Memastikan identitas trigonometri

Trigonometric Identities, or Trig Identities for short, are equalities that relate the six trigonometric functions (sine, cosine, tangent, cosecant, secant, and cotangent).

Let's start with the simplest: Reciprocal Identities.

$\sin(\theta)=\frac{1}{\csc(\theta)}$

$\cos(\theta)=\frac{1}{\sec(\theta)}$

$\tan(\theta)=\frac{1}{\cot(\theta)}$

$\csc(\theta)=\frac{1}{\sin(\theta)}$

$\sec(\theta)=\frac{1}{\cos(\theta)}$

$\cot(\theta)=\frac{1}{\tan(\theta)}$

As you can see, these six identities really just tell you what the reciprocal of each function is.

Try a quick problem.

Using the Reciprocal Identities, define $cosecant$ in terms of $sine$.

You should have gotten $\csc(\theta)=\frac{1}{\sin(\theta)}$.

Now let's move on to slightly more exciting identities! The Quotient Identities are pretty straightforward.

$\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$

$\cot(\theta)=\frac{\cos(\theta)}{\sin(\theta)}$

Next, we'll learn about the Pythagorean Identities.

$\sin^2(\theta)+\cos^2(\theta)=1$

$1+\tan^2(\theta)=\sec^2(\theta)$

$1+\cot^2(\theta)=\csc^2(\theta)$

These identities derive from the Pythagorean Theorem, which you can look up over here.

Now, let's talk about the Even-Odd Identities.

$\sin(-\theta)=-\sin(\theta)$

$\cos(-\theta)=\cos(\theta)$

$\tan(-\theta)=-\tan(\theta)$

$\csc(-\theta)=-\csc(\theta)$

$\sec(-\theta)=\sec(\theta)$

$\cot(-\theta)=-\cot(\theta)$

Using all the information we've gathered so far, try to solve the following problem.

Ok, you've got the first four trig identities. Now, here's the Co-Function Identities.

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

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

$\tan(\frac{\pi}{2}-\theta)=\cot(\theta)$

$\csc(\frac{\pi}{2}-\theta)=\sec(\theta)$

$\sec(\frac{\pi}{2}-\theta)=\csc(\theta)$

$\cot(\frac{\pi}{2}-\theta)=\tan(\theta)$

Here's one last one. The Double-Angle Identities can be used when you are trying to find the trig function of $(2\theta)$.

$\sin(2\theta)=2\cos(\theta)\sin(\theta)$

$\cos(2\theta)=\cos^2(\theta)-\sin^2(\theta)$

*Note that $\cos(2\theta)$ also equals $2\cos^2(\theta)-1$, which also equals $1-2\sin^2(\theta)$.

$\tan(2\theta)=\frac{2\tan(\theta)}{1-\tan^2(\theta)}$

All of these identities are very useful in solving complicated trigonometric equations or functions.

For example,

The correct answer for that problem is $\tan^2(\theta)$.

Here's how it's done:

1. We have $\csc^2(\theta)\tan^2(\theta)-1$. We can change $\csc^2(\theta)$ to $\frac{1}{\sin^2(\theta)}$ by using the Reciprocal Identities.

2. Now, we have $\frac{1}{\sin^2(\theta)}\tan^2(\theta)-1$. We can replace $\tan^2(\theta)$ with $(\frac{\sin(\theta)}{\cos(\theta)})^2$ with the Quotient Identities.

3. So far, we've simplified our original expression to $\frac{1}{\sin^2(\theta)}(\frac{\sin(\theta)}{\cos(\theta)})^2-1$. Now, we want to cancel out the $\sin^2(\theta)$ from the numerator and denominator.

4. This results in two terms: $\frac{1}{\cos^2(\theta)}-1$. We want to try and combine these two terms by multiplying the $1$ by $\frac{\cos^2(\theta)}{\cos^2(\theta)}$.

5. Almost done! By combining the two fractions we now get $\frac{1-\cos^2(\theta)}{\cos^2(\theta)}$

6. We can use our Pythagorean Identities for the next step to replace $1-\cos^2(\theta)$ with $\sin^2(\theta)$.

7. This leaves us with $\frac{\sin^2(\theta)}{\cos^2(\theta)}$ which can be simplified using the Quotient Identities to give us $\tan^2(\theta)$.

