Tính
P= sin 30độ - sin40 độ - sin50 độ + sin 60 độ
Q= \(c\text{os}^225-c\text{os}^235+c\text{os}^245-c\text{os}^255+c\text{os}^260\)
M= sin^2 10độ + sin^2 20độ + sin^2 30độ + ....+sin^2 80độ
Giải giúp mình
Những câu hỏi liên quan
chứng minh rằng:
a)\(\frac{c\text{os}a.cot\text{a}-sin\text{a}.t\text{ana}}{\frac{1}{sin\text{a}}-\frac{1}{c\text{os}a}}=1+sin\text{a}.c\text{os}a\)
b)\(\frac{c\text{os}a+sin\text{a}-1}{c\text{os}a-sin\text{a}+1}=\frac{sin\text{a}}{1+c\text{os}a}\)
c)\(\frac{sin\text{a}}{1+c\text{os}a}+\frac{1+c\text{os}a}{sin\text{a}}=\frac{2}{sin\text{a}}\)
chứng minh rằng
a)
frac{sintext{a}}{1+ctext{os}a}+cottext{a}frac{1}{sintext{a}}
b)frac{1}{ctext{os}a}-frac{ctext{os}a}{1+sintext{a}}ttext{ana}
c) frac{ttext{ana}-sintext{a}}{sin^3a}frac{1}{ctext{os}aleft(1+ctext{os}aright)}
d) frac{sintext{a}+ctext{os}a-1}{sintext{a}-ctext{os}a+1}frac{ctext{os}a}{1+sintext{a}}
Đọc tiếp
chứng minh rằng
a)
\(\frac{sin\text{a}}{1+c\text{os}a}+cot\text{a}=\frac{1}{sin\text{a}}\)
b)\(\frac{1}{c\text{os}a}-\frac{c\text{os}a}{1+sin\text{a}}=t\text{ana}\)
c) \(\frac{t\text{ana}-sin\text{a}}{sin^3a}=\frac{1}{c\text{os}a\left(1+c\text{os}a\right)}\)
d) \(\frac{sin\text{a}+c\text{os}a-1}{sin\text{a}-c\text{os}a+1}=\frac{c\text{os}a}{1+sin\text{a}}\)
Lời giải:
a)
\(\frac{\sin a}{1+\cos a}+\cot a=\frac{\sin a}{1+\cos a}+\frac{\cos a}{\sin a}=\frac{\sin ^2a+\cos^2a+\cos a}{\sin a(1+\cos a)}\)
\(=\frac{1+\cos a}{\sin a(1+\cos a)}=\frac{1}{\sin a}\) (đpcm)
b)
\(\frac{1}{\cos a}-\frac{\cos a}{1+\sin a}=\frac{1+\sin a-\cos ^2a}{\cos a(1+\sin a)}=\frac{(1-\cos ^2a)+\sin a}{\cos a(\sin a+1)}\)
\(=\frac{\sin^2a+\sin a}{\cos a(\sin a+1)}=\frac{\sin a(\sin a+1)}{\cos a(\sin a+1)}=\frac{\sin a}{\cos a}=\tan a\) (đpcm)
c)
\(\frac{\tan a-\sin a}{\sin ^3a}=\frac{\frac{\sin a}{\cos a}-\sin a}{\sin ^3a}=\frac{\frac{1}{\cos a}-1}{\sin ^2a}=\frac{1-\cos a}{\cos a\sin ^2a}=\frac{1-\cos a}{\cos a(1-\cos ^2a)}=\frac{1}{\cos a(1+\cos a)}\)
d)
\(\frac{\sin a+\cos a-1}{\sin a-\cos a+1}=\frac{(\sin a+\cos a-1)(\sin a+\cos a+1)}{(\sin a-\cos a+1)(\sin a+\cos a+1)}=\frac{(\sin a+\cos a)^2-1}{(\sin a+1)^2-\cos ^2a}\)
\(=\frac{\sin ^2a+\cos ^2a+2\sin a\cos a-1}{\sin ^2a+1+2\sin a-\cos ^2a}=\frac{1+2\sin a\cos a-1}{\sin ^2a+1+2\sin a-(1-\sin ^2a)}\)
\(=\frac{2\sin a\cos a}{2\sin ^2a+2\sin a}=\frac{2\sin a\cos a}{2\sin a(\sin a+1)}=\frac{\cos a}{1+\sin a}\) (đpcm)
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Mấu chốt trong các bài này là việc sử dụng công thức $\sin ^2a+\cos ^2a=1$
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Lời giải:
a)
\(\frac{\sin a}{1+\cos a}+\cot a=\frac{\sin a}{1+\cos a}+\frac{\cos a}{\sin a}=\frac{\sin ^2a+\cos^2a+\cos a}{\sin a(1+\cos a)}\)
\(=\frac{1+\cos a}{\sin a(1+\cos a)}=\frac{1}{\sin a}\) (đpcm)
b)
\(\frac{1}{\cos a}-\frac{\cos a}{1+\sin a}=\frac{1+\sin a-\cos ^2a}{\cos a(1+\sin a)}=\frac{(1-\cos ^2a)+\sin a}{\cos a(\sin a+1)}\)
\(=\frac{\sin^2a+\sin a}{\cos a(\sin a+1)}=\frac{\sin a(\sin a+1)}{\cos a(\sin a+1)}=\frac{\sin a}{\cos a}=\tan a\) (đpcm)
c)
