Problem 2

Question

Show that for any $x \in \mathbb{R}$ $$ \sin \left(\frac{\pi}{2}-x\right)=\cos (x) $$ and $$ \cos \left(\frac{\pi}{2}-x\right)=\sin (x) $$

Step-by-Step Solution

Verified

Answer

The identities are proven using complementary angle theorem.

1Step 1: Understanding the Problem

We need to verify the trigonometric identities involving complementary angles. Specifically, we need to prove that $ \sin \left(\frac{\pi}{2}-x\right)=\cos (x) $ and $ \cos \left(\frac{\pi}{2}-x\right)=\sin (x) $. These identities relate the sine and cosine of complementary angles.

2Step 2: Reviewing Complementary Angle Theorem

Recall that complementary angles in trigonometry have the property that their sine and cosine are swapped because $ \sin(\theta) = \cos\left(\frac{\pi}{2} - \theta\right) $ and $ \cos(\theta) = \sin\left(\frac{\pi}{2} - \theta\right) $. We will use this property to prove our identities.

3Step 3: Deriving the First Identity

Start by expressing $ \sin\left(\frac{\pi}{2} - x\right) $. According to the complementary angle theorem, $ \sin\left(\frac{\pi}{2} - x\right) = \cos(x) $. Therefore, we have $ \sin\left(\frac{\pi}{2} - x\right) = \cos(x) $, which is exactly the first identity we are proving.

4Step 4: Deriving the Second Identity

Now, consider $ \cos\left(\frac{\pi}{2} - x\right) $. Again, using the complementary angle theorem, we can see that $ \cos\left(\frac{\pi}{2} - x\right) = \sin(x) $. Thus, it proves the second identity: $ \cos\left(\frac{\pi}{2} - x\right) = \sin(x) $.

Key Concepts

Complementary AnglesSine FunctionCosine Function

Complementary Angles

Complementary angles are two angles whose measures add up to 90 degrees or $ \frac{\pi}{2} $ radians. It's like having two pieces of a whole when it comes to right angles. In trigonometry, complementary angles have a special relationship because the sine of one angle equals the cosine of its complement, and vice versa. This is depicted by the identities:

$ \sin(\theta) = \cos\left(\frac{\pi}{2} - \theta\right) $
$ \cos(\theta) = \sin\left(\frac{\pi}{2} - \theta\right) $

Using these properties, we can verify the trigonometric identities we are given.
They show how wonderfully intertwined sine and cosine functions are when dealing with complementary angles.

Sine Function

The sine function is one of the basic trigonometric functions. It is often abbreviated as "sin". For a given angle $ \theta $, $ \sin(\theta) $ represents the ratio of the length of the side of the opposite angle to the length of the hypotenuse in a right-angled triangle.

Its value ranges between -1 and 1.
It is periodic with a period of $ 2\pi $.

In the context of this exercise, when we take $ \sin\left(\frac{\pi}{2} - x\right) $, it's as if we are exploring how a sine function reacts when its input is rearranged by a complementary angle.
By learning that $ \sin\left(\frac{\pi}{2} - x\right) = \cos(x) $, we reinforce our understanding of how complementary angles affect the sine function.

Cosine Function

The cosine function, often abbreviated as "cos", is also a fundamental trigonometric function. For an angle $ \theta $, $ \cos(\theta) $ is the ratio of the adjacent side over the hypotenuse in a right-angled triangle.

Its range is also between -1 and 1.
It has a period of $ 2\pi $, just like the sine function.

The cosine function exhibits a co-function identity when dealing with complementary angles. This means

$ \cos\left(\frac{\pi}{2} - x\right) = \sin(x) $

By verifying this identity, we see how the cosine of a complementary angle can mimic the behavior of the sine function.
This is a fundamental aspect of trigonometry that assists in solving many problems involving angles and lengths.

Problem 2

Problem 3

Other exercises in this chapter

Problem 2

Define $f:(0,+\infty) \rightarrow \mathbb{R}$ by $f(x)=x^{a},$ where $a \in \mathbb{R}, a \neq 0$ Show that $f^{\prime}(x)=a x^{a-1}$.

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Problem 2

Show that $$ \lim _{x \rightarrow 0^{+}} x^{\alpha} \log (x)=0 $$ for any rational number $\alpha>0$.

View solution

Problem 3

Suppose $a$ is a positive real number and $f: \mathbb{R} \rightarrow \mathbb{R}$ is defined by $f(x)=a^{x}$. Show that $f^{\prime}(x)=a^{x} \log (a)$.

View solution

Problem 3

Show that for any $x \in \mathbb{R}$ $$ \sin (2 x)=2 \sin (x) \cos (x) $$ and $$ \cos (2 x)=\cos ^{2}(x)-\sin ^{2}(x) $$

View solution