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NV
1 tháng 6 2020

\(sin\left(x-\frac{\pi}{2}\right)+sin\frac{13\pi}{2}=sin\left(x+\frac{\pi}{2}\right)\)

\(\Leftrightarrow-cosx+1=cosx\)

\(\Leftrightarrow2cosx=1\Rightarrow cosx=\frac{1}{2}\)

1 tháng 9 2021

y = \(\dfrac{sin^2x}{cosx\left(sinx-cosx\right)}+\dfrac{1}{4}\)

y = \(\dfrac{sin^2x}{sinx.cosx-cos^2x}+\dfrac{1}{4}=\dfrac{\dfrac{sin^2x}{cos^2x}}{\dfrac{sinx.cosx}{cos^2x}-1}+\dfrac{1}{4}\)

y = \(\dfrac{tan^2x}{tanx-1}+\dfrac{1}{4}\)

y = \(\dfrac{4tan^2x+tanx-1}{4tanx-4}\). Đặt t =  tanx. Do x ∈ \(\left(\dfrac{\pi}{4};\dfrac{\pi}{2}\right)\) nên t ∈ (1 ; +\(\infty\))\

Ta đươc hàm số f(t) = \(\dfrac{4t^2+t-1}{4t-4}\)

⇒ ymin = \(\dfrac{17}{4}\) khi t = 2. hay x = arctan(2) + kπ 

NV
1 tháng 6 2020

\(cot1,25.tan\left(4\pi+1,25\right)-sin\left(x+\frac{\pi}{2}\right).cos\left(6\pi-x\right)=0\)

\(\Leftrightarrow cot1,25.tan1,25-cosx.cos\left(-x\right)=0\)

\(\Leftrightarrow1-cos^2x=0\)

\(\Leftrightarrow sin^2x=0\Rightarrow sinx=0\Rightarrow tanx=0\)

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: \(cos\left(a-\dfrac{pi}{6}\right)\)

\(=cosa\cdot cos\left(\dfrac{pi}{6}\right)+sina\cdot sin\left(\dfrac{pi}{6}\right)\)

\(=\dfrac{-1}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{2}}{\sqrt{3}}\cdot\dfrac{1}{2}=\dfrac{-\sqrt{3}+\sqrt{2}}{2\sqrt{3}}\)

NV
3 tháng 6 2020

\(sinx+cosx=\sqrt{2}\left(\frac{\sqrt{2}}{2}sinx+\frac{\sqrt{2}}{2}cosx\right)=\sqrt{2}\left(sinx.cos\frac{\pi}{4}+cosx.sin\frac{\pi}{4}\right)=\sqrt{2}sin\left(x+\frac{\pi}{4}\right)\)

\(=\sqrt{2}cos\left(\frac{\pi}{2}-\left(x+\frac{\pi}{4}\right)\right)=\sqrt{2}cos\left(\frac{\pi}{4}-x\right)=\sqrt{2}cos\left(x-\frac{\pi}{4}\right)\)

\(sinx-cosx=\sqrt{2}\left(\frac{\sqrt{2}}{2}sinx-\frac{\sqrt{2}}{2}cosx\right)=\sqrt{2}\left(sinx.cos\frac{\pi}{4}-cosx.sin\frac{\pi}{4}\right)=\sqrt{2}sin\left(x-\frac{\pi}{4}\right)\)

\(=-\sqrt{2}sin\left(\frac{\pi}{4}-x\right)=-\sqrt{2}cos\left(\frac{\pi}{2}-\left(\frac{\pi}{4}-x\right)\right)=-\sqrt{2}cos\left(x+\frac{\pi}{4}\right)\)

\(sin^4x-cos^4x=\left(sin^2x-cos^2x\right)\left(sin^2x+cos^2x\right)+sin2x\)

\(=sin^2x-cos^2x+sin2x=sin2x-cos2x\)

\(=\sqrt{2}sin\left(2x-\frac{\pi}{4}\right)\)

Bạn ghi ko đúng đề

3 tháng 6 2020

cos4x - sin4x + sin2x

NV
2 tháng 6 2020

\(A=\frac{1}{2}+\frac{1}{2}cos2x+\frac{1}{2}+\frac{1}{2}cos\left(2x+\frac{4\pi}{3}\right)+\frac{1}{2}+\frac{1}{2}cos\left(2x-\frac{4\pi}{3}\right)\)

\(=\frac{3}{2}+\frac{1}{2}cos2x+cos2x.cos\frac{4\pi}{3}\)

\(=\frac{3}{2}+\frac{1}{2}cos2x-\frac{1}{2}cos2x=\frac{3}{2}\)

\(B=\frac{1}{2}-\frac{1}{2}cos2x+\frac{1}{2}-\frac{1}{2}cos\left(2x+\frac{4\pi}{3}\right)+\frac{1}{2}-\frac{1}{2}cos\left(2x-\frac{4\pi}{3}\right)\)

\(=\frac{3}{2}-\frac{1}{2}cos2x-cos2x.cos\frac{4\pi}{3}\)

\(=\frac{3}{2}-\frac{1}{2}cos2x+\frac{1}{2}cos2x=\frac{3}{2}\)

NV
2 tháng 7 2020

\(sin\left(x+\frac{\pi}{3}\right)-sin\left(x-\frac{\pi}{3}\right)=sin\left(6\pi+\frac{\pi}{2}\right)\)

\(\Leftrightarrow2cosx.sin\frac{\pi}{3}=sin\left(\frac{\pi}{2}\right)\)

\(\Leftrightarrow2cosx.\frac{\sqrt{3}}{2}=1\)

\(\Rightarrow cosx=\frac{1}{\sqrt{3}}\)

17 tháng 6 2016

điều kiện : cosx\(\ne\)\(\frac{1}{\sqrt{2}}\)=> x\(\ne\)\(\pm\)\(\frac{\pi}{4}\)+2k\(\pi\), k\(\in\)Z

pt<=> tử số =0

<=>cos2x-sin(3x-\(\frac{\pi}{4}\)+x+\(\frac{3\pi}{4}\))-sin(3x-\(\frac{\pi}{4}\)-x-\(\frac{3\pi}{4}\))-2=0

<=> cos2x-sin(x+\(\frac{\pi}{2}\))-sin(2x-\(\pi\))-2=0

<=> cos2x-cosx+sin2x-2sin2x-2cos2x=0

<=>-cos2x-coxs+2sinx.cosx-2sin2x=0

đến đây bạn nhóm lại ra nghiệm rồi kiểm tra đk là xong