Nice job! You're ready to use these Trigonometric Identities to solve other complex problems! :)

Trigonometric Identities

There are many trigonometric identities in the world, and I'm not going to list them all here. However, I'll continue with some of the most common and important trig identites that you must know.

$\tan(x) = \frac{\sin(x)}{\cos(x)}$

This is an IMPORTANT fact. Sometimes, you might want to deal with sin and cos because most of the below identities are in sin and cos.

$\sin^2(x) + \cos^2(x)$

What is the value of $\sin^2(x) + \cos^2(x)$, where $x \in \mathbb{R}$. Let's test with some simple values.

Correct.

In fact, $\sin^2(x) + \cos^2(x) = 1$ for all real $x$! This is quite a useful fact and often known as the Pythagorean identity. It gives us a way to express sin terms in terms of cos and vice versa.

$\sin(x) = \sin(180 - x)$, $\sin(x) = -\sin(-x)$ and $\cos(x) = \cos(-x)$.

I'll not go into further detail on this. They're not really hard to prove.

$\sin(x) = \cos(90 - x), \cos(x) = \sin(90 - x)$ and $\tan(90 - x)$.

These three identities are again useful. If you list the values of $\sin(30), \sin(45), \sin(60)$, you'll notice that they're equal to $\cos(60), \cos(45), \cos(30)$ respectively.

And indeed, $\sin(x) = \cos(90 - x)$ and $\cos(x) = \sin(90 - x)$ for all real $x$.

Ineed, $\tan(90 - x) = \frac{1}{\tan(x)}$. Again, the proof is not really hard. We can in fact rewrite it as $\tan(90 - x) = \cot(x)$. (Recall that $\cot(x) = \frac{1}{\tan(x)}$.)

Addition Formulas and Subtraction Formulas

This is the most important trig identity in this article.

Addition Formulas :

$\sin(a + b) = \sin(a)\cos(b) + \sin(b)\cos(a)$ $\cos(a + b) = \cos(a)\cos(b) - \sin(a)\sin(b)$

So, we have $\sin(a - b) = \sin(a)\cos(b) - \sin(b)\cos(a)$, and $\cos(a - b) = \cos(a)\cos(b) + \sin(a)\sin(b)$, which are also known as the subtraction formulas.

These two identites gives

Double Angle Formula, Triple Angle Formula and Half Angle Formula.

From the above, we obtain :

and

Now, you might ask why are there three formulas. Note that these three expressions are in fact equal, by the Pythagorean Identity.

Similarly, we can obtain the Triple-angle and Half-angle Formulas :

Triple Angle Formulas :

$\sin(3a) = 3\sin(a) - 4\sin^3(a)$

$\cos(3a) = 4\cos^3(a) - 3\cos(a)$

Half Angle Formulas :

$\sin^2(\frac{1}{2}a) = \frac{1 - \cos(a)}{2}$

$\cos^2(\frac{1}{2}a) = \frac{1 + \cos(a)}{2}$

There're more trig identities, but you can derive them yourself from the above identities. Happy triging!

$$\large{\textbf{Complex Numbers in Proving Trig Identities}}$$ $$\large{\textbf{Basic Facts}}$$

We define the complex numbers $\mathbb{C}$ as $$\mathbb{C}=\left\{a+bi~|~(a,b)\in\mathbb{R}^2\right\}, ~~~~~i^2=-1.$$ Here in a complex number $z=a+bi$, $a$ is called the real part, written $\Re(z)=a$ and $b$ is called the imaginary part, written $\Im(z)=b$. One good thing about dealing with complex numbers in equations is that one can equate the real and imaginary parts. For some real parameters $A,B,C,D$ we have $$A+Bi=C+Di ~~~\Longleftrightarrow ~~~A=C,~ B=D.$$ Another helpful thing is the distributivity or linearity of the $\Re$ or $\Im$ functions. We have $$\Re\sum=\sum\Re,~~~~~\Im\sum=\sum\Im.$$ $$$$

$$\large{\textbf{Euler's Formula}}$$

The following formula was given by Leonhard Euler and is a very useful one relating complex numbers and trigonometry. For any complex $x$ we have $$e^{ix}=\cos x+i\sin x.$$ There are many ways to prove this. One outline is included here: first, rewrite the equation as $$\dfrac{\cos x+i\sin x}{e^{ix}}=1,$$ consider the function $y=e^{-ix}(\cos x+i\sin x)$, differentiate it. The works are not shown here, but the derivative turns out to be $0$ so the function must be constant. Plug in $x=0$ to get $y=1$, hence $y=1$ for all $x$. Notice that setting $x=\pi$ we get a beautiful identity $e^{i\pi}+1=0$.

