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\(=\dfrac{cos102\cdot cot\left(-168\right)}{cos\left(-168\right)}\)
\(=cos102\cdot sin\left(-168\right)\)
\(=sin12\cdot sin168\)
\(=sin12\cdot sin12=sin^212^0\)
a)
Đặt \(A = \left( {2\sin {{30}^o} + \cos {{135}^o} - 3\tan {{150}^o}} \right).\left( {\cos {{180}^o} - \cot {{60}^o}} \right)\)
Ta có: \(\left\{ \begin{array}{l}\cos {135^o} = - \cos {45^o};\cos {180^o} = - \cos {0^o}\\\tan {150^o} = - \tan {30^o}\end{array} \right.\)
\( \Rightarrow A = \left( {2\sin {{30}^o} - \cos {{45}^o} + 3\tan {{30}^o}} \right).\left( { - \cos {0^o} - \cot {{60}^o}} \right)\)
Sử dụng bảng giá trị lượng giác của một số góc đặc biệt, ta có:
\(\left\{ \begin{array}{l}\sin {30^o} = \frac{1}{2};\tan {30^o} = \frac{{\sqrt 3 }}{3}\\\cos {45^o} = \frac{{\sqrt 2 }}{2};\cos {0^o} = 1;\cot {60^o} = \frac{{\sqrt 3 }}{3}\end{array} \right.\)
\( \Rightarrow A = \left( {2.\frac{1}{2} - \frac{{\sqrt 2 }}{2} + 3.\frac{{\sqrt 3 }}{3}} \right).\left( { - 1 - \frac{{\sqrt 3 }}{3}} \right)\)
\(\begin{array}{l} \Leftrightarrow A = - \left( {1 - \frac{{\sqrt 2 }}{2} + \sqrt 3 } \right).\left( {1 + \frac{{\sqrt 3 }}{3}} \right)\\ \Leftrightarrow A = - \frac{{2 - \sqrt 2 + 2\sqrt 3 }}{2}.\frac{{3 + \sqrt 3 }}{3}\\ \Leftrightarrow A = - \frac{{\left( {2 - \sqrt 2 + 2\sqrt 3 } \right)\left( {3 + \sqrt 3 } \right)}}{6}\\ \Leftrightarrow A = - \frac{{6 + 2\sqrt 3 - 3\sqrt 2 - \sqrt 6 + 6\sqrt 3 + 6}}{6}\\ \Leftrightarrow A = - \frac{{12 + 8\sqrt 3 - 3\sqrt 2 - \sqrt 6 }}{6}.\end{array}\)
b)
Đặt \(B = {\sin ^2}{90^o} + {\cos ^2}{120^o} + {\cos ^2}{0^o} - {\tan ^2}60 + {\cot ^2}{135^o}\)
Ta có: \(\left\{ \begin{array}{l}\cos {120^o} = - \cos {60^o}\\\cot {135^o} = - \cot {45^o}\end{array} \right. \Rightarrow \left\{ \begin{array}{l}{\cos ^2}{120^o} = {\cos ^2}{60^o}\\{\cot ^2}{135^o} = {\cot ^2}{45^o}\end{array} \right.\)
\( \Rightarrow B = {\sin ^2}{90^o} + {\cos ^2}{60^o} + {\cos ^2}{0^o} - {\tan ^2}60 + {\cot ^2}{45^o}\)
Sử dụng bảng giá trị lượng giác của một số góc đặc biệt, ta có:
\(\left\{ \begin{array}{l}\cos {0^o} = 1;\;\;\cot {45^o} = 1;\;\;\cos {60^o} = \frac{1}{2}\\\tan {60^o} = \sqrt 3 ;\;\;\sin {90^o} = 1\end{array} \right.\)
\( \Rightarrow B = {1^2} + {\left( {\frac{1}{2}} \right)^2} + {1^2} - {\left( {\sqrt 3 } \right)^2} + {1^2}\)
\( \Leftrightarrow B = 1 + \frac{1}{4} + 1 - 3 + 1 = \frac{1}{4}.\)
c
Đặt \(C = \cos {60^o}.\sin {30^o} + {\cos ^2}{30^o}\)
Sử dụng bảng giá trị lượng giác của một số góc đặc biệt, ta có:
\(\sin {30^o} = \frac{1}{2};\;\;\cos {30^o} = \frac{{\sqrt 3 }}{2};\;\cos {60^o} = \frac{1}{2}\;\)
\( \Rightarrow C = \frac{1}{2}.\frac{1}{2} + {\left( {\;\frac{{\sqrt 3 }}{2}} \right)^2} = \frac{1}{4} + \frac{3}{4} = 1.\)
a)\(sin^2\left(180^o-\alpha\right)+tan^2\left(180-\alpha\right).tan^2\left(270^o+\alpha\right)\)\(+sin\left(90^o+\alpha\right)cos\left(\alpha-360^o\right)\)
\(=sin^2\alpha+tan^2\alpha.cot^2\alpha+cos\alpha cos\alpha\)
\(=sin^2\alpha+cos^2\alpha+\left(tan\alpha cot\alpha\right)^2=1+1=2\).
