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 1 0.7
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 1 x is also known as arcsin x .

The same holds true for the other inverse trig functions.

Important Note:

Beware!!! In Algebra, x 1 means the reciprocal of x or 1 / x.

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 angle values for x and generate ratio values for y.

When we apply y = arcsin x, we plug in ratio values for x and generate angle values for y.

So, if x = 30 0, sin 30 0 = 0.5 which means that arcsin 0.5 = 30 0

We substitute an angle value for x in y = sin x; we get a ratio.

We substitute a ratio value for x in y = arcsin x; we get an angle.

y = sin x is identical to x = arcsin y

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 domain restricted to the interval , and this is what we invert.

So, y = arcsin x is a function with domain [ – 1, 1 ] and range . On the Cartesian plane, this interval includes angles in the 4th and 1st quadrants of the unit circle. 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 domain of , and
a
range of [ – 1, 1 ]. This is the one-to-one function we invert.

So, y = arccos x is a function with domain [ –1, 1 ] and range . On the Cartesian plane, this interval includes angles in the 1st and 2nd quads of the unit circle.

y = arctan x is a function with domain and range .
Remember that
y = tan x has vertical asymptotes (infinite discontinuities) at the extremes of a period, so we cannot include those values in the range of the inverse function. They weren't included in the domain of the original function. In the unit circle, this is 4th and 1st quads without the endpoints. (same as y = arcsin x)

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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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 hypotenuse = 2. Now, with the Pythagorean Theorem, we find the 3rd side. 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: .

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

Keep in mind that:

an inverse trig function defines an angle of a right triangle,
an inverse trig function has a restricted domain and range.
a trig function defines a ratio of sides in a right triangle.
trig function values for composite angles are found using identities.

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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) .

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 A opposite = x + 1, adjacent = 5 hypotenuse =  b) arcCos (–7/x) is an angle A adjacent = – 7, hypotenuse = x, opposite =  c) call angle A hypotenuse = (x – 5), opposite = – 7 so adjacent =   d) arcSin x is angle A hypotenuse = 1, opposite = x so adjacent =  e) arcTan x/2 is an angle A opposite = x, adjacent = 2 hypotenuse =  f) is angle A. opposite = x – 2, hypotenuse = 3 so adjacent =  .

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 A opposite = x, adjacent = 1 hypotenuse = cos 2A = cos ² A – sin ² A .

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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 cos (A + A) = cos A cos A – sin A sin A.
Change one of the A's in each term to a B, and there's the formula for cos (A + B).

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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. 