Note: If you are familiar with calculus, there is also another proof involving a manipulation of the Taylor Series of sine and cosine. $$$$

$$\large{\textbf{De Moivre's Formula}}$$

De Moivre's Formula helps ease out computations a lot. It states that for any complex $x$ and integer $n$ $$\left(\cos x+i\sin x\right)^n=\cos nx +i\sin nx.$$ The proof is one liner if we apply Euler's formula twice as below $$\left(\cos x+i\sin x\right)^n=\left(e^{ix}\right)^n=e^{i(nx)}=\cos nx +i\sin nx.$$ Notice that in the complex number $z=e^{ix}=\cos x+i\sin x$, we have $\Re(z)=\cos x$ and $\Im(z)=\sin x$. $$$$

$$\large{\textbf{Complex Approaches to Trig Problems!}}$$

$\textbf{1.}$ Derive the addition formulas of sine and cosine.

$\textbf{Solution}.$ Just notice that by applications of Euler's formula

Now equating real and imaginary part of both sides, we have $$\cos(x+y)=\cos x\cos y-\sin x \sin y$$ $$\sin(x+y)=\sin x \cos y+\cos x\sin y$$ deriving the formulas. $\square$ $$$$ $\textbf{2.}$ Prove that $$\sum_{k=0}^{n-1} \cos \dfrac{2k\pi}{n}=0 ~~~~~\forall~n\in\mathbb{N}-\{1\}.$$

$\textbf{Solution}.$ Let $z=e^{2\pi i/n}=\cos (2\pi/n)+i\sin (2\pi/n)$. Then we have $z^n=e^{2\pi i}=1\implies z^n-1=0$. Since $z\neq 1$ $$z^n-1=(z-1)\left(1+z+z^2+\cdot\cdot\cdot+z^{n-1}\right)=0$$ $$\therefore ~~\sum_{k=0}^{n-1} z^k=0~\implies \Re\sum_{k=0}^{n-1} z^k=0.$$ Now notice that by application of De Moivre's formula $$z^k=\left(\cos\dfrac{2\pi}{n}+i\sin\dfrac{2\pi}{n}\right)^k=\cos\dfrac{2k\pi}{n}+i\sin\dfrac{2k\pi}{n}$$ $$\therefore ~~ \Re\left(z^k\right)=\cos\dfrac{2k\pi}{n}.$$ Therefore by the linearity of $\Re$ function we finally have $$0=\Re\sum_{k=0}^{n-1} z^k=\sum_{k=0}^{n-1} \Re\left(z^k\right)=\sum_{k=0}^{n-1} \cos\dfrac{2k\pi}{n}. \square$$

$\textbf{3.}$ Prove the Pythagorean identity, $\cos^2{x} + \sin^2{x} = 1$.

$\textbf{Solution.}$ We can cleverly factor $\cos^2{x} + \sin^2{x}$ as a difference of squares: $(\cos{x} + i\sin{x})(\cos{x} - i\sin{x})$. From here, we use Euler's identity, so we have $\cos^2{x} + \sin^2{x}=(\cos{x} + i\sin{x})(\cos{x} - i\sin{x})=(e^{ix})(e^{-ix})=1.$ $\square$ $$$$

$$\large{\textbf{Two Problems To Try}}$$

$\textbf{1. (IMO 1963/5)}$ Prove that $$\cos\dfrac{\pi}{7}-\cos\dfrac{2\pi}{7}+\cos\dfrac{3\pi}{7}=\dfrac{1}{2}.$$

$\textbf{2. (Proofathon)}$ Show that for all odd $k\ge 3$ $$\cos\dfrac{\pi}{k}+\cos\dfrac{3\pi}{k}+\cdots+\cos\dfrac{(k-2)\pi}{k}=\dfrac{1}{2}.$$ ($\textbf{1}$ is a special case of $\textbf{2}$!)

Sometimes, where working with a trigonometric expression, it is useful to replace all the $\sin \theta$ and $\cos\theta$ with a rational expression in terms of a single variable. As it turn out, $\tan\frac{\theta}{2}$ is a convenient choice.

