Chapter 4.5, Problem 44E

Solution

a.

To prove: $\sin \left(t+\pi \right)=-\sin t$

Given information:

The unit circle $x^2+y^2=1$ .

Proof:

Graph:

  

Considering the right angle triangle BDA,

  $\begin{array}{c}\sin \theta =\frac{\text{Opposite}}{\text{Hypotenuse}}\\ \sin t=\frac{b}{h}\end{array}$

Considering the right angle triangle CFA,

  $\begin{array}{c}\sin \theta =\frac{\text{Opposite}}{\text{Hypotenuse}}\\ \sin \left(t+\pi \right)=\frac{-b}{h}\\ \sin \left(t+\pi \right)=-\sin t\end{array}$

Therefore, the identity $\sin \left(t+\pi \right)=-\sin t$ has been proved.

b.

To prove: $\cos \left(t+\pi \right)=-\cos t$

Given information:

The unit circle $x^2+y^2=1$ .

Proof:

Graph:

  

Considering the right angle triangle BDA,

  $\begin{array}{c}\cos \theta =\frac{\text{Adjacent}}{\text{Hypotenuse}}\\ \cos t=\frac{a}{h}\end{array}$

Considering the right angle triangle CFA,

  $\begin{array}{c}\cos \theta =\frac{\text{Adjacent}}{\text{Hypotenuse}}\\ \cos \left(t+\pi \right)=-\frac{a}{h}\\ \cos (t+\pi )=-\cos t\end{array}$

Therefore, the identity $\cos \left(t+\pi \right)=-\cos t$ has been proved.

c.

To prove: $\tan \left(t+\pi \right)=-\tan t$

Given information:

The unit circle $x^2+y^2=1$ .

Proof:

Using the tangent formula to get,

  $\tan \left(t+\pi \right)=\frac{\sin \left(t+\pi \right)}{\cos \left(t+\pi \right)}$

Taking the result of part (a) and part (b),

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

Substituting the values in above equation to get,

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

Therefore, the identity $\tan \left(t+\pi \right)=\tan t$ has been proved.

The equation for the period of the function,

  $y=a\tan \left(bx\right)$

From this the period of function is $\frac{\pi }{\left|b\right|}$ .

Therefore, it is not enough to conclude the period of tangent $\pi$ .