Question

Drawing The Graph Of Inverse Trigonometric Functions Sin-1x And Cos-1x And Sloving The Inequality Cos-1x>Sin-1x

Answer 1

Let x be any positive integer and y = 3.

Let x be any positive integer and y = 3.By Euclid’s division algorithm, then,

Let x be any positive integer and y = 3.By Euclid’s division algorithm, then,x = 3q + r for some integer q≥0 and r = 0, 1, 2, as r ≥ 0 and r < 3.

Let x be any positive integer and y = 3.By Euclid’s division algorithm, then,x = 3q + r for some integer q≥0 and r = 0, 1, 2, as r ≥ 0 and r < 3.Therefore, x = 3q, 3q+1 and 3q+2

Let x be any positive integer and y = 3.By Euclid’s division algorithm, then,x = 3q + r for some integer q≥0 and r = 0, 1, 2, as r ≥ 0 and r < 3.Therefore, x = 3q, 3q+1 and 3q+2Now as per the question given, by squaring both the sides, we get,

Let x be any positive integer and y = 3.By Euclid’s division algorithm, then,x = 3q + r for some integer q≥0 and r = 0, 1, 2, as r ≥ 0 and r < 3.Therefore, x = 3q, 3q+1 and 3q+2Now as per the question given, by squaring both the sides, we get,x2 = (3q)2 = 9q2 = 3 × 3q2

Let x be any positive integer and y = 3.By Euclid’s division algorithm, then,x = 3q + r for some integer q≥0 and r = 0, 1, 2, as r ≥ 0 and r < 3.Therefore, x = 3q, 3q+1 and 3q+2Now as per the question given, by squaring both the sides, we get,x2 = (3q)2 = 9q2 = 3 × 3q2Let 3q2 = m

Let x be any positive integer and y = 3.By Euclid’s division algorithm, then,x = 3q + r for some integer q≥0 and r = 0, 1, 2, as r ≥ 0 and r < 3.Therefore, x = 3q, 3q+1 and 3q+2Now as per the question given, by squaring both the sides, we get,x2 = (3q)2 = 9q2 = 3 × 3q2Let 3q2 = mTherefore, x2= 3m …..(1)

Let x be any positive integer and y = 3.By Euclid’s division algorithm, then,x = 3q + r for some integer q≥0 and r = 0, 1, 2, as r ≥ 0 and r < 3.Therefore, x = 3q, 3q+1 and 3q+2Now as per the question given, by squaring both the sides, we get,x2 = (3q)2 = 9q2 = 3 × 3q2Let 3q2 = mTherefore, x2= 3m …..(1)x2 = (3q + 1)2 = (3q)2+12+2×3q×1 = 9q2 + 1 +6q = 3(3q2+2q) +1

Let x be any positive integer and y = 3.By Euclid’s division algorithm, then,x = 3q + r for some integer q≥0 and r = 0, 1, 2, as r ≥ 0 and r < 3.Therefore, x = 3q, 3q+1 and 3q+2Now as per the question given, by squaring both the sides, we get,x2 = (3q)2 = 9q2 = 3 × 3q2Let 3q2 = mTherefore, x2= 3m …..(1)x2 = (3q + 1)2 = (3q)2+12+2×3q×1 = 9q2 + 1 +6q = 3(3q2+2q) +1Substitute, 3q2+2q = m, to get,

Let x be any positive integer and y = 3.By Euclid’s division algorithm, then,x = 3q + r for some integer q≥0 and r = 0, 1, 2, as r ≥ 0 and r < 3.Therefore, x = 3q, 3q+1 and 3q+2Now as per the question given, by squaring both the sides, we get,x2 = (3q)2 = 9q2 = 3 × 3q2Let 3q2 = mTherefore, x2= 3m …..(1)x2 = (3q + 1)2 = (3q)2+12+2×3q×1 = 9q2 + 1 +6q = 3(3q2+2q) +1Substitute, 3q2+2q = m, to get,x2= 3m + 1 ……………………………. (2)

Let x be any positive integer and y = 3.By Euclid’s division algorithm, then,x = 3q + r for some integer q≥0 and r = 0, 1, 2, as r ≥ 0 and r < 3.Therefore, x = 3q, 3q+1 and 3q+2Now as per the question given, by squaring both the sides, we get,x2 = (3q)2 = 9q2 = 3 × 3q2Let 3q2 = mTherefore, x2= 3m …..(1)x2 = (3q + 1)2 = (3q)2+12+2×3q×1 = 9q2 + 1 +6q = 3(3q2+2q) +1Substitute, 3q2+2q = m, to get,x2= 3m + 1 ……………………………. (2)x2= (3q + 2)2 = (3q)2+22+2×3q×2 = 9q2 + 4 + 12q = 3 (3q2 + 4q + 1)+1

