Biết sin a=\(\dfrac{5}{13}\);cos b=\(\dfrac{3}{5}\); \(\dfrac{\text{π}}{2}\)<a<π; 0<b<\(\dfrac{\text{π}}{2}\). Hãy tính sin(a+b)
Tính sin 2a , cos 2a , tan 2a, biết \(cos a = \dfrac{-5}{13} , ( π < a < \dfrac{3π}{2}) \)
\(\pi< a< \frac{3\pi}{2}\Rightarrow sina< 0\)
\(\Rightarrow sina=-\sqrt{1-cos^2a}=-\frac{12}{13}\)
\(sin2a=2sina.cosa=\frac{120}{169}\)
\(cos2a=2cos^2a-1=-\frac{119}{169}\)
\(tan2a=\frac{sin2a}{cos2a}=-\frac{120}{119}\)
Cho góc α
thỏa mãn `π\2`<α<π,cosα=−\(\dfrac{1}{\sqrt{3}}\). Tính giá trị của các biểu thức sau:
a) sin(α+\(\dfrac{\text{π}}{6}\))
b) cos(α+$\frac{\text{π}}{6}$)
c) sin(α−$\frac{\text{π}}{3}$)
d) cos(α−$\frac{\text{π}}{6}$)
a: pi/2<a<pi
=>sin a>0
\(sina=\sqrt{1-\left(-\dfrac{1}{\sqrt{3}}\right)^2}=\dfrac{\sqrt{2}}{\sqrt{3}}\)
\(sin\left(a+\dfrac{pi}{6}\right)=sina\cdot cos\left(\dfrac{pi}{6}\right)+sin\left(\dfrac{pi}{6}\right)\cdot cosa\)
\(=\dfrac{\sqrt{3}}{2}\cdot\dfrac{\sqrt{2}}{\sqrt{3}}+\dfrac{1}{2}\cdot-\dfrac{1}{\sqrt{3}}=\dfrac{\sqrt{6}-2}{2\sqrt{3}}\)
b: \(cos\left(a+\dfrac{pi}{6}\right)=cosa\cdot cos\left(\dfrac{pi}{6}\right)-sina\cdot sin\left(\dfrac{pi}{6}\right)\)
\(=\dfrac{-1}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{2}-\dfrac{\sqrt{2}}{\sqrt{3}}\cdot\dfrac{1}{2}=\dfrac{-\sqrt{3}-\sqrt{2}}{2\sqrt{3}}\)
c: \(sin\left(a-\dfrac{pi}{3}\right)\)
\(=sina\cdot cos\left(\dfrac{pi}{3}\right)-cosa\cdot sin\left(\dfrac{pi}{3}\right)\)
\(=\dfrac{\sqrt{2}}{\sqrt{3}}\cdot\dfrac{1}{2}+\dfrac{1}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{2}=\dfrac{\sqrt{2}+\sqrt{3}}{2\sqrt{3}}\)
d: \(cos\left(a-\dfrac{pi}{6}\right)\)
\(=cosa\cdot cos\left(\dfrac{pi}{6}\right)+sina\cdot sin\left(\dfrac{pi}{6}\right)\)
\(=\dfrac{-1}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{2}}{\sqrt{3}}\cdot\dfrac{1}{2}=\dfrac{-\sqrt{3}+\sqrt{2}}{2\sqrt{3}}\)
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)
\(B=\dfrac{3cot^2a-1}{cot^2a+2}=...\)
Cho \(sina=\dfrac{3}{5},cosb=-\dfrac{5}{13}\)và \(\dfrac{\pi}{2}< a,b< \pi\)
Tính \(cos\dfrac{a}{2};sin\dfrac{b}{2};tan\left(a+b\right);sin\left(a-b\right)\)
GIÚP VỚI MÌNH ĐANG CẦN GẤP
pi/2<a,b<pi
=>cos a<0; cos b<0; sin a>0; sin b>0
\(cosa=-\sqrt{1-\left(\dfrac{3}{5}\right)^2}=-\dfrac{4}{5};sina=\sqrt{1-\left(-\dfrac{5}{13}\right)^2}=\dfrac{12}{13}\)
tan a=-3/5:4/5=-3/4; tan b=12/13:(-5/13)=-12/5
\(tan\left(a+b\right)=\dfrac{tana+tanb}{1-tana\cdot tanb}\)
\(=\dfrac{-\dfrac{3}{4}+\dfrac{-12}{5}}{1-\dfrac{-3}{4}\cdot\dfrac{-12}{5}}=\dfrac{63}{16}\)
sin(a-b)=sina*cosb-sinb*cosa
\(=\dfrac{3}{5}\cdot\dfrac{-5}{13}-\dfrac{-4}{5}\cdot\dfrac{12}{13}=\dfrac{-15+48}{65}=\dfrac{33}{65}\)
Rút gọn:
C= \(sin^2\dfrac{\pi}{3}+sin^2\dfrac{5\pi}{6}+sin^2\dfrac{\pi}{9}+sin^2\dfrac{11\pi}{18}+sin^2\dfrac{13\pi}{18}+sin^2\dfrac{2\pi}{9}\)
D=\(cos\left(x-\dfrac{\pi}{3}\right).cos\left(x+\dfrac{\pi}{4}\right)+cos\left(x+\dfrac{\pi}{6}\right).cos\left(x+\dfrac{3\pi}{4}\right)\)
a) Biết Sin α.cos α=\(\dfrac{12}{25}\). Tính tỉ số lượng giác của góc α
b) Biết Sin α=\(\dfrac{3}{5}\). Tính A=5.Sin2α + 6cos2α
c) Biết cot α=\(\dfrac{4}{3}\). Tính D=\(\dfrac{Sin\alpha+cos\alpha}{Sin\alpha-cos\alpha}\)
b) Ta có: \(\sin^2\alpha+\cos^2\alpha=1\)
\(\Leftrightarrow\cos^2\alpha=\dfrac{16}{25}\)
hay \(\cos\alpha=\dfrac{4}{5}\)
Ta có: \(A=5\cdot\sin^2\alpha+6\cdot\cos^2\alpha\)
\(=5\cdot\left(\dfrac{3}{5}\right)^2+6\cdot\left(\dfrac{4}{5}\right)^2\)
\(=5\cdot\dfrac{9}{25}+6\cdot\dfrac{16}{25}\)
\(=\dfrac{141}{25}\)
c) Ta có: \(\tan\alpha=\dfrac{1}{\cot\alpha}=\dfrac{1}{\dfrac{4}{3}}=\dfrac{3}{4}\)
\(D=\dfrac{\sin\alpha+\cos\alpha}{\sin\alpha-\cos\alpha}\)
\(=\dfrac{\dfrac{9}{16}+\dfrac{16}{9}}{\dfrac{9}{16}-\dfrac{16}{9}}=-\dfrac{337}{175}\)
Rút gọn cac biểu thức sau:
\(A=sin\left(\dfrac{5\pi}{2}-\alpha\right)+cos\left(13\pi+\alpha\right)-3sin\left(\alpha-5\pi\right)\)
\(B=sin\left(x+\dfrac{85\pi}{2}\right)+cos\left(2017\pi+x\right)+sin^2\left(33\pi+x\right)+sin^2\left(x-\dfrac{5\pi}{2}\right)+cos\left(x+\dfrac{3\pi}{2}\right)\)\(C=sin\left(x+\dfrac{2017\pi}{2}\right)+2sin^2\left(x-\pi\right)+cos\left(x+2019\pi\right)+cos2x+sin\left(x+\dfrac{9\pi}{2}\right)\)
\(A=sin\left(\dfrac{\pi}{2}-\alpha+2\pi\right)+cos\left(\pi+\alpha+12\pi\right)-3sin\left(\alpha-\pi-4\pi\right)\)
\(=sin\left(\dfrac{\pi}{2}-\alpha\right)+cos\left(\pi+\alpha\right)-3sin\left(\alpha-\pi\right)\)
\(=cos\alpha-cos\alpha+3sin\left(\pi-\alpha\right)\)\(=3sin\alpha\)
\(B=sin\left(x+\dfrac{\pi}{2}+42\pi\right)+cos\left(x+\pi+2016\pi\right)+sin^2\left(x+\pi+32\pi\right)+sin^2\left(x-\dfrac{\pi}{2}-2\pi\right)+cos\left(x-\dfrac{\pi}{2}+2\pi\right)\)
\(=sin\left(x+\dfrac{\pi}{2}\right)+cos\left(x+\pi\right)+sin^2\left(x+\pi\right)+sin^2\left(x-\dfrac{\pi}{2}\right)+cos\left(x-\dfrac{\pi}{2}\right)\)
\(=cosx-cosx+sin^2x+cos^2x+sinx\)
\(=1+sinx\)
\(C=sin\left(x+\dfrac{\pi}{2}+1008\pi\right)+2sin^2\left(\pi-x\right)+cos\left(x+\pi+2018\pi\right)+cos2x+sin\left(x+\dfrac{\pi}{2}+4\pi\right)\)
\(=sin\left(x+\dfrac{\pi}{2}\right)+2sin^2\left(\pi-x\right)+cos\left(x+\pi\right)+cos2x+sin\left(x+\dfrac{\pi}{2}\right)\)
\(=cosx+2sin^2x-cosx+1-2sin^2x+cosx\)
\(=1+cosx\)
Rút gọn :
a) \(\cos\dfrac{x}{5}\cos\dfrac{2x}{5}\cos\dfrac{4x}{5}\cos\dfrac{8x}{5}\)
b) \(\sin\dfrac{x}{7}+2\sin\dfrac{3x}{7}+\sin\dfrac{5x}{7}\)
a)\(\eqalign{ & A\sin {x \over 5} = \sin {x \over 5}\cos {x \over 5}\cos {{2x} \over 5}\cos {{4x} \over 5}\cos {{8x} \over 5} \cr & = {1 \over 2}\sin {{2x} \over 5}\cos {{2x} \over 5}\cos {{4x} \over 5}\cos {{8x} \over 5} \cr & = {1 \over 4}\sin {{4x} \over 5}\cos {{4x} \over 5}\cos {{8x} \over 5} = {1 \over 8}\sin {{8x} \over 5}\cos {{8x} \over 5} \cr & = {1 \over {16}}\sin {{16x} \over 5} \cr} \)
Suy ra biểu thức rút gọn \(A =\sin{{16x} \over 5}:16\sin {x \over 5}\)
b)\(\eqalign{ & B = \sin {x \over 7} + 2\sin {{3x} \over 7} + \sin {{5x} \over 7} = 2\sin {{3x} \over 7} + (\sin {x \over 7} + \sin {{5x} \over 7}) \cr & = 2\sin {{3x} \over 7} + 2\sin {1 \over 2}({{5x} \over 7} + {x \over 7})cos{1 \over 2}({{5x} \over 7} - {x \over 7}) \cr & = 2\sin {{3x} \over 7}(1 + \cos {{2x} \over 7}) = 4\sin {{3x} \over 7}{\cos ^2}{x \over 7} \cr}\)
Giải các pt
a) \(\sqrt{2}\sin\left(2x+\dfrac{\pi}{4}\right)=3\sin x+\cos x+2\)
b) \(\dfrac{\left(2-\sqrt{3}\right)\cos x-2\sin^2\left(\dfrac{x}{2}-\dfrac{\pi}{4}\right)}{2\cos x-1}=1\)
c) \(2\sqrt{2}\cos\left(\dfrac{5\pi}{12}-x\right)\sin x=1\)
a.
\(\sqrt{2}sin\left(2x+\dfrac{\pi}{4}\right)=3sinx+cosx+2\)
\(\Leftrightarrow sin2x+cos2x=3sinx+cosx+2\)
\(\Leftrightarrow2sinx.cosx-3sinx+2cos^2x-cosx-3=0\)
\(\Leftrightarrow sinx\left(2cosx-3\right)+\left(cosx+1\right)\left(2cosx-3\right)=0\)
\(\Leftrightarrow\left(2cosx-3\right)\left(sinx+cosx+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=\dfrac{3}{2}\left(vn\right)\\sinx+cosx+1=0\end{matrix}\right.\)
\(\Rightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=-1\)
\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=-\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow...\)
b.
ĐKXĐ: \(cosx\ne\dfrac{1}{2}\Rightarrow\left[{}\begin{matrix}x\ne\dfrac{\pi}{3}+k2\pi\\x\ne-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\dfrac{\left(2-\sqrt{3}\right)cosx-2sin^2\left(\dfrac{x}{2}-\dfrac{\pi}{4}\right)}{2cosx-1}=1\)
\(\Rightarrow\left(2-\sqrt{3}\right)cosx+cos\left(x-\dfrac{\pi}{2}\right)=2cosx\)
\(\Leftrightarrow-\sqrt{3}cosx+sinx=0\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{3}\right)=0\)
\(\Rightarrow x-\dfrac{\pi}{3}=k\pi\)
\(\Rightarrow x=\dfrac{\pi}{3}+k\pi\)
Kết hợp ĐKXĐ \(\Rightarrow x=\dfrac{4\pi}{3}+k2\pi\)
c.
\(2\sqrt{2}cos\left(\dfrac{5\pi}{12}-x\right)sinx=1\)
\(\Leftrightarrow\sqrt{2}\left(sin\left(\dfrac{5\pi}{12}\right)+sin\left(2x-\dfrac{5\pi}{12}\right)\right)=1\)
\(\Leftrightarrow sin\left(2x-\dfrac{5\pi}{12}\right)=\dfrac{-\sqrt{6}+\sqrt{2}}{2}\)
\(\Leftrightarrow sin\left(2x-\dfrac{5\pi}{12}\right)=sin\left(-\dfrac{\pi}{12}\right)\)
\(\Leftrightarrow...\)