Trigonometric functions and their properties

With \(\frac{3\pi}{2}\), \(\sin \theta\) changes to a cofunction and the sign is negative in the third quadrant. So the answer is \(-\cos \theta\).

With \(\frac{3\pi}{2}\), \(\cos \theta\) changes to the cofunction \(\sin \theta\). In the third quadrant, \(\cos \theta\) is negative.

With \(\frac{3\pi}{2}\), \(\tan \theta\) changes to \(\cot \theta\). In the third quadrant, \(\tan \theta\) is positive.

\(\sin\left(\frac{3\pi}{2}+\theta\right)=-\cos \theta\). In a \(\frac{3\pi}{2}\) form, \(\sin \theta\) changes to its cofunction.

\(\cos\left(\frac{3\pi}{2}+\theta\right)=\sin \theta\). In the fourth quadrant, \(\cos \theta\) is taken as positive.

With \(\frac{3\pi}{2}\), \(\tan \theta\) changes to \(\cot \theta\), and the sign is negative in the fourth quadrant. Therefore the answer is \(-\cot \theta\).

The period of (\sin \theta) is (2\pi), so adding (2\pi) does not change its value. Use periods to simplify large angles.

The cosine function has period \(2\pi\). Therefore, adding \(2\pi\) returns the point to the same position on the unit circle, giving \(\cos(\theta+2\pi)=\cos\theta\). In contrast, adding \(\pi\) gives \(\cos(\theta+\pi)=-\cos\theta\). Exam tip: for both sine and cosine, adding \(2\pi\) leaves the value unchanged.

The period of (\tan \theta) is (\pi), so (\tan(\theta+\pi)=\tan \theta). For (\tan \theta), (2\pi) is not needed.

The period of (\cot \theta) is (\pi). Therefore (\cot(\theta+\pi)) remains (\cot \theta).

The period of (\sec \theta) is (2\pi) because it is the reciprocal of (\cos \theta). Hence adding (2\pi) keeps the value same.

The period of (\cosec \theta) is (2\pi) because it is the reciprocal of (\sin \theta). The period gives the answer directly.

\(\sin(\theta+\pi)=-\sin \theta\). Adding \(\pi\) changes the sign of \(\sin \theta\).

\(\cos(\theta+\pi)=-\cos \theta\). Adding half a cycle changes the sign of \(\cos \theta\).

(\sin(\theta-\pi)=-\sin \theta). Subtracting (\pi) also changes the sign of (\sin \theta).

\(\cos(\theta-\pi)=-\cos \theta\). A difference of \(\pi\) changes the sign of \(\cos \theta\).

The period of (\tan \theta) is (\pi), so (\tan(\theta-\pi)=\tan \theta). Subtracting the period keeps the value same.

\(\sin(2\pi-\theta)=-\sin \theta\). In the fourth quadrant, \(\sin \theta\) is negative.

\(\cos(2\pi-\theta)=\cos \theta\). In the fourth quadrant, \(\cos \theta\) is positive.

\(\tan(2\pi-\theta)=-\tan \theta\). In the fourth quadrant, \(\tan \theta\) is negative.