Trigonometry: Inverse Trig Functions |

**Inverse Trig Functions**

Sometimes, when we're solving right triangles, the given information includes only the side measures but nothing about the size of the acute angles in the triangle. In such a case we have to use **inverse trig functions **to solve for one of the angles.

With **regular trig functions**, we use the sine function to find the **ratio** of the length of the opposite side to the length of the hypotenuse for a given angle.

If however, we need **to find the angle's measure** when we know its sine, we use the **inverse function** known as **arcsin A **or **sin ^{ – 1} A**.

Say we know that **sin A = 0.7** . To find **angle A** we enter **sin ^{ }**

and the calculator displays 42.427 in degree mode and 0.7754 in radian mode.

This means that:

**an Inverse Trig Function Value is an Angle.**

The function **sin ^{ }**

The same holds true for the other inverse trig functions.

**Important Note:**

**Beware!!!** In Algebra, *x*^{ }^{ –}** ^{ 1}** means the

In the notation for inverse trig functions, **the exponent** **(–1)** **does not mean reciprocal**!

It **indicates the inverse function** like *f *^{– 1}* *(*x*) does.

The reciprocal of the **sine function** is the **cosecant function** not the **arcsin function**.

The **inverse** of the **sine function** is the **arcsin function**.

Because of this confusion, it is best to always use **arcsin**, **arccos**, **arctan**, etc. to indicate the inverse trig functions.

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When we apply *y* = sin ** x**, we plug in

When we apply *y*** = arcsin **** x**, we plug in

So, if *x* = **30 ^{ 0}**,

We **substitute** an **angle value for **** x** in

We **substitute** **a ratio value for **** x** in

** y = sin x **is identical to

**A Fundamental Problem of Inverse Functions**

When we invert a function, its **range becomes the domain** of the inverse.

If the **original function is many-to-one**, the range elements of the inverse are paired with many domain elements, so **the inverse is not a function**.

For instance, arctan 1 = 45° but it's also = 225° since the tangent of both angles is 1.

Therefore, we have to **restrict the domain** of the trig functions before we consider their inverse functions.

**To make a many-to-one function one-to-one**

so that its *inverse is a function*,

we **restrict or limit** the **domain** to

**eliminate repetition** of the range elements (*y*'s)*.*

In order to better understand and visualize the situation, let's look at a sine graph.

All trig functions are periodic -- the *y*'s repeat themselves periodically.

What **we want** is a chunk or **interval** of this graph **which includes**, only once, all the *y-values* **from** the minimum of **–1 to the maximum of +1**, and we'd like that interval to include or be **near** **the origin**.

From the graph it's easy to see that the interval from will do it.

We now call this function **Sin x** (note the uppercase "S") with range unchanged but

So, ** y = arcsin x** is a function with domain [ – 1, 1 ] and range . On the Cartesian plane, this interval includes angles in the

With the cosine function, we select the interval from 0 to , since it includes all the values of cos *x* only once. The **invertible**, one-to-one **function** ** y = Cos x**, has a

a range of [ – 1, 1 ]. This is the

So, ** y = arccos x** is a function with domain [ –1, 1 ] and range . On the Cartesian plane, this interval includes angles in the

** y = arctan x** is a function with domain and range .

Remember that

For *y* = Sec *x,** *we restrict the domain to the **1st and 3rd quads**, so, ** y = arcsec x** is a weird function with domain , and range

**Graphs of the Inverse Trig Functions**

(pi/2 = 1.571)

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**Examples** (answers in radians)

**ex. 1:** however, the restrictions on the domain of ** y = arcSin x** rule out the second solution.

**ex. 2:** however the restrictions on the domain of ** y = arcSin x** rule out the second solution.

**ex. 3:** however the restrictions on the domain of ** y = arcCos x** rule out the second solution.

**ex. 4:** Find .

Let arctan (2/3) = **A**. We know that tan A = 2/3 so let's make a picture of angle **A**.

Since we want **sec A**, (hypotenuse over adjacent), we get .

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**ex. 5:**

Let arcsin (*x*/2) = A. We know that sin A = *x*/2 so let's make a picture of angle A which lives in a right triangle with **opposite side = x** and

Since we want cos A, (adjacent over hypotenuse ), we get .

When the question involves composite angles like **(A + B**), **2A** or **½A**, we use trig identities.

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**ex. 6:** Find a value for sin (arcsin ½ + arccos 0).

We'll let **A** = **arcsin ½** and **B = ** **arccos 0**, so the question asks for **sin (A + B)**.

**Sin A = ½,** which makes angle **A = 30° **or ** ** radians.

**Cos B** = 0, which makes angle** B = ****90° **or** ** radians.

The question now becomes ** ** or **sin (30° + 90°) = sin 120 °**.

Since sine values are positive in both the 1st and 2nd quads, **sin 120° = sin 60°**.

From the 30°, 60°, 90°, right triangle, we know that** **.

In this case, we were able to add the 2 angles together to make a "*known*" angle. When that's impossible, we make a picture of the 2 triangles with angles **A** and **B** so we can see the value of all trig functions for the 2 angles -- then we use trig identities to solve.

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**ex. 7:** Find a value for .

Again, we'll call the 1st angle** A** and the 2nd angle **B**.

Here are the images of the right triangles these ratios generate.

Since we need to find cos (A + B) we'll use **cos (A + B) = cos A cos B – sin A sin B**.

