find the exact real value of arcsin 1 2

Solve Inverse Pure mathematics Functions Questions

Problems on inverse trigonometric functions are solved and detailed solutions are presented. Also exercises with answers are presented at the end of this page. We first review much of the theorems and properties of the inverse functions.

Theorem

1. y = arcsin x is equivalent to goof y = x
with -1 ≤ x ≤ 1 and - π / 2 ≤ y ≤ π / 2

y = arcos x is equivalent to cos y = x
with -1 ≤ x ≤ 1 and 0 ≤ y ≤ π

y = arctan x is equal to tan y = x
with - π / 2 < y < π / 2

Question 1

Find the photographic valuate of
1.     arcsin(- √3 / 2)
2.     arctan(- 1 )
3.     arccosine(- 1 / 2)

Solution to question 1

1. arcsin(- √3 / 2)
Let y = arcsin(- √3 / 2). According to theorem 1 above, this is combining weight to
sin y = - √3 / 2 , with - π / 2 ≤ y ≤ π / 2
From table of special angles sin (π /3) = √3 / 2.

We also hump that sin(-x) = - sin x. So

sin (- π / 3) = - √3 / 2
Comparing the last expression with the equation sin y = - √3 / 2, we close that
y = - π / 3

2. arctan(- 1 )

Let y = inverse tangent(- 1 ). According to 3 above
burn y = - 1 with - π / 2 < y < π / 2
From table of special angles suntan (π / 4) = 1.

We also know that topaz(- x) = - bronze x. Thus

tan (-π / 4) = - 1
Liken the final financial statement with tan y = - 1 to obtain
y = - π/4

3. arccosine(- 1 / 2)

Let y = arccosine(- 1 / 2). According to theorem 2 above
cos y = - 1 / 2 with 0 ≤ y ≤ π
From table of peculiar angles cos (π / 3) = 1 / 2
We also hump that cos(π - x) = - cos x. So
cos (π - π/3) = - 1 / 2
Compare the last statement with cos y = - 1 / 2 to obtain
y = π - π / 3 = 2 π / 3

Doubtfulness 2

Simplify cos(arcsin x )

Resolution to question 2:
Lashkar-e-Toiba z = cos ( arcsin x ) and y = arcsin x so that z = cos y. Accordant to theorem 1 above y = arcsin x may also represent written as
trespass y = x with - π / 2 ≤ y ≤ π / 2
Also
sin²y + cos²y = 1
Substitute sin y by x and figure out for cos y to obtain
romaine lettuce y = + OR - √ (1 - x²)
But - π / 2 ≤ y ≤ π / 2 so that cos y is positive
z = romaine y = cos(arcsin x) = √ (1 - x ²)

Question 3

Simplify csc ( arctan x )

Solution to question 3
Let z = csc ( arctan x ) and y = arctan x then that z = csc y = 1 / sin y. Using theorem 3 in a higher place y = inverse tangent x may also be written as
sunburn y = x with - π / 2 < y < π / 2
Also
tan²y = sin²y / cos²y = sin²y / (1 - sin²y)
Figure out the above for sin y
sin y = + or - √ [ tan²y / (1 + tan²y) ]
= + or - | tan y | / √ [ (1 + bronze²y) ]
For - π / 2 < y ≤ 0 sin y is negative and bronze y is also negative so that | tan y | = - tan y and
sin y = - ( - tan y ) / √ [ (1 + suntan²y) ] = tan y / √ [ (1 + tan²y) ]
For 0 ≤ y < π/2 sin y is positive and tan y is also positive so that | tan y | = tan y and
sin y = tan y / √ [ (1 + tan²y) ]
Eventually
z = csc ( arctan x ) = 1 / goof y = √ [ (1 + x²) ] / x

Question 4

Valuate the following
1. arcsin( sin (7 π / 4))
2. arccos( cos lettuce (4 π / 3 ))

Solution to question 4

1. inverse sine( sin ( y ) ) = y only for - π / 2 ≤ y ≤ π / 2. So we prime transform the given look noting that sin (7 π / 4) = sin (-π / 4) arsenic follows
arcsin( sin (7 π / 4)) = arcsin( wickedness (- π / 4))
- π / 4 was chosen because it satisfies the discipline - π / 2 ≤ y ≤ π / 2. Hence
arcsin( sin (7 π / 4)) = - π / 4

arccos( cos ( y ) ) = y only for 0 ≤ y ≤ π . We prototypal transform the given expression noting that cos lettuce (4 π / 3) = romaine lettuce (2 π / 3) as follows
arccos( cos (4 π / 3)) = arccos( romaine lettuce (2 π / 3))
2 π / 3 was chosen because it satisfies the condition 0 ≤ y ≤ π . Which gives
arccos( cos (4 π / 3)) = 2 π / 3