Để \(cos\left(x-\dfrac{\Omega}{4}\right);sinx;cos\left(x+\dfrac{\Omega}{4}\right)\) là ba số hạng liên tiếp của cấp số nhân thì \(sin^2x=cos\left(x-\dfrac{\Omega}{4}\right)\cdot cos\left(x+\dfrac{\Omega}{4}\right)\)

=>\(sin^2x=\sqrt{2}\left(cosx-sinx\right)\cdot\sqrt{2}\left(cosx+sinx\right)\)

=>\(sin^2x=2cos^2x-2sin^2x\)

=>\(3\cdot sin^2x=2\cdot cos^2x\)

=>\(\dfrac{sin^2x}{cos^2x}=\dfrac{2}{3}\)

=>\(tan^2x=\dfrac{2}{3}\)

=>\(\left[{}\begin{matrix}tanx=\dfrac{\sqrt{6}}{3}\\tanx=-\dfrac{\sqrt{6}}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=arctan\left(\dfrac{\sqrt{6}}{3}\right)+k\Omega\\x=arctan\left(-\dfrac{\sqrt{6}}{3}\right)+k\Omega\end{matrix}\right.\)

\(y=\sin^4x+\cos^4x\\ =1-2\sin^2x\cdot\cos^2x\\ =1-\dfrac{1}{2}\sin^22x\\ 0\le\sin^22x\le1\\ \Leftrightarrow\dfrac{1}{2}\le y\le1\\ y_{min}=\dfrac{1}{2}\Leftrightarrow\sin^22x=1\Leftrightarrow x=\dfrac{k\pi}{2}\pm\dfrac{\pi}{4}\\ y_{max}=1\Leftrightarrow\sin^22x=0\Leftrightarrow x=k\pi\)

\(y=3\sin x+4\cos x\\ =5\left(\dfrac{3\sin x}{5}+\dfrac{4\cos x}{5}\right)\\ =5\cos\left(x-a\right),\forall\cos a=\dfrac{4}{5},\sin a=\dfrac{3}{5}\\ -1\le\cos\left(x-a\right)\le1\\ \Leftrightarrow-5\le y\le5\\ y_{min}=-5\Leftrightarrow\cos\left(x-a\right)=-1\\ y_{max}=5\Leftrightarrow\cos\left(x-a\right)=1\)

\(y=sin^4x+cos^4x\)

Ta có: \(0\le sin^4x\le1\)

\(0\le cos^4x\le1\)

\(0\le sin^4x+cos^4x\le2\)

Vây GTNN là 0, GTLN là 2

y=3sinx+4cosx

\(-3\le3sinx\le3\\ -4\le4cosx\le4\\ -7\le3sinx+4cosx\le7\)

Vậy GTNN là -7, GTLN là 7

Chắc là \(0< a< \dfrac{\pi}{2}\)?

\(0< a< \dfrac{\pi}{2}\Rightarrow sina;cosa>0\)

\(\left\{{}\begin{matrix}sina=\sqrt{3}cosa\\sin^2a+cos^2a=1\end{matrix}\right.\) \(\Rightarrow\left(\sqrt{3}cosa\right)^2+cos^2a=1\)

\(\Rightarrow4cos^2a=1\Rightarrow cosa=\dfrac{1}{2}\)

\(\Rightarrow sina=\sqrt{3}cosa=\dfrac{\sqrt{3}}{2}\)

Vì π/2 < a < π nên tan a < 0, do đó tan a = -2.

Áp dụng công thức

Đáp án là B.

\(B=cos\frac{\pi}{7}.cos\left(\pi-\frac{4\pi}{7}\right).cos\left(\pi-\frac{2\pi}{7}\right)\)

\(B=cos\frac{\pi}{7}.cos\frac{2\pi}{7}.cos\frac{4\pi}{7}\)

\(B.sin\frac{\pi}{7}=sin\frac{\pi}{7}.cos\frac{\pi}{7}.cos\frac{2\pi}{7}.cos\frac{4\pi}{7}\)

\(B.sin\frac{\pi}{7}=\frac{1}{2}sin\frac{2\pi}{7}.cos\frac{2\pi}{7}.cos\frac{4\pi}{7}\)

\(B.sin\frac{\pi}{7}=\frac{1}{4}sin\frac{4\pi}{7}.cos\frac{4\pi}{7}=\frac{1}{8}sin\frac{8\pi}{7}\)

\(B.sin\frac{\pi}{7}=\frac{1}{8}sin\left(\pi+\frac{\pi}{7}\right)=-\frac{1}{8}sin\frac{\pi}{7}\)

\(\Rightarrow B=-\frac{1}{8}\)

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) √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).