Rút gọn :
\(A=\frac{1+2\sin\alpha\cos\alpha}{\sin\alpha+\cos\alpha}\)
Những câu hỏi liên quan
rút gọn biểu thức sau:
b, \(\frac{\left(\cos\alpha-\sin\alpha\right)^2-\left(\cos\alpha-\sin^2\alpha\right)}{\cos\alpha.\sin\alpha}\)
c,\(C=\sin^6\alpha+\cos^6\alpha+3\sin^6\alpha.\cos^2\alpha\)
trang 23 SGK Toán 11 tập 1 - Chân trời sáng tạo
16 tháng 7 2023 lúc 17:04
Rút gọn các biểu thức sau:
a, \(\sqrt 2 \sin \left( {\alpha + \frac{\pi }{4}} \right) - cos\alpha \),
b, \({\left( {cos\alpha + \sin \alpha } \right)^2} - \sin 2\alpha \)
\(a,\sqrt{2}sin\left(\alpha+\dfrac{\pi}{4}\right)-cos\alpha\\ =\sqrt{2}\left(sin\alpha cos\dfrac{\pi}{4}+cos\alpha sin\dfrac{\pi}{4}\right)-cos\alpha\\ =\sqrt{2}\left(sin\alpha\cdot\dfrac{\sqrt{2}}{2}+cos\alpha\cdot\dfrac{\sqrt{2}}{2}\right)-cos\alpha\\ =\sqrt{2}\cdot sin\alpha\cdot\dfrac{\sqrt{2}}{2}+\sqrt{2}\cdot cos\alpha\cdot\dfrac{\sqrt{2}}{2}-cos\alpha\\ =sin\alpha+cos\alpha-cos\alpha\\ =sin\alpha\)
\(b,\left(cos\alpha+sin\alpha\right)^2-sin2\alpha\\ =cos^2\alpha+sin^2\alpha=2cos\alpha sin\alpha-2sin\alpha cos\alpha\\ =sin^2\alpha+cos^2\alpha\\ =1\)
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25 tháng 7 2023 lúc 11:00
Cho góc nhọn α
a) Rút gọn biểu thức S=\(\cos^2\alpha+tg^2.\cos^2\alpha\)
b) Chứng minh:
\(\dfrac{\left(\sin\alpha+\cos\alpha\right)^2-\left(\sin\alpha-\cos\alpha\right)^2}{\sin\alpha.\cos\alpha}=4\)
Help me plsssssssssss
\(\dfrac{\left(sina+cosa\right)^2-\left(sina-cosa\right)^2}{sina.cosa}=4\\ VT=\dfrac{sin^2a+2sinacosa+cos^2a-sin^2a+2sinacosa-cos^2a}{sinacosa}\\ =\dfrac{4sinacosa}{sinacosa}=4=VP\)
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a: \(S=cos^2a\left(1+tan^2a\right)=cos^2a\cdot\dfrac{1}{cos^2a}=1\)
b: \(VP=\dfrac{1+sin2a-1+sin2a}{\dfrac{1}{2}\cdot sin2a}=\dfrac{2\cdot sin2a}{\dfrac{1}{2}\cdot sin2a}=4=VT\)
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a) S= \(cos^2a\left(tg^2a+1\right)=cos^2a.\dfrac{1}{cos^2a}=1\)
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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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RÚT GỌN : \(\frac{1+2sin\alpha+cos\alpha}{cos^2\alpha-sin^2\alpha}\)
2) Rút gọna)1-sin^22b)left(1-cosalpharight)left(1+cosalpharight)c)1+sin^2alpha+cos^2alphad)sinalpha-sinalpha.cos^2alphae)sin^2alpha+cos^2alpha+2sin^2alpha.cos^2alphaf)tan^2alpha-sin^2alpha.tan^2alphag)cos^2alpha+tan^2alpha.cos^2alphah)tan^2alphaleft(2cos^2alpha+sin^2alpha-1right)