Recall from the Double Angle Identity section that: $$\cos \theta=2\cos^2\frac{\theta}{2}-1$$ Since $\cos^2\frac{\theta}{2}+\sin^2\frac{\theta}{2}=1$, this can also be written as: $$\cos \theta=\frac{2\cos^2\frac{\theta}{2}}{\cos^2\frac{\theta}{2}+\sin^2\frac{\theta}{2}}-1=\frac{2}{1+\tan^2\frac{\theta}{2}}-1=\frac{1-\tan^2\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}}$$ Now $\sin\theta$ can be dealt with in a similar fashion. Try to derive an expression for $\sin\theta$ in terms of $\tan\frac{\theta}{2}$

We start in the same way. Recall that: $$\sin\theta=2\sin\frac{\theta}{2}\cos\frac{\theta}{2}$$ This can be rewritten using $\sin\frac{\theta}{2}=\tan\frac{\theta}{2}\cos\frac{\theta}{2}$: $$\sin\theta=2\tan\frac{\theta}{2}\cos^2\frac{\theta}{2}$$ Now we finish exactly as before: $$\sin\theta=(2\tan\frac{\theta}{2})\frac{\cos^2\frac{\theta}{2}}{\cos^2\frac{\theta}{2}+\sin^2\frac{\theta}{2}}=\frac{2\tan\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}}$$ Using these two expressions, we can derive similar expressions for $\csc\theta, \sec\theta, \tan\theta, \cot\theta$. The result is the following. If $\tan\frac{\theta}{2}=t$, then: $$\sin\theta=\frac{2t}{1+t^2}$$ $$\cos\theta=\frac{1-t^2}{1+t^2}$$ $$\tan\theta=\frac{2t}{1-t^2}$$ $$\cot\theta=\frac{1-t^2}{2t}$$ $$\sec\theta=\frac{1+t^2}{1-t^2}$$ $$\csc\theta=\frac{1+t^2}{2t}$$ Using these identities, any trigonometric identity with a single argument can be reduced to a polynomial verification in one variable. Additionally, this characterization can be used to prove trigonometric inequalities.

We will now work a brief example illustrating the former use.

Let $\theta\in (0, \frac{\pi}{2})$ be an angle. Simplify the following: $$\left(\frac{\csc\theta-\sin\theta}{\sin\theta+\cos\theta+1}\right)\left(\frac{\sec\theta-\cos\theta}{\sin\theta+\cos\theta-1}\right)$$

We let $t=\tan\frac{\theta}{2}$. Then we can simply plug in the above identities into the expression: $$\left(\frac{\frac{1+t^2}{2t}-\frac{2t}{1+t^2}}{\frac{2t}{1+t^2}+\frac{1-t^2}{1+t^2}+1}\right)\left(\frac{\frac{1+t^2}{1-t^2}-\frac{1-t^2}{1+t^2}}{\frac{2t}{1+t^2}+\frac{1-t^2}{1+t^2}-1}\right)$$ Simplifying a little, this becomes: $$\left(\frac{(1+t^2)^2-(2t)^2}{2t(2t+2)}\right)\left(\frac{(1+t^2)^2-(1-t^2)^2}{(1-t^2)(2t-2t^2)}\right)$$ Using the fact that $(1-t^2)^2+(2t)^2=(1+t^2)^2$, this can be further simplified: $$\left(\frac{(2t)^2(1-t^2)^2}{2(2t)(t+1)(1-t^2)(2t)(1-t)}\right)$$ $$=\left(\frac{4t^2(1+t)^2(1-t)^2}{8t^2(1+t)^2(1-t)^2}\right)$$ $$=\frac{1}{2}$$ And so the expression is simply $\frac{1}{2}$.

Some of us have seen the formula $e^{i \theta} = \cos \theta + i \sin \theta$ before (if you haven't, check it out here!). However, this doesn't really help us find the value of $\cos \theta$ or $\sin \theta$ individually.

The best we can do is $\Re [e^{i \theta}] = \cos \theta$ and $\Im [e^{i \theta}] = i \sin \theta$. (For a complex number $z = a+bi$, we denote $\Re(z) = a$, the real part, and $\Im(z) = b$, the imaginary part.)

If we plugged another value of $\theta$ into the equation, maybe we could get a system and solve for $\cos \theta, \sin \theta$.

If we plug in $- \theta$, we get the equation $e^{-i \theta} = \cos (-\theta) + i \sin (-\theta)$.

