Question

A Reduction Formulas, A reduction formula is one that can be used to “reduce” the number of terms in the input for a trigonometric function. Explain how the figure shows that the following reduction formulas are valid:

$$\begin{gathered} \sin \left(t+\pi \right)=-\sin t\cos \left(t+\pi \right)=-\cos t \\ \tan \left(t+\pi \right)=\tan t \end{gathered}$$

Short answer

The given formula has been proved.

Step-by-step solution

Given information.

The figure is given as,

Explanation.

Consider the given information.

Proof 1: By using the quadrant.

Since the given angle $\left(\pi +t\right)$will lie in the third quadrant. In the third quadrant, sine and cosine are negative and tangent are positive. Thus,

$$\begin{gathered} \sin \left(t+\pi \right)=-\sin t \\ \cos \left(t+\pi \right)=-\cos t \\ \tan \left(t+\pi \right)=\tan t \end{gathered}$$

Proof 2: By using the formulas.

Now, apply the sine formula over the right side $\sin \left(t+\pi \right)$.

$\sin \left(t+\pi \right)=\sin t\cos \pi +\sin \pi \cos t$

$$\begin{gathered} =\sin t\left(-1\right)+\left(0\right)\cos t \\ =-\sin t \end{gathered}$$

And,

$\cos \left(t+\pi \right)=\cos t\cos \pi -\sin t\sin \pi$

$$\begin{gathered} =\cos t\left(-1\right)-\sin t\left(0\right) \\ =-\cos t \end{gathered}$$

And,

$\tan \left(t+\pi \right)=\frac{\tan t+\tan \pi }{1-\tan t\tan \pi }$

$$\begin{gathered} =\frac{\tan t+0}{1-\tan t\left(0\right)} \\ =\frac{\tan t}{1} \\ =\tan t \end{gathered}$$

Hence proved.