Let x be any positive integer and y = 3.By Euclid’s division algorithm, then,x = 3q + r for some integer q≥0 and r = 0, 1, 2, as r ≥ 0 and r < 3.Therefore, x = 3q, 3q+1 and 3q+2Now as per the question given, by squaring both the sides, we get,x2 = (3q)2 = 9q2 = 3 × 3q2Let 3q2 = mTherefore, x2= 3m …..(1)x2 = (3q + 1)2 = (3q)2+12+2×3q×1 = 9q2 + 1 +6q = 3(3q2+2q) +1Substitute, 3q2+2q = m, to get,x2= 3m + 1 ……………………………. (2)x2= (3q + 2)2 = (3q)2+22+2×3q×2 = 9q2 + 4 + 12q = 3 (3q2 + 4q + 1)+1Again, substitute, 3q2+4q+1 = m, to get,

Let x be any positive integer and y = 3.By Euclid’s division algorithm, then,x = 3q + r for some integer q≥0 and r = 0, 1, 2, as r ≥ 0 and r < 3.Therefore, x = 3q, 3q+1 and 3q+2Now as per the question given, by squaring both the sides, we get,x2 = (3q)2 = 9q2 = 3 × 3q2Let 3q2 = mTherefore, x2= 3m …..(1)x2 = (3q + 1)2 = (3q)2+12+2×3q×1 = 9q2 + 1 +6q = 3(3q2+2q) +1Substitute, 3q2+2q = m, to get,x2= 3m + 1 ……………………………. (2)x2= (3q + 2)2 = (3q)2+22+2×3q×2 = 9q2 + 4 + 12q = 3 (3q2 + 4q + 1)+1Again, substitute, 3q2+4q+1 = m, to get,x2= 3m + 1… (3)

Let x be any positive integer and y = 3.By Euclid’s division algorithm, then,x = 3q + r for some integer q≥0 and r = 0, 1, 2, as r ≥ 0 and r < 3.Therefore, x = 3q, 3q+1 and 3q+2Now as per the question given, by squaring both the sides, we get,x2 = (3q)2 = 9q2 = 3 × 3q2Let 3q2 = mTherefore, x2= 3m …..(1)x2 = (3q + 1)2 = (3q)2+12+2×3q×1 = 9q2 + 1 +6q = 3(3q2+2q) +1Substitute, 3q2+2q = m, to get,x2= 3m + 1 ……………………………. (2)x2= (3q + 2)2 = (3q)2+22+2×3q×2 = 9q2 + 4 + 12q = 3 (3q2 + 4q + 1)+1Again, substitute, 3q2+4q+1 = m, to get,x2= 3m + 1… (3)Hence, from equation 1, 2 and 3, we can say that, the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

Answer 2

The value of x is given by

$\displaystyle \sf{ x \in \: \bigg[ - 1, \frac{1}{ \sqrt{2} } \bigg) }$

Given :

$\sf {sin}^{ - 1} x \: \: and \: \: {cos}^{ - 1} x$

To find :

1. Draw the graph of the inverse Trigonometric function

2. To solve the inequality

$\sf {cos}^{ - 1} x \: > \: {sin}^{ - 1} x$

Solution :

Step 1 of 2 :

Draw the graph of the inverse Trigonometric functions

The graph of the inverse Trigonometric function is referred to the attachment

$\sf Green \: line \: : {sin}^{ - 1} x$

$\sf Red \: line : \: {cos}^{ - 1} x$

Domain of each function is [ - 1 , 1 ]

Step 2 of 2 :

Solve the inequality

$\sf {cos}^{ - 1} x \: > \: {sin}^{ - 1} x$

$\displaystyle \sf\implies 2{cos}^{ - 1} x > {sin}^{ - 1} x +{cos}^{ - 1} x \:$

$\displaystyle \sf\implies 2{cos}^{ - 1} x > \frac{\pi}{2}$

$\displaystyle \sf\implies {cos}^{ - 1} x > \frac{\pi}{4}$

$\displaystyle \sf\implies x < cos \frac{\pi}{4} \: \: \bigg( \because \: {cos}^{ - 1} x \: is \: decreasing \: )$

$\displaystyle \sf\implies x < \frac{1}{ \sqrt{2} }$

$\because \: \: \displaystyle \sf{ Domain \: of \: {cos}^{ - 1} x \: \: is \: \: \bigg[ - 1, 1 \bigg]}$

Hence the required solution set is

$\displaystyle \sf{ x \in \: \bigg[ - 1, \frac{1}{ \sqrt{2} } \bigg) }$

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