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\(sina\sqrt{1+\frac{sin^2a}{cos^2a}}=sina\sqrt{\frac{cos^2a+sin^2a}{cos^2a}}=\frac{sina}{\left|cosa\right|}=\pm tana\)
\(\frac{1-cos^2x}{1-sin^2x}+tanx.cotx=\frac{sin^2x}{cos^2x}+\frac{sinx}{cosx}.\frac{cosx}{sinx}=tan^2x+1=\frac{1}{cos^2x}\)
\(\frac{1-4sin^2xcos^2x}{\left(sinx+cosx\right)^2}=\frac{\left(1-2sinx.cosx\right)\left(1+2sinx.cosx\right)}{sin^2x+cos^2x+2sinx.cosx}=\frac{\left(1-sin2x\right)\left(1+2sinx.cosx\right)}{1+2sinx.cosx}=1-2sinx\)
\(sin\left(90-x\right)+cos\left(180-x\right)+sin^2x\left(1+tan^2x\right)-tan^2x\)
\(=cosx-cosx+sin^2x.\frac{1}{cos^2x}-tan^2x=tan^2x-tan^2x=0\)
Câu 2:
\(A=2\cdot\dfrac{1}{2}+3\cdot\dfrac{1}{2}+1=1+1+1=3\)
Bài 3:
\(cos^2a=1-\left(\dfrac{12}{13}\right)^2=\dfrac{25}{169}\)
mà cosa>0
nên cosa=5/13
=>tan a=12/5; cot a=5/12
Câu 4: \(sin^2a=1-\dfrac{1}{4}=\dfrac{3}{4}\)
mà sina <0
nên sin a=-căn 3/2
=>tan a=-căn 3
\(A=-\dfrac{\sqrt{3}}{2}+\dfrac{1}{2}\cdot\left(-\sqrt{3}\right)=-\sqrt{3}\)
\(A=\frac{1}{2}+\frac{1}{2}cos2x+\frac{1}{2}+\frac{1}{2}cos\left(2x+\frac{4\pi}{3}\right)+\frac{1}{2}+\frac{1}{2}cos\left(2x-\frac{4\pi}{3}\right)\)
\(=\frac{3}{2}+\frac{1}{2}cos2x+cos2x.cos\frac{4\pi}{3}\)
\(=\frac{3}{2}+\frac{1}{2}cos2x-\frac{1}{2}cos2x=\frac{3}{2}\)
\(B=\frac{1}{2}-\frac{1}{2}cos2x+\frac{1}{2}-\frac{1}{2}cos\left(2x+\frac{4\pi}{3}\right)+\frac{1}{2}-\frac{1}{2}cos\left(2x-\frac{4\pi}{3}\right)\)
\(=\frac{3}{2}-\frac{1}{2}cos2x-cos2x.cos\frac{4\pi}{3}\)
\(=\frac{3}{2}-\frac{1}{2}cos2x+\frac{1}{2}cos2x=\frac{3}{2}\)
ta có : \(sin136^0=sin\left(180-136\right)^0=sin44^0\left(đpcm\right)\)
ta có : \(cos136^0=-cos\left(180-136\right)^0=-cos44^0\left(đpcm\right)\)
\(\sin^4x.\sin^2x+\cos^4x.\cos^2x-\left(\sin^4x+\cos^4x+\dfrac{1}{2}\sin^4x+\dfrac{1}{2}\cos^4x-\dfrac{3}{2}\right)-1=-\sin^4x.\left(1-\sin^2x\right)-cos^4x.\left(1-\cos^2x\right)-\dfrac{1}{2}\left(\sin^4x+\cos^4x\right)+\dfrac{1}{2}=-\left(\sin^4x.\cos^2x+\cos^4x.\sin^2x\right)-\dfrac{1}{2}\left(\left(\sin^2x+\cos^2x\right)^2-2\sin^2x.\cos^2x\right)+\dfrac{1}{2}=-\left(\sin^2x.\cos^2x.\left(\sin^2x+\cos^2x\right)\right)-\dfrac{1}{2}.\left(1-2\sin^2x.\cos^2x\right)+\dfrac{1}{2}=-\sin^2x.\cos^2x+\sin^2x.\cos^2x-\dfrac{1}{2}+\dfrac{1}{2}=0\)
Từ M kẻ MP ⊥ Ox, MQ ⊥ Oy
=> = cosα;
=
= sinα;
Trong tam giác vuông MPO:
MP2+ PO2 = OM2 => cos2 α + sin2 α = 1
B=1-sin2a+cos2a
\(=\sin^2a+\cos^2a-\sin^2a+\cos^2a=2\cos^2a\)
C= 1-sina.cosa.tana
\(=1-\sin a.\cos a.\frac{\sin a}{\cos a}=1-\sin^2a=\cos^2a\)
biết có vậy thôi à
Chọn D.
Ta có:
A = sin2410+ sin2450+ sin2490+ sin2450= (sin2410+ sin2490)+( sin2450+ sin2450)
A = (sin2410 + cos2490) + ( sin2450 + cos2450) = 1 + 1 = 2.