tính giá trị biểu thức
\(A=\cos^{^2}\alpha-tg60^0+cotg45^0-2\sin30^0+\cos^220^0tg^0\alpha\)
Mình không biết viết độ ở đâu nên ghi mũ 0 là độ nhé
Rút gọn các biểu thức (không dùng bảng số và máy tính)
a) \(\sin^2\left(180^0-\alpha\right)+\tan^2\left(180^0-\alpha\right).\tan^2\left(270^0+\alpha\right)+\sin\left(90^0+\alpha\right)\cos\left(\alpha-360^0\right)\)
b) \(\dfrac{\cos\left(\alpha-180^0\right)}{\sin\left(180^0-\alpha\right)}+\dfrac{\tan\left(\alpha-180^0\right)\cos\left(180^0+\alpha\right)\sin\left(270^0+\alpha\right)}{\tan\left(270^0+\alpha\right)}\)
c) \(\dfrac{\cos\left(-288^0\right)\cot72^0}{\tan\left(-162^0\right)\sin108^0}-\tan18^0\)
d) \(\dfrac{\sin20^0\sin30^0\sin40^0\sin50^0\sin60^0\sin70^0}{\cos10^0\cos50^0}\)
a)\(sin^2\left(180^o-\alpha\right)+tan^2\left(180-\alpha\right).tan^2\left(270^o+\alpha\right)\)\(+sin\left(90^o+\alpha\right)cos\left(\alpha-360^o\right)\)
\(=sin^2\alpha+tan^2\alpha.cot^2\alpha+cos\alpha cos\alpha\)
\(=sin^2\alpha+cos^2\alpha+\left(tan\alpha cot\alpha\right)^2=1+1=2\).
\(\dfrac{cos\left(\alpha-180^o\right)}{sin\left(180^o-\alpha\right)}+\dfrac{tan\left(\alpha-180^o\right)cos\left(180^o+\alpha\right)sin\left(270^o+\alpha\right)}{tan\left(270^o+\alpha\right)}\)
\(=\dfrac{cos\left(180^o-\alpha\right)}{sin\left(180^o-\alpha\right)}+\dfrac{-tan\left(180^o-\alpha\right).cos\alpha.sin\left(90^o+\alpha\right)}{-tan\left(90^o+\alpha\right)}\)
\(=tan\left(180^o-\alpha\right)+\dfrac{tan\alpha.cos\alpha.cos\alpha}{cot\alpha}\)
\(=-tan\alpha+tan^2\alpha cos^2\alpha\)
\(=tan\alpha\left(-1+tan\alpha cos^2\alpha\right)\)
\(=tan\alpha\left(sin\alpha cos\alpha-1\right)\).
c) \(\dfrac{cos\left(-288^o\right)cot72^o}{tan\left(-162^o\right)sin108^o}-tan18^o\)
\(=\dfrac{cos72^ocot72^o}{tan18^o.sin72^o}-tan18^o\)
\(=\dfrac{cos^272^o.cos18^o}{sin72^osin18^o.sin72^o}-tan18^o\)
\(=cot^272^ocot18^o-tan18^o\)
\(=tan^218^ocot18^o-tan18^o\)
\(=tan18^o-tan18^o=0\).
Cho góc \(\alpha \;\;({0^o} < \alpha < {180^o})\) thỏa mãn \(\tan \alpha = 3\)
Tính giá trị biểu thức: \(P = \frac{{2\sin \alpha - 3\cos \alpha }}{{3\sin \alpha + 2\cos \alpha }}\)
\(P=\dfrac{2sin\alpha-3cos\alpha}{3sin\alpha+2cos\alpha}\\ =\dfrac{\dfrac{2sin\alpha}{cos\alpha}-\dfrac{3cos\alpha}{cos\alpha}}{\dfrac{3sin\alpha}{cos\alpha}+\dfrac{2cos\alpha}{cos\alpha}}\\ =\dfrac{2tan\alpha-3}{3tan\alpha+2}=\dfrac{2.3-3}{3.3+2}=\dfrac{3}{11}\)
Ta có: \(1 + {\tan ^2}\alpha = \frac{1}{{{{\cos }^2}\alpha }}\quad (\alpha \ne {90^o})\)
\( \Rightarrow \frac{1}{{{{\cos }^2}\alpha }} = 1 + {3^2} = 10\)
\( \Leftrightarrow {\cos ^2}\alpha = \frac{1}{{10}} \Leftrightarrow \cos \alpha = \pm \frac{{\sqrt {10} }}{{10}}\)
Vì \({0^o} < \alpha < {180^o}\) nên \(\sin \alpha > 0\).
Mà \(\tan \alpha = 3 > 0 \Rightarrow \cos \alpha > 0 \Rightarrow \cos \alpha = \frac{{\sqrt {10} }}{{10}}\)
Lại có: \(\sin \alpha = \cos \alpha .\tan \alpha = \frac{{\sqrt {10} }}{{10}}.3 = \frac{{3\sqrt {10} }}{{10}}.\)
\( \Rightarrow P = \dfrac{{2.\frac{{3\sqrt {10} }}{{10}} - 3.\frac{{\sqrt {10} }}{{10}}}}{{3.\frac{{3\sqrt {10} }}{{10}} + 2.\frac{{\sqrt {10} }}{{10}}}} = \dfrac{{\frac{{\sqrt {10} }}{{10}}\left( {2.3 - 3} \right)}}{{\frac{{\sqrt {10} }}{{10}}\left( {3.3 + 2} \right)}} = \dfrac{3}{{11}}.\)
Mấy bạn giải giúp mình nha
Cho tan\(\alpha\)=2 và \(-\pi< \alpha< 0\).Tính giá trị biểu thức \(P=4\sin2\alpha-\cos\alpha\)
Cho \(0< \alpha\text{ ≤}\frac{\pi}{2}\)và \(\cos^2\alpha=\cos\alpha\). Tính giá trị biểu thức \(A=cot\frac{\alpha}{8}\)
Cho\(cos4x+sin2x=1\). Tính giá trị biểu thức \(B=tan^22x+cot^22x\)
Cho \(\sin\alpha+\cos\alpha=\frac{\sqrt{6}}{2},a\in\left(0;\frac{\pi}{4}\right)\)
Tính giá trị biểu thức: \(P=\cos\left(\alpha+\frac{\pi}{4}\right)+\sqrt{2\left(1-\sin\alpha\cos\alpha+\sin\alpha-\cos\alpha\right)}\)
a) Biến đổi \(\sin\alpha-1\)thành tích
b) Rút gọn biểu thức \(P=\dfrac{\cos\alpha+2\cos3\alpha+\cos5a}{\sin\alpha+2\sin3\alpha+\sin5a}\)
c) Tính giá trị biểu thức \(P=\sin30.\cos60+\sin60.\cos30\)
d) Giá đúng của \(cos\dfrac{2\pi}{7}+\cos\dfrac{4\pi}{7}+\cos\dfrac{6\pi}{7}\)
e) Giá trị đúng của \(\tan\dfrac{\pi}{24}+\tan\dfrac{7\pi}{24}\)
a/\(sina-1=2sin\dfrac{a}{2}.cos\dfrac{a}{2}-sin^2\dfrac{a}{2}-cos^2\dfrac{a}{2}=-\left(sin\dfrac{a}{2}-cos\dfrac{a}{2}\right)^2\)
b/\(P=\dfrac{cosa+cos5a+2cos3a}{sina+sin5a+2sin3a}=\dfrac{2cos3a.cos2a+2cos3a}{2sin3a.cos2a+2sin3a}=\dfrac{2cos3a\left(cos2a+1\right)}{2sin3a\left(cos2a+1\right)}=cot3a\)
c/\(P=sin\left(30+60\right)=sin90=1\)
d/
\(A=cos\dfrac{2\pi}{7}+cos\dfrac{6\pi}{7}+cos\dfrac{4\pi}{7}\Rightarrow A.sin\dfrac{\pi}{7}=sin\dfrac{\pi}{7}.cos\dfrac{2\pi}{7}+sin\dfrac{\pi}{7}cos\dfrac{4\pi}{7}+sin\dfrac{\pi}{7}.cos\dfrac{6\pi}{7}\)
\(=\dfrac{1}{2}sin\dfrac{3\pi}{7}-\dfrac{1}{2}sin\dfrac{\pi}{7}+\dfrac{1}{2}sin\dfrac{5\pi}{7}-\dfrac{1}{2}sin\dfrac{3\pi}{7}+\dfrac{1}{2}sin\dfrac{7\pi}{7}-\dfrac{1}{2}sin\dfrac{5\pi}{7}\)
\(=-\dfrac{1}{2}sin\dfrac{\pi}{7}\Rightarrow A=-\dfrac{1}{2}\)
e/
\(tan\dfrac{\pi}{24}+tan\dfrac{7\pi}{24}=\dfrac{sin\dfrac{\pi}{24}}{cos\dfrac{\pi}{24}}+\dfrac{sin\dfrac{7\pi}{24}}{cos\dfrac{7\pi}{24}}=\dfrac{sin\dfrac{\pi}{24}cos\dfrac{7\pi}{24}+sin\dfrac{7\pi}{24}cos\dfrac{\pi}{24}}{cos\dfrac{\pi}{24}.cos\dfrac{7\pi}{24}}\)
\(=\dfrac{sin\left(\dfrac{\pi}{24}+\dfrac{7\pi}{24}\right)}{\dfrac{1}{2}cos\dfrac{\pi}{4}+\dfrac{1}{2}cos\dfrac{\pi}{3}}=\dfrac{2sin\dfrac{\pi}{3}}{cos\dfrac{\pi}{4}+cos\dfrac{\pi}{3}}=\dfrac{\sqrt{3}}{\dfrac{\sqrt{2}}{2}+\dfrac{1}{2}}=\dfrac{2\sqrt{3}}{\sqrt{2}+1}\)
sina - 1 = sina - sin\(\dfrac{\pi}{2}\)
Cho biểu thức A= 1-2sinα.cosα/sin2α - cos2α với α ≠ 450
a) Chứng minh A = sinα - cosα / sinα + cosα
b) Tính giá trị của biểu thức A biết tanα = 1/3
\(A=\frac{1-2sina.cosa}{sin^2a-cos^2a}=\frac{sin^2a+cos^2a-2sina.cosa}{\left(sina-cosa\right)\left(sina+cosa\right)}=\frac{\left(sina-cosa\right)^2}{\left(sina-cosa\right)\left(sina+cosa\right)}=\frac{sina-cosa}{sina+cosa}\)
b/ \(A=\frac{\frac{sina}{cosa}-\frac{cosa}{cosa}}{\frac{sina}{cosa}+\frac{cosa}{cosa}}=\frac{tana-1}{tana+1}=\frac{\frac{1}{3}-1}{\frac{1}{3}+1}=-\frac{1}{2}\)
00< góc alpha < 900 . CMR : P=sin6alpha + cos6alpha + 3sin2alpha + cos2alpha không phụ vào giá trị góc alpha
để mình làm cho
\(P=\sin^6_a+\cos^6_a+3\sin_a^2+\cos^2_a=\left(\sin^2_a+\cos^2_a\right)\left(\sin^4_a-\sin^2_a\cos^2_a+\cos^4_a\right)\) \(+3.\sin^2_a.\cos^2_a\)
\(=\sin^4_a+2\sin^2_a.\cos^2_a+\cos^4_a=\left(\sin^2_a+\cos^2_a\right)^2=1\)
đề đoạn cuối phải là nhân chứ không phải +
Rút gọn
\(1,D=cos^220^0+cos^230^0+cos^240^0+cos^250^0+cos^260^0+cos^270^0\)
\(2,E=sin^25^0+sin^225^0+sin^245^0+sin^265^0+sin^285^0\)
\(3,F=sin^6\alpha+cos^6\alpha+3sin^2\alpha.cos^2\alpha\)
Bài 1 :
\(D=cos^220^0+cos^230^0+cos^240^0+cos^250^0+cos^260^0+cos^270^0\)
\(=\left(cos^220^0+cos^270^0\right)+\left(cos^230^0+cos^260^0\right)+\left(cos^240^0+cos^250^0\right)\)
\(=1+1+1=3\)
Bài 2 :
\(E=sin^25^0+sin^225^0+sin^245^0+sin^265^0+sin^285^0\)
\(=\left(sin^25^0+sin^285^0\right)+\left(sin^225^0+sin^265^0\right)+sin^245^0\)
\(=1+1+\dfrac{1}{2}=\dfrac{5}{2}\)
Bài 3 :
\(F=sin^6\alpha+cos^6\alpha+3sin^2\alpha.cos^2\alpha\)
\(=1-3sin^2\alpha.cos^2\alpha+3sin^2a.cos^2\alpha\)
\(=1\)
Không dùng bảng số và máy tính, rút gọn các biểu thức :
a) \(A=\tan18^0\tan288^0+\sin32^0\sin148^0-\sin302^0\sin122^0\)
b) \(B=\dfrac{1+\sin^4\alpha-\cos^4\alpha}{1-\sin^6\alpha-\cos^6\alpha}\)
\(A=tan18^otan288+sin32^osin148^o-sin302^osin122^o\)
\(=tan18^o.tan\left(-72^o\right)+sin32^o.sin32^o+sin58^o.sin58^o\)
\(=-tan18^o.cot18^o+sin^232^o+sin^258^o\)
\(=-1+sin^232^o+cos^232^2=-1+1=0\).
b) \(B=\dfrac{1+sin^4\alpha-cos^4\alpha}{1-sin^6\alpha-cos^6\alpha}\)
\(=\dfrac{1+\left(sin^2\alpha+cos^2\alpha\right)\left(sin^2\alpha-cos^2\alpha\right)}{1-\left(sin^6\alpha+cos^6\alpha\right)}\)
\(=\dfrac{1+sin^2\alpha-cos^2\alpha}{1-\left(sin^2\alpha+cos^2\alpha\right)\left(sin^2\alpha-sin\alpha cos\alpha+cos^2\alpha\right)}\)
\(=\dfrac{sin^2\alpha+1-cos^2\alpha}{1-\left(1-sin\alpha.cos\alpha\right)}\)
\(=\dfrac{sin^2\alpha+sin^2\alpha}{sin\alpha cos\alpha}\)
\(=\dfrac{2sin^2\alpha}{sin\alpha cos\alpha}=\dfrac{2sin\alpha}{cos\alpha}=2tan\alpha\).