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Ta có \sin 5\alpha -2\sin \alpha \left({\cos} 4\alpha +\cos 2\alpha \right)=\sin 5\alpha -2\sin \alpha .\cos 4\alpha -2\sin \alpha .\cos 2\alphasin5α−2sinα(cos4α+cos2α)=sin5α−2sinα.cos4α−2sinα.cos2α
=\sin 5\alpha -\left(\sin 5\alpha -\sin 3\alpha \right)-\left(\sin 3\alpha -\sin \alpha \right)=sin5α−(sin5α−sin3α)−(sin3α−sinα)
=\sin \alpha .=sinα.
Vậy \sin 5\alpha -2\sin \alpha \left({\cos} 4\alpha +\cos 2\alpha \right)=\sin \alphasin5α−2sinα(cos4α+cos2α)=sinα
a) Theo định lý sin: \(\frac{a}{{\sin A}} = \frac{b}{{\sin B}} \to b = \frac{{a.\sin B}}{{\sin A}}\) thay vào \(S = \frac{1}{2}ab.\sin C\) ta có:
\(S = \frac{1}{2}ab.\sin C = \frac{1}{2}a.\frac{{a.\sin B}}{{\sin A}}.sin C = \frac{{{a^2}\sin B\sin C}}{{2\sin A}}\) (đpcm)
b) Ta có: \(\hat A + \hat B + \hat C = {180^0} \Rightarrow \hat A = {180^0} - {75^0} - {45^0} = {60^0}\)
\(S = \frac{{{a^2}\sin B\sin C}}{{2\sin A}} = \frac{{{{12}^2}.\sin {{75}^0}.\sin {{45}^0}}}{{2.\sin {{60}^0}}} = \frac{{144.\frac{1}{2}.\left( {\cos {{30}^0} - \cos {{120}^0}} \right)}}{{2.\frac{{\sqrt 3 }}{2}\;}} = \frac{{72.(\frac{{\sqrt 3 }}{2}-\frac{{-1 }}{2}})}{{\sqrt 3 }} = 36+12\sqrt 3 \)
Ta có: \(A + B + C = {180^0}\)(tổng 3 góc trong một tam giác)
\(\begin{array}{l} \Rightarrow A = {180^0} - \left( {B + C} \right)\\ \Leftrightarrow \sin A = \sin \left( {{{180}^0} - \left( {B + C} \right)} \right)\\ \Leftrightarrow \sin A = \sin \left( {B + C} \right) = \sin B.\cos C + \sin C.\cos B\end{array}\)
Đkxđ: \(x\in R\).
\(cos2x-cos3x+cos4x=0\Leftrightarrow\left(cos2x+cos4x\right)-cos3x=0\)
\(\Leftrightarrow2cos3x.cosx-cos3x=0\)
\(\Leftrightarrow cos3x\left(2cos2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos3x=0\\2cos2x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}cos3x=0\\cos2x=\dfrac{1}{2}\end{matrix}\right.\)
\(cos3x=0\Leftrightarrow3x=\dfrac{\pi}{2}+k\pi\Leftrightarrow x=\dfrac{\pi}{6}+\dfrac{k\pi}{3}\)
\(cos2x=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{\pi}{3}+k2\pi\\2x=-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k\pi\\x=-\dfrac{\pi}{6}+k\pi\end{matrix}\right.\)
\(\dfrac{sinB}{sinC}=2cosA\Leftrightarrow sinB=2cosA.sinC\)
\(\Leftrightarrow sinB=sin\left(A+C\right)+sin\left(C-A\right)\)
\(\Leftrightarrow sinB=sin\left(\pi-\left(A+C\right)\right)+sin\left(C-A\right)\)
\(\Leftrightarrow sinB=sinB+sin\left(C-A\right)\)
\(\Leftrightarrow sin\left(C-A\right)=0\) (1)
Do A, C là số đo các góc trong tam giác nên từ (1) suy ra:
\(C=A\) hay tam giác ABC cân.
Chứng minh:
Không mất tính tổng quát, giả sử \(\Delta ABC\) có \(\widehat{A}=90^0\).
Khi đó ta có \(sinB=cosC\)
\(\Rightarrow sin^2A+sin^2B+sin^2C=1+cos^2C+sin^2C=2\)
1.
a.
\(\Leftrightarrow sin\left(3x-30^0\right)=sin\left(45^0\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-30^0=45^0+k360^0\\3x-30^0=135^0+k360^0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{75^0}{3}+k120^0\\x=\frac{165^0}{3}+k120^0\end{matrix}\right.\)
b.
\(sin\left(5x-\frac{\pi}{3}\right)=sin\left(2\pi-\frac{\pi}{4}-2x\right)\)
\(\Leftrightarrow sin\left(5x-\frac{\pi}{3}\right)=sin\left(-\frac{\pi}{4}-2x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-\frac{\pi}{3}=-\frac{\pi}{4}-2x+k2\pi\\5x-\frac{\pi}{3}=\frac{5\pi}{4}+2x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{84}+\frac{k2\pi}{7}\\x=\frac{19\pi}{36}+\frac{k2\pi}{3}\end{matrix}\right.\)
c.
\(4x-\frac{\pi}{3}=k\pi\)
\(\Leftrightarrow x=\frac{\pi}{12}+\frac{k\pi}{4}\)
d.
\(sin\left(2x+\frac{\pi}{6}\right)=-1\)
\(\Leftrightarrow2x+\frac{\pi}{6}=-\frac{\pi}{2}+k2\pi\)
\(\Leftrightarrow x=-\frac{\pi}{3}+k\pi\)
Do \(x\in\left(-\frac{\pi}{4};2\pi\right)\Rightarrow-\frac{\pi}{4}< -\frac{\pi}{3}+k\pi< 2\pi\)
\(\Rightarrow\frac{1}{12}< k< \frac{7}{3}\Rightarrow k=\left\{1;2\right\}\)
\(\Rightarrow x=\left\{\frac{2\pi}{3};\frac{5\pi}{3}\right\}\)
e.
\(sin\left(x+\frac{\pi}{6}\right)=\frac{\sqrt{2}}{2}\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{6}=\frac{\pi}{4}+k2\pi\\x+\frac{\pi}{6}=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{12}+k2\pi\\x=\frac{7\pi}{12}+k2\pi\end{matrix}\right.\) \(\Rightarrow x=\left\{\frac{\pi}{12};\frac{7\pi}{12}\right\}\)
ta có
sinA + sinB – sinC = 4sin (A/2) sin(B/2) cos(C/2) (2)
suy ra điều phải chứng minh.