tan ⁡ ( 180 ° + θ ) = \tan\left(180\degree+\theta\right)= tan ( 1 8 0 ° + θ ) =

− tan ⁡ θ -\tan\theta − tan θ

− cot ⁡ θ -\cot\theta − cot θ

cot ⁡ ( − 180 ° + θ ) = \cot\left(-180\degree+\theta\right)= cot ( − 1 8 0 ° + θ ) =

− tan ⁡ θ -\tan\theta − tan θ

− cot ⁡ θ -\cot\theta − cot θ

cot ⁡ ( − θ ) = \cot\left(-\theta\right)= cot ( − θ ) =

− cot ⁡ θ -\cot\theta − cot θ

− tan ⁡ θ -\tan\theta − tan θ

tan ⁡ ( 3 π 2 − θ ) = \tan\left(\frac{3\pi}{2}-\theta\right)= tan ( 2 3 π − θ ) =

− tan ⁡ θ -\tan\theta − tan θ

− cot ⁡ θ -\cot\theta − cot θ

sin ⁡ ( π 2 + θ ) = \sin\left(\frac{\pi}{2}+\theta\right)= sin ( 2 π + θ ) =

− sin ⁡ θ -\sin\theta − sin θ

− cos ⁡ θ -\cos\theta − cos θ

sin ⁡ ( π − θ ) = \sin\left(\pi-\theta\right)= sin ( π − θ ) =

− sin ⁡ θ -\sin\theta − sin θ

− cos ⁡ θ -\cos\theta − cos θ

sin ⁡ ( 90 ° + θ ) = \sin\left(90\degree+\theta\right)= sin ( 9 0 ° + θ ) =

− sin ⁡ θ -\sin\theta − sin θ

− cos ⁡ θ -\cos\theta − cos θ

cos ⁡ ( − 180 ° + θ ) = \cos\left(-180\degree+\theta\right)= cos ( − 1 8 0 ° + θ ) =

− cos ⁡ θ -\cos\theta − cos θ

− sin ⁡ θ -\sin\theta − sin θ

sec ⁡ ( 180 ° − θ ) = \sec\left(180\degree-\theta\right)= sec ( 1 8 0 ° − θ ) =

− sec ⁡ θ -\sec\theta − sec θ

− csc ⁡ θ -\csc\theta − csc θ

csc ⁡ ( 180 ° + θ ) = \csc\left(180\degree+\theta\right)= csc ( 1 8 0 ° + θ ) =

− csc ⁡ θ -\csc\theta − csc θ

− sec ⁡ θ -\sec\theta − sec θ

csc ⁡ ( 90 ° + θ ) = \csc\left(90\degree+\theta\right)= csc ( 9 0 ° + θ ) =

− sec ⁡ θ -\sec\theta − sec θ

− csc ⁡ θ -\csc\theta − csc θ

sec ⁡ ( 270 ° + θ ) = \sec\left(270\degree+\theta\right)= sec ( 2 7 0 ° + θ ) =

− sec ⁡ θ -\sec\theta − sec θ

− csc ⁡ θ -\csc\theta − csc θ

第一冊2-2化任意角為銳角三角函數

Authored by 游佳穆游佳穆

Mathematics

10th - 12th Grade

Used 100+ times

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20 questions

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MULTIPLE CHOICE QUESTION

1 min • 1 pt

已知 $\left(\csc\theta,\cot\theta\right)$ 在第二象限，則角 $\theta$ 在哪一象限？

一

二

三

四

MULTIPLE CHOICE QUESTION

1 min • 1 pt

設 $\tan\theta=-\frac{3}{4}$ ，且 $\cos\theta<0$ ，則角 $\theta$ 在哪一象限？

一

二

三

四

MULTIPLE CHOICE QUESTION

1 min • 1 pt

設 $\sec\theta>0$ ，且 $\tan\theta<0$ ，則角 $\theta$ 在哪一象限？

一

二

三

四

MULTIPLE CHOICE QUESTION

1 min • 1 pt

試問 $\theta$ 在第三象限，下列三角函數值何者為正？

$\tan\theta$

$\sec\theta$

$\csc\theta$

$\sin\theta$

MULTIPLE CHOICE QUESTION

1 min • 1 pt

$\tan\left(180\degree+\theta\right)=$

$\tan\theta$

$-\tan\theta$

$\cot\theta$

$-\cot\theta$

MULTIPLE CHOICE QUESTION

1 min • 1 pt

$\cot\left(-180\degree+\theta\right)=$

$\tan\theta$

$-\tan\theta$

$\cot\theta$

$-\cot\theta$

MULTIPLE CHOICE QUESTION

1 min • 1 pt

$\cot\left(-\theta\right)=$

$\tan\theta$

$-\tan\theta$

$\cot\theta$

$-\cot\theta$

MULTIPLE CHOICE QUESTION

1 min • 1 pt

$\tan\left(\frac{3\pi}{2}-\theta\right)=$

$\tan\theta$

$-\tan\theta$

$\cot\theta$

$-\cot\theta$

MULTIPLE CHOICE QUESTION

1 min • 1 pt

$\sin\left(\frac{\pi}{2}+\theta\right)=$

$\sin\theta$

$-\sin\theta$

$\cos\theta$

$-\cos\theta$

10.

MULTIPLE CHOICE QUESTION

1 min • 1 pt

$\sin\left(\pi-\theta\right)=$

$\sin\theta$

$-\sin\theta$

$\cos\theta$

$-\cos\theta$

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第一冊2-2化任意角為銳角三角函數

已知 (csc⁡θ,cot⁡θ)\left(\csc\theta,\cot\theta\right)(cscθ,cotθ) 在第二象限，則角 θ\thetaθ 在哪一象限？

設 tan⁡θ=−34\tan\theta=-\frac{3}{4}tanθ=−43​ ，且 cos⁡θ<0\cos\theta<0cosθ<0 ，則角 θ\thetaθ 在哪一象限？

設 sec⁡θ>0\sec\theta>0secθ>0 ，且 tan⁡θ<0\tan\theta<0tanθ<0 ，則角 θ\thetaθ 在哪一象限？

試問 \thetaθ 在第三象限，下列三角函數值何者為正？

tan⁡(180°+θ)=\tan\left(180\degree+\theta\right)=tan(180°+θ)=

cot⁡(−180°+θ)=\cot\left(-180\degree+\theta\right)=cot(−180°+θ)=

cot⁡(−θ)=\cot\left(-\theta\right)=cot(−θ)=

tan⁡(3π2−θ)=\tan\left(\frac{3\pi}{2}-\theta\right)=tan(23π​−θ)=

sin⁡(π2+θ)=\sin\left(\frac{\pi}{2}+\theta\right)=sin(2π​+θ)=

sin⁡(π−θ)=\sin\left(\pi-\theta\right)=sin(π−θ)=

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已知 $\left(\csc\theta,\cot\theta\right)$ 在第二象限，則角 $\theta$ 在哪一象限？

設 $\tan\theta=-\frac{3}{4}$ ，且 $\cos\theta<0$ ，則角 $\theta$ 在哪一象限？

設 $\sec\theta>0$ ，且 $\tan\theta<0$ ，則角 $\theta$ 在哪一象限？

試問 $\theta$ 在第三象限，下列三角函數值何者為正？

$\tan\left(180\degree+\theta\right)=$

$\cot\left(-180\degree+\theta\right)=$

$\cot\left(-\theta\right)=$

$\tan\left(\frac{3\pi}{2}-\theta\right)=$

$\sin\left(\frac{\pi}{2}+\theta\right)=$

$\sin\left(\pi-\theta\right)=$