Grade 11 Mathematics Note

Inverse Circular Functions.

Trigonometric functions defined by y=sin x, y=cos x, y=tan x, etc are known as the sine, cosine, tangent, etc. functions respectively. The inverse trigonometric functions are the inverse functions of the trigonometric functions.

Specifically, they are the inverses of the sine, cosine, tangent, cotangent, secant, and cosecant functions. They are used to obtain an angle from any of the angle's trigonometric ratios. Inverse trigonometric functions are widely used in engineering, navigation, physics, and geometry.

Example:

The inverse sine function is defined by y = sin-1x.

Now, for y = sin-1x.

Domain = {x: - 1 ≤ x ≤ 1}

And range = {y : −π2−π2 ≤ y ≤ π2π2}

For y = cos-1x

Domain = {x: -1 ≤ x ≤ 1}

And Range = {y: 0 ≤ y ≤ π}

And for y = tan-1x.

Domain = {x:-∞ < x < ∞} and range = {y: π2π2 ≤ y ≤ π2π2}

Thus, for the above domain and range the functions sin-1x, cos-1x and tan-1x becomes the functions of x.

 

Since none of the six trigonometric functions are one-to-one, they are restricted in order to have inverse functions. Therefore the ranges of the inverse functions are proper subsets of the domains of the original functions. The principal inverses are listed in the following table.

Name

Usual notation

Definition

Domain of x for real result

Range of usual principal value
(radians)

Range of usual principal value
(degrees)

arcsine

y = arcsin(x)

x = sin(y)

−1 ≤ x ≤ 1

−π/2 ≤ y ≤ π/2

−90° ≤ y ≤ 90°

arccosine

y = arccos(x)

x = cos(y)

−1 ≤ x ≤ 1

0 ≤ y ≤ π

0° ≤ y ≤ 180°

arctangent

y = arctan(x)

x = tan(y)

all real numbers

−π/2 < y < π/2

−90° < y < 90°

arccotangent

y = arccot(x)

x = cot(y)

all real numbers

0 < y < π

0° < y < 180°

arcsecant

y = arcsec(x)

x = sec(y)

x ≤ −1 or 1 ≤ x

0 ≤ y < π/2 or π/2 < y ≤ π

0° ≤ y < 90° or 90° < y ≤ 180°

arccosecant

y = arccsc(x)

x = csc(y)

x ≤ −1 or 1 ≤ x

−π/2 ≤ y < 0 or 0 < y ≤ π/2

−90° ≤ y < 0° or 0° < y ≤ 90°

 

Examples:

Sin-1 1 = π2π2

Sin-1 = (−12)(−12) = –sin-1(12)(12) = −π6−π6

Arc cot(-1) = cos-1(-1) s= 3π43π4

Arc tan(−13√)(−13) = tan-1(−13√)(−13) = -tan-1(13√)(13) = −π6−π6.

 

Solve:

3tan-112+3√12+3 - tan-11x1x = tan-11313.

Soln:

The given equation is:

3tan-112+3√12+3 - tan-11x1x = tan-11313.

Or, 3tan-1 (2 – 3√3) – tan-11x1x = tan-113.13.        [3tan-1x = tan-1(3x−x31−3x2)(3x−x31−3x2)]

Or, tan-1{(3(2−3√)−(2−3√)2)1−3(2−3√)2}{(3(2−3)−(2−3)2)1−3(2−3)2} – tan-11313= tan-11x1x

Or, tan-1{123√−20123√−20}{123−20123−20} - tan-11313 = tan-11x1x.

Or, tan-1(1−131+1.13)(1−131+1.13) = tan-11x1x

Or, tan-1(23.34)(23.34) = tan-11x1x.

Or, tan-11212 = tan-11x1x.

Or, 1212 = 1x1x

So, x = 2.

 

Prove:

Cos-1x = 2sin-11−x2−−−√1−x2 = 2cos-11+x2−−−√1+x2

Let, cos-1 = θ.

So, cosθ = x

Or, 2cos2θ2θ2 - 1 = x

Or, cos2θ2θ2 = 1+x21+x2

Or, cos θ2θ2 = 1+x2−−−√1+x2

Or, θ2θ2 = cos-11+x2−−−√1+x2.

Or, θ = cos-11+x2−−−√1+x2.

So, cos-1x = cos-11+x2−−−√1+x2 …(i)

Again,

Cosθ = x

Or 1 – sin2θ2θ2= x

Or, 2sin2θ2θ2= 1 – x

Or, sinθ2θ2= 1−x2−−−√1−x2

Or, θ2θ2 = sin-11−x2−−−√1−x2

Or, θ = 2sin-11−x2−−−√1−x2   

So, cos-1x = 2sin-11−x2−−−√1−x2 …(ii)

Hence, from (i) and (ii), we have,

Cos-1x = 2sin-1x 1−x2−−−√1−x2 = 2cos-11+x2−−−√1+x2.

 

Proof:

Let sin-1x à x = sinA

Sin-1y = B à y sinB

And sin-1z = C à z = sinC.

Then the given question changed to:

If A + B + C = π2π2, prove that,

Sin2A + sin2B + sin2C = 1 – 2sinA.sinB.sinC.

From the given part, A + B = π2π2 – C.

Sin(A + B) = cosC

And cos (A + B) = sinC.

Now, L.H.S.= 1212(2sin2A + 2sin2B) + sin2C.

= 1212 [(1 – cos2A) + (1 – cos2B)] + sin2C.

= 1212 [2 – cos2A – cos2B] + sin2c.

= 1 – 1212 (cos2A + cos2B) + sin2C.

= 1 – 1212 2cos(A + B).cos(A – B) + sin2C.

= 1 – sinC [cos(A – B) – cos(A + B)]

= 1 – sinC [2sinA−B+A+B2.sinA+B−A+B2][2sin⁡A−B+A+B2.sin⁡A+B−A+B2]

= 1 – sinC [2sinA.sinB]

= 1 – 2sinA.sinB.sinC = R.H.S>

Hence, x2 + y2 + z2 + 2xyz = 1.


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