Đọc tiếp
2) Rút gọn
a)\(1-\sin^22\)
b)\(\left(1-\cos\alpha\right)\left(1+\cos\alpha\right)\)
c)\(1+\sin^2\alpha+\cos^2\alpha\)
d)\(\sin\alpha-\sin\alpha.\cos^2\alpha\)
e)\(\sin^2\alpha+\cos^2\alpha+2\sin^2\alpha.\cos^2\alpha\)
f)\(\tan^2\alpha-\sin^2\alpha.\tan^2\alpha\)
g)\(\cos^2\alpha+\tan^2\alpha.\cos^2\alpha\)
h)\(\tan^2\alpha\left(2\cos^2\alpha+\sin^2\alpha-1\right)\)
\(a,1-sin^2\alpha=cos^2\alpha\)
\(b,\left(1-cos\alpha\right)\left(1+cos\alpha\right)=1-cos^2\alpha=sin^2\alpha\)
\(c,1+sin^2\alpha+cos^2\alpha=1+1=2\)
\(d,sin\alpha-sin\alpha.cos^2\alpha=sin\alpha.\left(1-cos^2\alpha\right)=sin\alpha.sin^2\alpha=sin^3\alpha\)
\(e,sin^2\alpha+cos^2\alpha+2sin^2\alpha.cos^2\alpha\)
\(=1+2sin^2\alpha.cos^2\alpha\)
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rút gọn:
a/\(\cos^2\alpha\left(1+\cos^2\alpha\right)\) b/\(\frac{1+\sin\alpha}{1-\sin\alpha}-\frac{1-\sin\alpha}{1+\sin\alpha}\)
Biết tan α=3. Tính giá trị các biểu thức sau:
a)\(\frac{\sin\alpha-\cos\alpha}{\sin\alpha+\cos\alpha}\)
b)\(\frac{2\sin\alpha+3\cos\alpha}{3\sin\alpha-5\cos\alpha}\)
c)\(\frac{1+2\cos^2\alpha}{\sin^2\alpha-\cos^2\alpha}\)
d)\(\frac{\sin^4\alpha+\cos^4\alpha}{1+\sin^2\alpha}\)
\(\frac{1}{cos^2a}=1+tan^2a\Rightarrow cos^2a=\frac{1}{1+tan^2a}=\frac{1}{10}\)
a/ \(\frac{sina-cosa}{sina+cosa}=\frac{\frac{sina}{cosa}-\frac{cosa}{cosa}}{\frac{sina}{cosa}+\frac{cosa}{cosa}}=\frac{tana-1}{tana+1}=\frac{3-1}{3+1}\)
b/ \(\frac{2sina+3cosa}{3sina-5cosa}=\frac{3tana+3}{3tana-5}=\frac{3.3+3}{3.3-5}\)
c/ \(\frac{1+2cos^2a}{1-cos^2a-cos^2a}=\frac{1+2cos^2a}{1-2cos^2a}=\frac{1+2.\frac{1}{10}}{1-2.\frac{1}{10}}\)
d/ \(\frac{\left(1-cos^2a\right)^2+\left(cos^2a\right)^2}{1+1-cos^2a}=\frac{\left(1-\frac{1}{10}\right)^2+\left(\frac{1}{10}\right)^2}{2-\frac{1}{10}}\)
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Rút gọn:
\(\dfrac{2\cos^2\alpha-1}{\sin\alpha+\cos\alpha}\)
\(=\dfrac{2cos^2a-\left(sin^2a+cos^2a\right)}{sina+cosa}=\dfrac{cos^2a-sin^2a}{sina+cosa}=\dfrac{\left(cosa-sina\right)\left(cosa+sina\right)}{sina+cosa}=cosa-sina\)
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\(\dfrac{2\cos^2\alpha-1}{\sin\alpha+\cos\alpha}\)
\(=\dfrac{2\cos^2\alpha-\sin^2\alpha-\cos^2\alpha}{\cos\alpha+\sin\alpha}\)
\(=\cos\alpha-\sin\alpha\)
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