First of all, we have to understand what exactly is inverse cosine and how to write them in mathematics. Trigonometric functions can be found for well-defined intervals, even though they are not one-to-one functions. Some of the more important inverse trigonometric functions are inverse tan, inverse sine, and inverse cosine. Let us look into the inverse cosine in this article.
Definition and Representation of Inverse Cosine
Presume we are given that cos(a) = x. This implies cos-1 (x)=a. (This reads as cos inverse of x).That is inverse of cosine of a number x is a; where a is the angle for which the cosine gave the number x. To be more clear let us take an example. What is cos-1 (0.5)?
We know ,cos (π/3) = 0.5. That is cos-1 (0.5) =π/3.
As stated earlier, to have inverses, the functions should have a well-defined interval. In the case of inverse of cosine of x, x needs to lie between -1 and 1 and the angle a should lie between 0 and π.
Hence we can say the following,
The principal domain of inverse cosine is [-1,1]
The range corresponding to the principle domain is [0,π]
Using this information we can find values for some inverse cosines. They are
Cos-1 (-1)=π
Cos-1 (0)=π/2
Cos-1 (1)=0
Graphical Representation
First, let us look at a graphical representation of cos x. This will make it easier to appreciate the difference between the cosine function and the inverse cosine function.
[Image will be Uploaded Soon]
Now let us look at the graph inverse trigonometric function including inverse cosine.
Some Important Properties of Inverse Cosine
Just like the trigonometric functions, inverse trigonometric functions have important properties and formulas which are essential for understanding these functions thoroughly.
Let us look at some of the properties using some examples.
Example 1:
Think we are given a question to find the answer to sec -1 (2). Can we do it using the formulas we have learned so far?
We know that cos(θ) = 1/sec(θ).
Similarly,
cos-1 (x) = sec-1(1/x), x∈[-1,1] – {0}
Hence, we can easily find the answer to our question.
As sec-1 (2) = cos-1(1/2) which is equal to cos-1(0.5 )
= π/3, we can write sec-1(2) = π/3.
Example 2:
Suppose we need to find the value of cos-1 (-1/2). How do we do this problem?
We know the value of cos-1(1/2) as π/3. We also know that this means cos(π/3) =1/2.
If we multiply both sides by -1,
we get –cos(π/3) = -1/2.
Hence we can write
cos-1 (-1/2) = -π/3.
But our range is from 0 to π and –π/3 is outside our range.
However from trigonometry we know –π/3 = π-π/3 = 2π/3.
That is we have found out the answer to cos-1(-1/2) = π-cos-1(1/2) = 2π/3.
We can generalize this into a rule as:
Cos-1 (-x) = π-cos-1(x)
Example 3:
Now let us look into two very important property.
1)Cos-1 (cos(θ)) = θ,if θ∈[0,π]
2)Cos(cos-1(x)) = x, if x ∈ [-1,1]
These properties though important are very easy to understand. Let us look at the first property.
Suppose cos(θ) = x. This means cos-1(x) = θ.
Hence we can write Cos-1 (cos(θ)) = θ,if θ∈[0,π]. Similarly, we can explain the second property too.
Let us try and solve some problems applying these properties. Find the value of tan(cos-1(3/5)).
Example 4: Let us suppose cos-1(3/5) = θ; This means cos(θ)=3/5. So if we draw a triangle as shown below,
5
4 33 \[\angle5\]
3
We can find the third side using Pythagoras theorem as \[\sqrt{5^{2}-3^{2}}\] = \[\sqrt{16}\] = 4.
Now our question is to just find out tan(θ).
That is, tan(θ)=tan(cos-1(3/5)) = 3/4
And we have obtained our answer as 3/4.
Now let us look at some properties that can make problems like this easier for us.
Example 4:
1)Cos-1(sin(θ)) = π/2-θ; if θ∈[0,π]
2)Sin-1(cos(θ)) = π/2-θ; if θ ∈ [-π/2,π/2]
3)Sin-1(x) =} \[\ cos^{-1}\] (\[\sqrt{1-x^{2}}\]),0≤x≤1
\[\ -cos^{-1}\] (\[\sqrt{1-x^{2}}\]),-1≤x≤0
The first and second properties are not that hard to explain. We know sin(θ) = cos(π/2-θ) and vice versa. Applying our first property from the third section, we can easily see that Cos-1(sin(θ)) = π/2-θ and vice versa. To explain the third property we have to recall some of the properties of trigonometric functions.
Let sin(θ) = x. That is sin-1(x) =θ; We know,
sin2(θ) +cos2(θ)=1; That is, cos2 (θ)=1-sin2 (θ) = 1-x2.
Substituting for θ, cos2(sin-1(x)) = 1-x2; taking square roots,we get, Sin-1(x) = cos-1(±(\[\sqrt{1-x^{2}}\])), and hence we arrive at our result.
We can find out yet another property from this discussion.
4)Sin-1(x)+cos-1(x)=π/2
This can be easily explained using the points discussed while understanding our third property.
Some Important Identities
The identities of inverse cosines are analogous to that of the cosine function. Suppose we have to find cos-1(1/\[\sqrt{2}\])+cos-1(1/\[\sqrt{2}\]). This is where the first identity comes to our aid.
Cos-1(x)+cos-1(y)=cos-1 (xy-\[\sqrt{1-x^{2}}\]\[\sqrt{1-y^{2}}\] , if x,y>0 and x2+y2≤1.
Cos-1(x)-cos-1(y)=cos-1 (xy+\[\sqrt{1-x^{2}}\]\[\sqrt{1-y^{2}}\], if x,y>0 and x2+y2≤1
So the answer to our problem is cos-1(1/2-\[\sqrt{1/2}\]\[\sqrt{1/2}\] = cos-1(½-½) = cos-1(0) ; That is the answer is π/2.
Now analogous to the cos(2θ) and cos (3θ), we have identities for 2cos-1(x) and 3cos-1(x).
2cos-1(x)=cos-1(2x2-1), if x ∈ [-1,1]
3cos-1(x)=cos-1(4x3-3x),if x ∈ [1/2,1]
The most important thing to pay attention to the domain and range in which the said properties and formulas work. As seen in the article, the properties, identities, and formulas can change when the domain and range are changed. It is important to learn to adapt our formulas according to the domain and range.
