)\)
\(=sin^2\alpha.\dfrac{sin^2\alpha}{cos^2\alpha}=sin^2\alpha.tan^2\alpha\left(đpcm\right)\)
Sao ko chuyển về cái kia nó dễ hiểu hơn :v AHihi
\(sin^2a=\left(1-cosa\right)\left(1+cosa\right)\Leftrightarrow sin^2a=1-cos^2a\Leftrightarrow sin^2a+cos^2a=1\)
Lời giải: