Question

Establish each identity.

$\sin \left(\theta +k\mathrm{\pi }\right)={(-1)}^k\sin \theta ,k$any integer

Short answer

The given equation is identity.

Step-by-step solution

Step 1. Given information

The given equation is$\sin \left(\theta +k\mathrm{\pi }\right)={(-1)}^k\sin \theta ,k$any integer.

Step 2. Use L.H.S. to solve

Use the sum formula for the sine function,

$\sin \left(\theta +k\mathrm{\pi }\right)=\sin \theta ·\cos \left(k\mathrm{\pi }\right)+\cos \theta ·\sin \left(k\mathrm{\pi }\right)$

Step 3. Values of coskπ where k can be any integer.

When $k=1$,

role="math" localid="1646548827797" $$\begin{gathered} \cos \left(k\mathrm{\pi }\right) \\ =\mathrm{cos\pi } \\ =-1 \end{gathered}$$

When $k=2$,

role="math" localid="1646549059159" $$\begin{gathered} \cos \left(k\mathrm{\pi }\right) \\ =\cos 2\mathrm{\pi } \\ ={(-1)}^2 \\ =1 \end{gathered}$$

When $k=3$,

role="math" localid="1646549217406" $$\begin{gathered} \cos \left(k\mathrm{\pi }\right) \\ =\cos 3\mathrm{\pi } \\ ={(-1)}^3 \\ =-1 \end{gathered}$$

Hence,$\cos \left(k\mathrm{\pi }\right)={(-1)}^k$

Step 4. Values of sinkπ where k can be any integer.

When $k=1$,

$$\begin{gathered} \sin \left(k\mathrm{\pi }\right) \\ =\mathrm{sin\pi } \\ =0 \end{gathered}$$

When $k=2$,

$$\begin{gathered} \sin \left(k\mathrm{\pi }\right) \\ =\sin 2\mathrm{\pi } \\ =0 \end{gathered}$$

Hence,$\sin \left(k\mathrm{\pi }\right)=0$for any value of$k$.

Step 5. Rewrite the expression

$$\begin{gathered} \sin \left(\theta +k\mathrm{\pi }\right) \\ =\sin \theta ·\cos \left(k\mathrm{\pi }\right)+\cos \theta ·\sin \left(k\mathrm{\pi }\right) \\ =\sin \theta ·{(-1)}^k+\cos \theta ·0 \\ ={(-1)}^k·\sin \theta \end{gathered}$$

Since $L.H.S.=R.H.S.$, the given equation is identity.

Therefore, the given identity is established.