\(\frac{\tan a-\sin a}{\sin ^3a}=\frac{\frac{\sin a}{\cos a}-\sin a}{\sin ^3a}=\frac{\frac{1}{\cos a}-1}{\sin ^2a}=\frac{1-\cos a}{\cos a\sin ^2a}=\frac{1-\cos a}{\cos a(1-\cos ^2a)}=\frac{1}{\cos a(1+\cos a)}\)
d)
\(\frac{\sin a+\cos a-1}{\sin a-\cos a+1}=\frac{(\sin a+\cos a-1)(\sin a+\cos a+1)}{(\sin a-\cos a+1)(\sin a+\cos a+1)}=\frac{(\sin a+\cos a)^2-1}{(\sin a+1)^2-\cos ^2a}\)
\(=\frac{\sin ^2a+\cos ^2a+2\sin a\cos a-1}{\sin ^2a+1+2\sin a-\cos ^2a}=\frac{1+2\sin a\cos a-1}{\sin ^2a+1+2\sin a-(1-\sin ^2a)}\)
\(=\frac{2\sin a\cos a}{2\sin ^2a+2\sin a}=\frac{2\sin a\cos a}{2\sin a(\sin a+1)}=\frac{\cos a}{1+\sin a}\) (đpcm)
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\(sin^3\alpha+c\text{os}^3\alpha+\left(sin\alpha+c\text{os}\alpha\right).sin\alpha.c\text{os}\alpha-c\text{os}\alpha.\)
=(sin a+cos a)(sin^2.a-sina.cosa+cos^2a)+(sina+cosa)sina.cosa-cos a
=(sin a+cos a)(1-sina.cosa+sina.cosa)-cosa
=sina+cosa-cosa
=sina
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Thu gọn các biểu thức sau:
a. \(sin^6a+c\text{os}^6a+3sin^2a.c\text{os}^2a\)
b.\(sin^4a-c\text{os}^4a-\left(sina+c\text{os}a\right)\left(sina-c\text{os}a\right)\)
c.\(c\text{os}^2a+tan^2a.c\text{os}^2a\)
d.\(c\text{os}^2a+tan^2a.c\text{os}^2a\)
a) \(sin^6x+cos^6x+3sin^2x.cos^2x\)
\(=\left(sin^2x+cos^2x\right)\left(sin^4x-sin^2x.cox^2x+cos^4x\right)+3sin^2x.cos^2x\)
\(=sin^4x-sin^2x.cox^2x+cos^4x+3sin^2x.cos^2x\)
\(=sin^4x+2sin^2x.cox^2x+cos^4x=\left(sin^2x+cos^2x\right)^2=1\text{}\text{}\)
b) \(sin^4x-cos^4x-\left(sinx+cosx\right)\left(sinx-cosx\right)\)
\(=\left(sin^2x+cos^2x\right)\left(sin^2x-cos^2x\right)-\left(sin^2x-cos^2x\right)\)
\(=1\left(sin^2x-cos^2x\right)-\left(sin^2x-cos^2x\right)=0\)
c) \(cos^2x+tan^2x.cos^2x\)
\(=cos^2x+\dfrac{sin^2x}{cos^2x}.cos^2x=sin^2x+cos^2x=1\)
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Chứng minh đạo hàm của y không phụ thuộc vào x:
y=\(\dfrac{Sin^6x+C\text{os}^6x-1}{Sin^4x+C\text{os}^4x-1}\)
\(y=\dfrac{\left(sin^2x+cos^2x\right)^2-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)-1}{\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x-1}\)
\(=\dfrac{1-3sin^2x.cos^2x-1}{1-2sin^2x.cos^2x-1}=\dfrac{3}{2}\) ko phụ thuộc x
Nên \(y'=0\) không phụ thuộc x
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\(sin^4x+c\text{os}^4x=\frac{3}{4}+\frac{1}{4}c\text{os}4x\)
\(sin^4x+cos^4x=\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x\)
\(=1-\frac{1}{2}\left(2sinx.cosx\right)^2=1-\frac{1}{2}sin^22x\)
\(=1-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2}cos4x\right)=\frac{3}{4}+\frac{1}{4}cos4x\)
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\(sin^6a+c\text{os}^6a+3\text{s}in^2a.c\text{os}^2a=\)
Cho
1.tan α=\(\dfrac{1}{3} \) tính A=\(\dfrac{2\sin^2x+5}{4\cos^2x-3}\)
2.cot α=\(\dfrac{2}{5}\) tính B=\(\dfrac{3\cos^2x-\sin^2x}{c\text{os}^2x+2\sin^2x}\)
\(A=\dfrac{2tan^2a+\dfrac{5}{cos^2a}}{4-\dfrac{3}{cos^2a}}=\dfrac{2tan^2a+5\left(1+tan^2a\right)}{4-3\left(1+tan^2a\right)}=...\) (bạn tự thay số bấm máy nhé)
\(B=\dfrac{3cot^2a-1}{cot^2a+2}=...\)
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Cho góc nhọn \(\alpha\).Rút gọn biểu thức: \(\sin^6\alpha+c\text{os}^6\alpha+3sin^2\alpha-c\text{os}^2\alpha\)