Question

a. Use the identity $\sin (a+h)=\sin a\cos h+\cos a\sin h$ with the fact that $\underset{x\to 0}{lim}\sin x=0$ to prove that $\underset{x\to a}{lim}\sin x=\sin a$ thereby establishing that $\sin x$ is continuous for all $x$. (Hint: Let $h=x-a$ so that $x=a+h$ and note that $h\to 0$ as $x\to a$.) b. Use the identity $\cos (a+h)=\cos a\cos h-\sin a\sin h$ with the fact that $\underset{x\to 0}{lim}\cos x=1$ to prove that $\underset{x\to a}{lim}\cos x=\cos a$.

Short answer

Question: Prove that sine and cosine functions are continuous for all x using given trigonometric identities and limits. Answer: We showed that the limit of the sine function as x approaches a is equal to sine of a, and the limit of the cosine function as x approaches a is equal to cosine of a, using the given trigonometric identities and limits. Therefore, both sine and cosine functions are continuous for all x.

Step-by-step solution

a. Proving sine function continuity at x=a

Using the identity $\sin (a+h)=\sin a\cos h+\cos a\sin h$, we need to prove that $\underset{x\to a}{lim}\sin x=\sin a$ by letting $h=x-a$ and utilizing the information that $\underset{x\to 0}{lim}\sin x=0$. First, let's rewrite the expression using the provided hint: $\sin (x)=\sin (a+h)=\sin a\cos (x-a)+\cos a\sin (x-a)$. Now we can analyze the limit by breaking it down into two parts: $\underset{x\to a}{lim}\sin x=\underset{x\to a}{lim}[\sin a\cos (x-a)+\cos a\sin (x-a)]$. Use the limit properties by splitting the limit into two parts: $\underset{x\to a}{lim}\sin a\cos (x-a)+\underset{x\to a}{lim}\cos a\sin (x-a)$. As $x\to a$, we have $(x-a)\to 0$. So, we can rewrite our expression using this fact: $\sin a\underset{x\to a}{lim}\cos (x-a)+\cos a\underset{x\to a}{lim}\sin (x-a)$. Now, we know that $\underset{x\to 0}{lim}\cos x=1$ and $\underset{x\to 0}{lim}\sin x=0$. So, our expression becomes: $\sin a(1)+\cos a(0)=\sin a$. Thus, we have proved that $\underset{x\to a}{lim}\sin x=\sin a$, and the sine function is continuous for all x.

b. Proving cosine function continuity at x=a

Using the identity $\cos (a+h)=\cos a\cos h-\sin a\sin h$, we need to prove that $\underset{x\to a}{lim}\cos x=\cos a$ by letting $h=x-a$ and using the fact that $\underset{x\to 0}{lim}\cos x=1$. First, rewrite the expression using the provided hint: $\cos (x)=\cos (a+h)=\cos a\cos (x-a)-\sin a\sin (x-a)$. Now we can analyze the limit by breaking it down into two parts: $\underset{x\to a}{lim}\cos x=\underset{x\to a}{lim}[\cos a\cos (x-a)-\sin a\sin (x-a)]$. Use the limit properties by splitting the limit into two parts: $\underset{x\to a}{lim}\cos a\cos (x-a)-\underset{x\to a}{lim}\sin a\sin (x-a)$. As $x\to a$, we have $(x-a)\to 0$. So, we can rewrite our expression using this fact: $\cos a\underset{x\to a}{lim}\cos (x-a)-\sin a\underset{x\to a}{lim}\sin (x-a)$. Now, we know that $\underset{x\to 0}{lim}\cos x=1$ and $\underset{x\to 0}{lim}\sin x=0$. So, our expression becomes: $\cos a(1)-\sin a(0)=\cos a$. Thus, we have proved that $\underset{x\to a}{lim}\cos x=\cos a$, and the cosine function is continuous for all x.

Key concepts

Sine Function

The sine function, often denoted as $\sin (x)$, is a fundamental concept in trigonometry. It describes a periodic wave, which is crucial for understanding various physical phenomena. In our context, the sine function is continuous, meaning it has no sudden jumps or breaks in its graph.
The sine function is defined by how it relates to the sides of a right triangle. In a unit circle, the sine of an angle $\theta$ is the y-coordinate of the point on the unit circle corresponding to that angle.
Continuity in mathematics means that as you approach a point $x=a$, the function $\sin (x)$ approaches $\sin (a)$. This is validated by the identity $\sin (a+h)=\sin a\cos h+\cos a\sin h$, along with the known limits $\underset{x\to 0}{lim}\sin x=0$ and $\underset{x\to 0}{lim}\cos x=1$.

• By calculating these limits, we find that $\underset{x\to a}{lim}\sin x=\sin a$, proving that the sine function is smooth and unbroken at every point.

Cosine Function

Just like the sine function, the cosine function is a pillar of trigonometry, expressed as $\cos (x)$. It illustrates another periodic wave, often representing horizontal or vertical displacement, and is crucial in phenomena like oscillations and rotations.
For a given angle $\theta$ in a right triangle or unit circle, the cosine is represented by the x-coordinate. It's an even function, meaning $\cos (-x)=\cos (x)$.
Using continuity, we show that $\cos (x)$ has no interruptions in its progression. The identity $\cos (a+h)=\cos a\cos h-\sin a\sin h$ coupled with the limit laws $\underset{x\to 0}{lim}\cos x=1$ and $\underset{x\to 0}{lim}\sin x=0$ supports finding that $\underset{x\to a}{lim}\cos x=\cos a$.

• This confirms that for every $a$, the cosine function smoothly continues as x approaches a.

Limit Properties

Limit properties form the backbone of calculus and analysis. They help in understanding how functions behave as variable inputs get infinitesimally close to a point. For trigonometric functions, limit properties help ensure continuity.
A limit, like $\underset{x\to a}{lim}f(x)=L$, implies that as $x$ approaches $a$, $f(x)$ gets arbitrarily close to $L$.
For sine and cosine functions, critical limits are:$$\underset{x\to 0}{lim}\sin x=0$$, stating that sine diminishes to zero as its input approaches zero, and $$\underset{x\to 0}{lim}\cos x=1$$, indicating cosine holds steady at 1.

• These limits are pivotal when proving the continuity of sine and cosine functions using identities. By breaking these down with identity transformations, we apply these limit properties, attaining results like $\underset{x\to a}{lim}\sin x=\sin a$ and $\underset{x\to a}{lim}\cos x=\cos a$.