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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: \(cos\left(a-\dfrac{pi}{6}\right)\)
\(=cosa\cdot cos\left(\dfrac{pi}{6}\right)+sina\cdot sin\left(\dfrac{pi}{6}\right)\)
\(=\dfrac{-1}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{2}}{\sqrt{3}}\cdot\dfrac{1}{2}=\dfrac{-\sqrt{3}+\sqrt{2}}{2\sqrt{3}}\)
\(\left|z\right|=\sqrt{\left(1+\cos a\right)^2+\sin^2a}=\sqrt{2\left(1+\cos a\right)}=\sqrt{4\cos^2\frac{a}{2}}=2\left|\cos\frac{a}{2}\right|\)
a) Nếu \(a\in\left(0,\pi\right)\Rightarrow\frac{a}{2}\in\left(0,\frac{\pi}{2}\right)\), P nằm góc phần tư thứ nhất.
Do đó
\(\theta=arctan\frac{\sin a}{1+\cos a}=arctan\left(tan\frac{a}{2}\right)=\frac{a}{2}\)
\(z=2\cos\frac{a}{2}\left(\cos\frac{a}{2}+i\sin\frac{a}{2}\right)\)
b)
Nếu \(a\in\left(\pi,2\pi\right)\Rightarrow\frac{a}{2}\in\left(\frac{\pi}{2},\pi\right)\), P nằm góc phần tư thứ tư.
Do đó
\(\theta=arctan\left(tan\frac{a}{2}\right)+2\pi=\frac{a}{2}-\pi+2\pi=\frac{a}{2}+\pi\)
\(z=-2\cos\frac{a}{2}\left[\cos\left(\frac{a}{2}+\pi\right)+i\sin\left(\frac{a}{2}+\pi\right)\right]\)
c) Nếu \(a=\pi\) thì \(z=0\)
\(\cos a=\dfrac{-12}{13}\)
\(\sin b=\dfrac{4}{5}\)
\(\sin\left(a+b\right)=\sin a\cos b+\sin b\cos a\)
\(=\dfrac{5}{13}\cdot\dfrac{3}{5}+\dfrac{4}{5}\cdot\dfrac{-12}{13}=\dfrac{-45}{65}=\dfrac{-9}{13}\)
a) √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).
Chọn A.
Ta có: