Chứng minh biểu thức sau không phụ thuộc vào x (giúp mình vớiiii)\(\frac{2cos^2x-1}{2cot\left(\frac{5\pi}{4}+x\right).sin^2\left(\frac{\pi}{4}+x\right)}\)
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\(P=sin^4x+\left(sin^2\left(x+\frac{\pi}{4}\right)\right)^2+cos^4x+\left(cos^2\left(x+\frac{\pi}{4}\right)\right)^2\)
\(=\left(\frac{1}{2}-\frac{1}{2}cos2x\right)^2+\left(\frac{1}{2}-\frac{1}{2}cos\left(2x+\frac{\pi}{2}\right)\right)^2+\left(\frac{1}{2}+\frac{1}{2}cos2x\right)^2+\left(\frac{1}{2}+\frac{1}{2}cos\left(2x+\frac{\pi}{4}\right)\right)^2\)
\(=\frac{1}{4}-\frac{1}{2}cos2x+\frac{1}{4}cos^22x+\frac{1}{4}+\frac{1}{2}sin2x+\frac{1}{4}sin^22x+\frac{1}{4}+\frac{1}{2}cos2x+\frac{1}{4}cos^22x+\frac{1}{4}-\frac{1}{2}sin2x+\frac{1}{4}sin^22x\)
\(=1+\frac{1}{2}\left(sin^22x+cos^22x\right)=\frac{3}{2}\)
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mình ghi đáp án cho cái lượng giác này thui nhé
\(=\frac{3}{2}\)
A=sin2x+sin2x\(\left(\frac{2\pi}{3}+x\right)\)+sin2\(\left(\frac{2\pi}{3}-x\right)\)
\(A=\sin^2x+\left[\sin\left(\frac{2\pi}{3}+x\right)+\sin\left(\frac{2\pi}{3}-x\right)\right]^2-2\sin\left(\frac{2\pi}{3}-x\right).\sin\frac{2\pi}{3}+x\)
\(A=\sin^2x+4\left[\frac{\sin2\pi}{3}.\sin x\right]^2-\left[\frac{\sin4\pi}{3}+\sin2x\right]\)
\(A=\sin^2x+\sin x^2-\left[\sin2x-\frac{1}{2}\right]\)
\(A=2\sin x^2-\left[2\sin^2x-\frac{3}{2}\right]\)
\(A=\frac{3}{2}\)
vậy biểu thức trên ko phụ thuộc vào biến số x
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Ta có \(cos^2\left(\frac{\pi}{4}-x\right)=sin^2\left(\frac{\pi}{2}-\left(\frac{\pi}{4}-x\right)\right)=sin^2\left(x+\frac{\pi}{4}\right)\)
\(\Rightarrow\frac{1-sin^2x}{2cot\left(\frac{\pi}{4}+x\right).cos^2\left(\frac{\pi}{4}-x\right)}=\frac{cos^2x}{2cot\left(\frac{\pi}{4}+x\right).sin^2\left(\frac{\pi}{4}+x\right)}=\frac{cos^2x}{2.cos\left(\frac{\pi}{4}+x\right).sin\left(\frac{\pi}{4}+x\right)}\)
\(=\frac{cos^2x}{sin\left(\frac{\pi}{2}+2x\right)}=\frac{cos^2x}{cos2x}\)???
Đến đây thì đoán là bạn ghi sai đề, tử số phải là \(cos^2x-sin^2x\) chứ ko phải \(1-sin^2x\) vì \(cos^2x-sin^2x=cos2x\) mới rút gọn hết với mẫu
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\(=cos\left(x-\frac{\pi}{3}\right)cos\left(x+\frac{\pi}{4}\right)+sin\left(\frac{\pi}{2}-x-\frac{\pi}{6}\right)sin\left(\frac{\pi}{2}-x-\frac{3\pi}{4}\right)\)
\(=cos\left(x-\frac{\pi}{3}\right)cos\left(x+\frac{\pi}{4}\right)+sin\left(\frac{\pi}{3}-x\right)sin\left(-x-\frac{\pi}{4}\right)\)
\(=cos\left(x-\frac{\pi}{3}\right)cos\left(x+\frac{\pi}{4}\right)+sin\left(x-\frac{\pi}{3}\right)sin\left(x+\frac{\pi}{4}\right)\)
\(=cos\left(x-\frac{\pi}{3}-x-\frac{\pi}{4}\right)=cos\left(-\frac{7\pi}{12}\right)=cos\frac{7\pi}{12}=\frac{\sqrt{2}-\sqrt{6}}{4}\)
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1,\(A=3\left(sin^4x+cos^4x\right)-2\left(sin^2x+cos^2x\right)\left(sin^4x-sin^2x.cos^2x+cos^4x\right)\)
\(=3\left(sin^4x+cos^4x\right)-2\left(sin^4x-sin^2x.cos^4x+cos^4x\right)\)
\(=sin^4x+2sin^2x.cos^2x+cos^4x=\left(sin^2x+cos^2x\right)^2=1\)
Vậy...
2,\(B=cos^6x+2sin^4x\left(1-sin^2x\right)+3\left(1-cos^2x\right)cos^4x+sin^4x\)
\(=-2cos^6x+3sin^4x-2sin^6x+3cos^4x\)
\(=-2\left(sin^2x+cos^2x\right)\left(sin^4x-sin^2x.cos^2x+cos^4x\right)+3\left(cos^4x+sin^4x\right)\)
\(=-2\left(sin^4x-sin^2x.cos^2x+cos^4x\right)+3\left(cos^4x+sin^4x\right)\)\(=cos^4x+sin^4x+2sin^2x.cos^2x=1\)
Vậy...
3,\(C=\dfrac{1}{2}\left[cos\left(-\dfrac{7\pi}{12}\right)+cos\left(2x-\dfrac{\pi}{12}\right)\right]+\dfrac{1}{2}\left[cos\left(-\dfrac{7\pi}{12}\right)+cos\left(2x+\dfrac{11\pi}{12}\right)\right]\)
\(=cos\left(-\dfrac{7\pi}{12}\right)+\dfrac{1}{2}\left[cos\left(2x-\dfrac{\pi}{12}\right)+cos\left(2x+\dfrac{11\pi}{12}\right)\right]\)\(=\dfrac{-\sqrt{6}+\sqrt{2}}{4}+\dfrac{1}{2}\left[cos\left(2x-\dfrac{\pi}{12}\right)+cos\left(2x-\dfrac{\pi}{12}+\pi\right)\right]\)
\(=\dfrac{-\sqrt{6}+\sqrt{2}}{4}+\dfrac{1}{2}\left[cos\left(2x-\dfrac{\pi}{12}\right)-cos\left(2x-\dfrac{\pi}{12}\right)\right]\)\(=\dfrac{-\sqrt{6}+\sqrt{2}}{4}\)
Vậy...
4, \(D=cos^2x+\left(-\dfrac{1}{2}cosx-\dfrac{\sqrt{3}}{2}sinx\right)^2+\left(-\dfrac{1}{2}.cosx+\dfrac{\sqrt{3}}{2}.sinx\right)^2\)
\(=cos^2x+\dfrac{1}{4}cos^2x+\dfrac{\sqrt{3}}{4}cosx.sinx+\dfrac{3}{4}sin^2x+\dfrac{1}{4}cos^2x-\dfrac{\sqrt{3}}{4}cosx.sinx+\dfrac{3}{4}sin^2x\)
\(=\dfrac{3}{2}\left(cos^2x+sin^2x\right)=\dfrac{3}{2}\)
Vậy...
5, Xem lại đề
6,\(F=-cosx+cosx-tan\left(\dfrac{\pi}{2}+x\right).cot\left(\pi+\dfrac{\pi}{2}-x\right)\)
\(=tan\left(\pi-\dfrac{\pi}{2}-x\right).cot\left(\dfrac{\pi}{2}-x\right)\)\(=tan\left(\dfrac{\pi}{2}-x\right).cot\left(\dfrac{\pi}{2}-x\right)\)\(=cotx.tanx=1\)
Vậy...
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\(\frac{1-cosx+cos2x}{sin2x-sinx}=\frac{1-cosx+2cos^2x-1}{2sinx.cosx-sinx}=\frac{cosx\left(2cosx-1\right)}{sinx\left(2cosx-1\right)}=\frac{cosx}{sinx}=cotx\)
\(A=sin\left(\frac{\pi}{4}+x\right)-sin\left(\frac{\pi}{2}-\frac{\pi}{4}+x\right)=sin\left(\frac{\pi}{4}+x\right)-sin\left(\frac{\pi}{4}+x\right)=0\)
\(\frac{2cos^2x-1}{2cot\left(\pi+\frac{\pi}{4}+x\right).sin^2\left(\frac{\pi}{4}+x\right)}=\frac{cos2x}{2cot\left(\frac{\pi}{4}+x\right).sin^2\left(\frac{\pi}{4}+x\right)}=\frac{cos2x}{2cos\left(\frac{\pi}{4}+x\right).sin\left(\frac{\pi}{4}+x\right)}\)
\(=\frac{cos2x}{sin\left(\frac{\pi}{2}+2x\right)}=\frac{cos2x}{cos2x}=1\)