\(\dfrac{cos\left(\alpha-180^o\right)}{sin\left(180^o-\alpha\right)}+\dfrac{tan\left(\alpha-180^o\right)cos\left(180^o+\alpha\right)sin\left(270^o+\alpha\right)}{tan\left(270^o+\alpha\right)}\)
\(=\dfrac{cos\left(180^o-\alpha\right)}{sin\left(180^o-\alpha\right)}+\dfrac{-tan\left(180^o-\alpha\right).cos\alpha.sin\left(90^o+\alpha\right)}{-tan\left(90^o+\alpha\right)}\)
\(=tan\left(180^o-\alpha\right)+\dfrac{tan\alpha.cos\alpha.cos\alpha}{cot\alpha}\)
\(=-tan\alpha+tan^2\alpha cos^2\alpha\)
\(=tan\alpha\left(-1+tan\alpha cos^2\alpha\right)\)
\(=tan\alpha\left(sin\alpha cos\alpha-1\right)\).
\(A=cos3a+2cos\left(\pi-3a\right)sin^2\left(\dfrac{\pi}{4}-1,5a\right)\)
\(=cos3a-2cos3a\dfrac{1-cos\left(\dfrac{\pi}{2}-3a\right)}{2}\)
\(=cos3a-cos3a\left(1-sin3a\right)\)
\(=cos3a-cos3a+cos3asin3a=\dfrac{1}{2}sin6a\)
\(=\dfrac{1}{2}sin\left(6\dfrac{5\pi}{6}\right)=\dfrac{1}{2}sin\left(4\pi+\pi\right)=\dfrac{1}{2}sin\pi=0\)
Vì a=\(\dfrac{5\pi}{6}\) nên: \(3a=\dfrac{5\pi}{2}\) => \(\cos3a=0\)
\(\pi-3a=\pi-\dfrac{5\pi}{2}=\dfrac{-3\pi}{2}\)
=> \(\cos\left(\pi-3a\right)=0\)
Do tam giác ABC vuông tại A và \(\widehat{B}=30^o\) \(\Rightarrow C=60^o\)
\(\Rightarrow\left(\overrightarrow{AB},\overrightarrow{BC}\right)=150^o;\)\(\left(\overrightarrow{BA},\overrightarrow{BC}\right)=30^o;\left(\overrightarrow{AC},\overrightarrow{CB}\right)=120^o\)
\(\left(\overrightarrow{AB},\overrightarrow{AC}\right)=90^o;\left(\overrightarrow{BC},\overrightarrow{BA}\right)=30^o\).Do vậy:
a) \(\cos\left(\overrightarrow{AB},\overrightarrow{BC}\right)+\sin\left(\overrightarrow{BA},\overrightarrow{BC}\right)+\tan\frac{\left(\overrightarrow{AC},\overrightarrow{CB}\right)}{2}\)
\(=\cos150^o+\sin30^o+\tan60^o\)
\(=-\frac{\sqrt{3}}{2}+\frac{1}{2}+\sqrt{3}\)
\(=\frac{\sqrt{3}+1}{2}\)
b) \(\sin\left(\overrightarrow{AB},\overrightarrow{AC}\right)+\cos\left(\overrightarrow{BC},\overrightarrow{AB}\right)+\cos\left(\overrightarrow{CA},\overrightarrow{BA}\right)\)
\(=\sin90^o+\cos30^o+\cos0^o\)
\(=1+\frac{\sqrt{3}}{2}\)
\(=\frac{2+\sqrt{3}}{2}\)
a) \(sin20^o+2sin40^o-sin100^o=sin20^o-sin100^o+2sin40^o\)
\(=2cos60^osin\left(-40^o\right)+2sin40^o\)\(=-2cos60^osin40^o+2sin40^o\)
\(=2sin40^o\left(-cos60^o+1\right)=2sin40^o.\left(-\dfrac{1}{2}+1\right)=sin40^o\)(đpcm).
b) \(\dfrac{sin\left(45^o+\alpha\right)-cos\left(45^o+\alpha\right)}{sin\left(45^o+\alpha\right)+cos\left(45^o+\alpha\right)}\)
\(=\dfrac{sin\left(45^o+\alpha\right)-sin\left(45^o-\alpha\right)}{sin\left(45^o+\alpha\right)+sin\left(45^o-\alpha\right)}=\dfrac{2cos45^o.sin\alpha}{2sin45^o.cos\alpha}\)
\(=tan\alpha\) (Đpcm).
a) \(A=sin\left(\dfrac{\pi}{4}+x\right)-cos\left(\dfrac{\pi}{4}-x\right)\)
\(\Leftrightarrow A=sin\dfrac{\pi}{4}.cosx+cos\dfrac{\pi}{4}.sinx-\left(cos\dfrac{\pi}{4}.cosx+sin\dfrac{\pi}{4}.sinx\right)\)
\(\Leftrightarrow A=sin\dfrac{\pi}{4}.cosx+cos\dfrac{\pi}{4}.sinx-cos\dfrac{\pi}{4}.cosx-sin\dfrac{\pi}{4}.sinx\)
\(\Leftrightarrow A=\dfrac{\sqrt{2}}{2}.cosx+\dfrac{\sqrt{2}}{2}.sinx-\dfrac{\sqrt{2}}{2}.cosx-\dfrac{\sqrt{2}}{2}.sinx\)
\(\Leftrightarrow A=0\)
b) \(B=cos\left(\dfrac{\pi}{6}-x\right)-sin\left(\dfrac{\pi}{3}+x\right)\)
\(\Leftrightarrow B=cos\dfrac{\pi}{6}.cosx+sin\dfrac{\pi}{6}.sinx-\left(sin\dfrac{\pi}{3}.cosx+cos\dfrac{\pi}{3}.sinx\right)\)
\(\Leftrightarrow B=cos\dfrac{\pi}{6}.cosx+sin\dfrac{\pi}{6}.sinx-sin\dfrac{\pi}{3}.cosx-cos\dfrac{\pi}{3}.sinx\)
\(\Leftrightarrow B=\dfrac{\sqrt{3}}{2}.cosx+\dfrac{1}{2}.sinx-\dfrac{\sqrt{3}}{2}.cosx-\dfrac{1}{2}.sinx\)
\(\Leftrightarrow B=0\)
c) \(C=sin^2x+cos\left(\dfrac{\pi}{3}-x\right).cos\left(\dfrac{\pi}{3}+x\right)\)
\(\Leftrightarrow C=sin^2x+\left(cos\dfrac{\pi}{3}.cosx+sin\dfrac{\pi}{3}.sinx\right).\left(cos\dfrac{\pi}{3}.cosx-sin\dfrac{\pi}{3}.sinx\right)\)
\(\Leftrightarrow C=sin^2x+\left(\dfrac{1}{2}.cosx+\dfrac{\sqrt{3}}{2}.sinx\right).\left(\dfrac{1}{2}.cosx-\dfrac{\sqrt{3}}{2}.sinx\right)\)
\(\Leftrightarrow C=sin^2x+\dfrac{1}{4}.cos^2x-\dfrac{3}{4}.sin^2x\)
\(\Leftrightarrow C=\dfrac{1}{4}.sin^2x+\dfrac{1}{4}.cos^2x\)
\(\Leftrightarrow C=\dfrac{1}{4}\left(sin^2x+cos^2x\right)\)
\(\Leftrightarrow C=\dfrac{1}{4}\)
d) \(D=\dfrac{1-cos2x+sin2x}{1+cos2x+sin2x}.cotx\)
\(\Leftrightarrow D=\dfrac{1-\left(1-2sin^2x\right)+2sinx.cosx}{1+2cos^2a-1+2sinx.cosx}.cotx\)
\(\Leftrightarrow D=\dfrac{2sin^2x+2sinx.cosx}{2cos^2x+2sinx.cosx}.cotx\)
\(\Leftrightarrow D=\dfrac{2sinx\left(sinx+cosx\right)}{2cosx\left(cosx+sinx\right)}.cotx\)
\(\Leftrightarrow D=\dfrac{sinx}{cosx}.cotx\)
\(\Leftrightarrow D=tanx.cotx\)
\(\Leftrightarrow D=1\)
rút gọn biểu thức:
E=cos(\(\dfrac{3\pi}{3}-\alpha\))-sin(\(\dfrac{3\pi}{2}-\alpha\))+sin(\(\alpha+4\pi\))
Đặt \(\alpha=12^o\)
Ta có : \(B=\dfrac{cos\left(\dfrac{9}{2}\pi+\alpha\right).cot\left(-3\pi+\alpha\right)}{cos\left(-5\pi+\alpha\right)}\) \(=\dfrac{cos\left(\dfrac{\pi}{2}+\alpha\right).cot\left(\alpha-\pi\right)}{cos\left(\alpha-\pi\right)}\)
\(=\dfrac{-sin\alpha.-cot\left(\pi-\alpha\right)}{-cos\alpha}\) \(=\dfrac{-sin\alpha.cot\alpha}{-cos\alpha}=tan\alpha.cot\alpha=1\)