We substitute the values from the diagram to get:

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If you've forgotten your trig identitites (*who hasn't??*) click here.

**Practice**

Keep in mind that:

- an

an inverse trig function has a

a

trig function values for

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1) Find the exact value without a calculator or trig tables (*as if anyone does that anymore!*)

a) arcTan 1 |
b) arcCos 0 | c) sin [arcCos (2/5)] | d) cos [ arcSin 0 ] |

e) | f) sin [arcCos ] | g) sin (arcTan 3/4) | h) tan (arcSin 12/13) |

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2) Rewrite as an algebraic expression in *x*.

a) sec [arcTan (x + 1)/5] |
b) sin {arcCos (–7/x)} |

c) | d) cot (arcSin x) |

e) sec (arcTan x/2) |
f) |

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3) __SIMPLIFY__: (make diagrams!)

a) sin (2 arcCos 1/2) | b) sin (arcSin 1/2 – arcCos 4/5) |

c) cos (arcSin 4/5 – arcCos½) | d) |

e) cos (2 arcTan x) |

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If you've forgotten your trig identitites (*who hasn't??*) click here.

**SOLUTIONS**

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1) Find the exact value without a calculator or trig tables

a) arctan (1) is an angle A opposite side = 1, adjacent side = 1 which makes hypotenuse = angle A = 45° or radians. b) arcCos 0 = 90° or radians. |
c) sin [ arcCos (2/5) is an angle A adjacent side = 2, hypotenuse = 5 which makes opposite = sin A = . |

d) cos [ arcSin 0 ] = cos (0) = 1 | e) is an angle A opposite side = – 2, hypotenuse = which makes adjacent = 1 cos A = |

f) arcCos is an angle A adjacent side = hypotenuse = 2, which makes opposite = 1 This is the 30°, 60°, 90° triangle angle A = 30° so sin A = ½. |
g) arcTan ( ) is an angle A opposite side = 3, adjacent = 4 so hypotenuse = 5 sin A = 3/5. |

h) arcSin 12/13 is an angle A opposite side = 12, hypotenuse = 13 which makes adjacent = 5. tan A = 12/5. |

2) Rewrite as an algebraic expression in *x*.

a) arcTan (x + 1)/5 is an angle Aopposite = x + 1, adjacent = 5hypotenuse = |
b) arcCos (–7/x) is an angle Aadjacent = – 7, hypotenuse = x,opposite = |

c) call angle A hypotenuse = ( x – 5), opposite = – 7so adjacent = |
d) arcSin x is angle Ahypotenuse = 1, opposite = xso adjacent = |

e) arcTan x/2 is an angle Aopposite = x, adjacent = 2hypotenuse = |
f) is angle A. opposite = x – 2, hypotenuse = 3so adjacent = |

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3) __SIMPLIFY__: (make diagrams!)

a) arccos (1/2) is an angle A adjacent = 1, hypotenuse = 2 this is the 30°, 60°, 90° triangle so opposite = (this means A = 60°) we want sin (2)(60°) = sin 120° = |
b) Let arcsin ½ = A and arccos 4/5 = B adjacent(A) = , opposite(B) = 3 sin (A – B) = sinA cosB – cos A sin B substituting the values gives us: |

c) Let arcsin 4/5 = A and arccos 1/2 = B adjacent(A) = 3, opposite(B) = . cos (A – B) = cosA cosB + sin A sin B substituting the values gives us: |
d) is angle A we have opposite and hypotenuse from sin A the adjacent = 3 so tan A = p/3. |

e) arcTan x is an angle Aopposite = x, adjacent = 1hypotenuse = cos 2A = cos ² A – sin ² A |

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(*all content **© MathRoom Learning Service; 2004 - *).

**Trig Identities for Composite Angles**

Use these to find the trig function values for a sum, difference or multiple of angles.

sin (A + B) = sin A cos B + cos A sin B | cos (A + B) = cos A cos B – sin A sin B |

sin (A – B) = sin A cos B – cos A sin B | cos (A – B) = cos A cos B + sin A sin B |

sin 2A = 2 sin A cos A | cos 2A = cos^{ 2} A – sin^{ 2} A |

- = 1 – 2 sin
^{ 2} A | |

- = 2 cos
^{ 2} A – 1 | |

** Note the relation between the formulae for **sums** and those for **double angles**.

If we replace **B** with **A** in the sum formula, we get **2A** as the argument (angle).

So, knowing that **sin 2A = 2 sin A cos A**, we can extrapolate that

**sin (A + B)** is a sum of 2 terms, both of which have sin( ) and cos ( ).

Obviously, the blank brackets are filled with **A** and **B**, so

**sin (A + B)** must equal **sin A cos B + cos A sin B**.

And since addition is commutative, the order doesn't matter.

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Look at **cos 2A = cos ^{ 2} A – sin^{ 2} A**.

This means

Change

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**Note also the **pattern of signs** (*pluses and minuses*).

For **sin** and **tan**, the **signs correspond**.

Sin or tan of a sum (+) means there's a sum in the formula too.

The same is true for sin or tan of a difference (–).

Note that **sin (A – B) = sin A cos B – cos A sin B** with emphasis on the " – "

For **cosine** formulae, the** signs are always opposite**.

For cos of a sum (+), there's a difference (–) in the formula and vice versa.