Now notice that $\cos (- \theta) + i \sin (-\theta) = \cos \theta - i \sin \theta$! If we put together our two equations

$e^{i \theta} = \cos \theta + i \sin \theta$ and

$e^{-i \theta} = \cos \theta - i \sin \theta$, we get a system that allows us to solve for $\cos \theta$ and $\sin \theta$!

By adding the two equations, we find $\cos \theta = \dfrac{e^{i \theta} + e^{- i \theta}}{2}$.

These are two very useful formulas for $\cos, \sin$. Let's just do a sanity check and make sure that $\cos^2 \theta + \sin^2 \theta = 1$:

$\left( \dfrac{e^{i \theta} + e^{- i \theta}}{2} \right)^2 + \left( \dfrac{e^{i \theta} - e^{-i \theta}}{2i} \right)^2$ equals

$\dfrac{1}{4} (e^{2i \theta} + e^{-2 i \theta} + 2e^0 - e^{2i \theta} - e^{- 2 i \theta} + 2e^0) = 1$, as expected!

There's still a bit more we can do though: If $e^{i \theta} = z$, then $e^{- i \theta} = \frac{1}{z}$. Thus we may simplify the formulas to

$\cos \theta = \dfrac{1}{2} \left( z + \dfrac{1}{z} \right)$ and

$\sin \theta = \dfrac{1}{2i} \left(z - \dfrac{1}{z} \right)=-\dfrac{i}{2} \left(z - \dfrac{1}{z} \right)$

In this explanation we're going to prove the identity

$$\tan(2x) = \frac{2\sin(x)}{2\cos(x) - \sec(x)}.$$

In doing so we're going to highlight some of the techniques you can use to prove other identities. Every time we prove an identity, we can use it to derive other ones.

A good first attempt is to work from left-to-right, trying to turn the simpler-looking side of the identity into the more complicated-looking side. In this case the left-hand side is less intimating so we start there.

We know more about sines and cosines than tangents, and they are easier to manipulate, so we write

$$\tan(2x) = \frac{\sin(2x)}{\cos(2x)}.$$

The reason this is helpful is we know double-angle identities for sine and cosine:

$$\begin{align} \sin(2x) &= 2\sin(x)\cos(x) \\ \cos(2x) &= 2\cos^2(x) - 1 \end{align}$$

We substitute these into our previous equation:

$$\tan(2x) = \frac{2\sin(x)\cos(x)}{2\cos^2(x) - 1}$$

This fraction is starting to resemble the right-hand side of the identity we wanted to prove. The numerator we have now and the numerator we want differ by a factor of $\cos(x),$ so divide the numerator and denominator by $\cos(x):$

$$\tan(2x) = \frac{\frac{2\sin(x)\cancel{\cos(x)}}{\cancel{\cos(x)}}}{\frac{2\cos^2(x) - 1}{\cos(x)}} = \frac{2\sin(x)}{2\cos(x) - \sec(x)}.$$

This is the same as the right-hand side of the identity we wanted to prove, so we're all done! We've successfully proven a trig identity!

One of the great things about trig identities is that there are multiple ways to prove most, if not all of them. Perhaps you can look around the other explanations and get more ideas for how to prove this identity.

Verifying Trigonometric Identities

by Professor Dave Explains

In this video by Professor Dave Explains, you will be able to prove and verify the trigonometric identities.

Creating New Trig Identities from Old Ones

Trig identities are math rules that are always true for certain angle functions like sine, cosine, and tangent. We've already learned some of the basic ones, like:

$$\text{sin}^2(\theta) + \text{cos}^2(\theta)=1$$

You can use the identities you already know to create new ones!

As we can see in the above example, we can divide both sides by $\text{cos}^2(\theta)$. So, we can simplify to:

$$\text{tan}^2(\theta) + 1 = \text{sec}^2(\theta)$$

...and now you've just made a new identity! We used what we already knew and turned it into something new. You can also apply this new identity in real problems, like proving that two sides of an equation are equal.

You can use it to simplify an expression like $\text{sec}^2(\theta) + \text{tan}^2(\theta)$, which we now know equals $1$.

You can use it to work backward: if you see $\text{tan}^2(\theta)$, you can replace it with $\text{sec}^2(\theta)-1$ to make solving easier.

It's like having a toolbox - the more complex the tools, the more ways you can solve things.