Bài 1: \(sin^2x+cos^2x=1\)
=>\(sin^2x=1-\left(-\dfrac{1}{5}\right)=\dfrac{24}{25}\)
mà sin x>0(900<x<1800)
nên \(sinx=\dfrac{2\sqrt{6}}{5}\)
a: \(sin\left(x+\dfrac{\Omega}{6}\right)=sinx\cdot cos\left(\dfrac{\Omega}{6}\right)+cosx\cdot sin\left(\dfrac{\Omega}{6}\right)\)
\(=\dfrac{2\sqrt{6}}{5}\cdot\dfrac{\sqrt{3}}{2}+\dfrac{-1}{5}\cdot\dfrac{1}{2}=\dfrac{6\sqrt{2}-1}{10}\)
b: \(cos2x=2\cdot cos^2x-1=2\cdot\left(-\dfrac{1}{5}\right)^2-1\)
\(=\dfrac{2}{25}-1=-\dfrac{23}{25}\)
c:\(tanx=\dfrac{sinx}{cosx}=\dfrac{2\sqrt{6}}{5}:\dfrac{-1}{5}=-2\sqrt{6}\)
\(tan\left(x-\dfrac{\Omega}{3}\right)=\dfrac{tanx-tan\left(\dfrac{\Omega}{3}\right)}{1+tanx\cdot tan\left(\dfrac{\Omega}{3}\right)}\)
\(=\dfrac{-2\sqrt{6}-\sqrt{3}}{1+\left(-2\sqrt{6}\right)\cdot\sqrt{3}}=\dfrac{-2\sqrt{6}-\sqrt{3}}{1-2\sqrt{18}}=\dfrac{8\sqrt{6}+25\sqrt{3}}{71}\)
d: \(cos\left(2x+\dfrac{\Omega}{4}\right)=cos2x\cdot cos\left(\dfrac{\Omega}{4}\right)-sin2x\cdot sin\left(\dfrac{\Omega}{4}\right)\)
\(=-\dfrac{23}{25}\cdot\dfrac{\sqrt{2}}{2}-2\cdot sinx\cdot cosx\cdot\dfrac{\sqrt{2}}{2}\)
\(=\dfrac{-23\sqrt{2}}{50}-2\cdot\dfrac{2\sqrt{6}}{5}\cdot\dfrac{-1}{5}\cdot\dfrac{\sqrt{2}}{2}\)
\(=\dfrac{-23\sqrt{2}+8\sqrt{3}}{50}\)
Bài 3:
sinx+sin4x+sin7x
\(=\left(sinx+sin7x\right)+sin4x\)
\(=2\cdot sin\left(\dfrac{7x+x}{2}\right)\cdot cos\left(\dfrac{7x-x}{2}\right)+sin4x\)
\(=2\cdot sin4x\cdot cos3x+sin4x=sin4x\left(2\cdot cos3x+1\right)\)
cosx+cos4x+cos7x
=(cosx+cos7x)+cos4x
\(=2\cdot cos\left(\dfrac{7x+x}{2}\right)\cdot cos\left(\dfrac{7x-x}{2}\right)+cos4x\)
\(=2\cdot cos4x\cdot cos3x+cos4x=cos4x\left(2\cdot cos3x+1\right)\)
\(A=\dfrac{sinx+sin4x+sin7x}{cosx+cos4x+cos7x}\)
\(=\dfrac{sin4x\left(2\cdot cos3x+1\right)}{cos4x\left(2\cdot cos3x+1\right)}=\dfrac{sin4x}{cos4x}=tan4x\)
