Biến đổi thành tích 1 + cos π 2 + 3 a - sin 3 π 2 - 3 a + c o t 5 π 2 + 3 a
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\(\dfrac{3\pi}{2}< a< 2\pi\Rightarrow sina< 0\)
\(\Rightarrow sina=-\sqrt{1-cos^2a}=-\sqrt{1-\left(\dfrac{3}{5}\right)^2}=-\dfrac{4}{5}\)
\(\Rightarrow sin2a=2sina.cosa=2.\left(-\dfrac{4}{5}\right).\left(\dfrac{3}{5}\right)=-\dfrac{24}{25}\)
Câu sau có nhầm đề ko nhỉ?
\(sin\left(\pi-\dfrac{\pi}{3}\right)=sin\left(\dfrac{2\pi}{3}\right)=\dfrac{\sqrt{3}}{2}\)
\(\cos a=\dfrac{-12}{13}\)
\(\sin b=\dfrac{4}{5}\)
\(\sin\left(a+b\right)=\sin a\cos b+\sin b\cos a\)
\(=\dfrac{5}{13}\cdot\dfrac{3}{5}+\dfrac{4}{5}\cdot\dfrac{-12}{13}=\dfrac{-45}{65}=\dfrac{-9}{13}\)
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: \(cos\left(a-\dfrac{pi}{6}\right)\)
\(=cosa\cdot cos\left(\dfrac{pi}{6}\right)+sina\cdot sin\left(\dfrac{pi}{6}\right)\)
\(=\dfrac{-1}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{2}}{\sqrt{3}}\cdot\dfrac{1}{2}=\dfrac{-\sqrt{3}+\sqrt{2}}{2\sqrt{3}}\)
\(A=\frac{1}{2}+\frac{1}{2}cos2x+\frac{1}{2}+\frac{1}{2}cos\left(2x+\frac{4\pi}{3}\right)+\frac{1}{2}+\frac{1}{2}cos\left(2x-\frac{4\pi}{3}\right)\)
\(=\frac{3}{2}+\frac{1}{2}cos2x+cos2x.cos\frac{4\pi}{3}\)
\(=\frac{3}{2}+\frac{1}{2}cos2x-\frac{1}{2}cos2x=\frac{3}{2}\)
\(B=\frac{1}{2}-\frac{1}{2}cos2x+\frac{1}{2}-\frac{1}{2}cos\left(2x+\frac{4\pi}{3}\right)+\frac{1}{2}-\frac{1}{2}cos\left(2x-\frac{4\pi}{3}\right)\)
\(=\frac{3}{2}-\frac{1}{2}cos2x-cos2x.cos\frac{4\pi}{3}\)
\(=\frac{3}{2}-\frac{1}{2}cos2x+\frac{1}{2}cos2x=\frac{3}{2}\)
a) \(cos\left(4x+\dfrac{\pi}{3}\right)=\dfrac{\sqrt{3}}{2}\Rightarrow cos\left(4x+\dfrac{\pi}{3}\right)=cos\dfrac{\pi}{6}\)
\(\Rightarrow\left[{}\begin{matrix}4x+\dfrac{\pi}{3}=\dfrac{\pi}{6}+k2\pi\\4x+\dfrac{\pi}{3}=-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)
..... bạn tự tìm x nhé!
b)\(sin^2x-3sin3x+2=0\)\(\Rightarrow sin^2x-3\left(3sinx-4sin^3x\right)+2=0\)
\(\Rightarrow12sin^3x+sin^2x-9sinx+2=0\)
\(\Rightarrow\left[{}\begin{matrix}sinx=-1\\sinx=\dfrac{2}{3}\\sinx=\dfrac{1}{4}\end{matrix}\right.\) \(\Rightarrow\).... bạn tự tìm x nhé!
c)\(tan\left(2x+10^o\right)=\sqrt{3}\Rightarrow tan\left(2x+10^o\right)=tan60^o\)
\(\Rightarrow2x+10^o=60^o+k180^o\)
\(\Rightarrow x=25^o+k90^o\left(k\in Z\right)\)
d) \(tanx\cdot cot2x=1\)
Đk: \(\left\{{}\begin{matrix}cosx\ne0\\sin2x\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ne\dfrac{\pi}{2}+m\pi\\x\ne m\dfrac{\pi}{2}\end{matrix}\right.\)
Pt: \(\Rightarrow tanx=tan2x\Rightarrow x=2x+k\pi\)
\(\Rightarrow x=k\pi\)
Đối chiếu với đk trên thỏa mãn đk\(\Rightarrow x=k\pi\)
a) √2 cos(x - π/4)
= √2.(cosx.cos π/4 + sinx.sin π/4)
= √2.(√2/2.cosx + √2/2.sinx)
= √2.√2/2.cosx + √2.√2/2.sinx
= cosx + sinx (đpcm)
b) √2.sin(x - π/4)
= √2.(sinx.cos π/4 - sin π/4.cosx )
= √2.(√2/2.sinx - √2/2.cosx )
= √2.√2/2.sinx - √2.√2/2.cosx
= sinx – cosx (đpcm).