Derivation in inverse trigonometric functions

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Derivation :


Therefore , 



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Formula to find sum of the first n Natural numbers

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Hello, Everyone. In this, we are going to see a proof for the formula to find sum of the first n Natural numbers. Let S = 1 + 2 + 3 + 4 + ......... + (n-1) + n , where n is any natural number.  --------------- (1) The r th term from the starting of the series, in the above series is r. ------- (2) From the property of  addition of numbers, we can write S as shown below. S = n + (n-1) + (n-2) + ......... + 2 + 1 ---------- (3)  The r th term from the starting of the series, in the above series is n-(r-1) = n - r + 1 -------- (4) Let us add (1) and (3) such that the corresponding terms are added as shown in the below figure. From (2) and (4), The r th term from the starting of the series, in the resulting series obtained from addition of (1) and (3) is equal to r + (n - r + 1) = n + 1. From this, it is obvious that the r th term of the resulting series is independent of r. Therefore, as n is a constant, every term in the resulting series is equal to n + 1. Therefore, by add...
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Sum of cubes of the first n natural numbers - Formula derivation

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Hello, Everyone. In this, we are going to derive the formula to find the sum of cubes of the first n natural numbers. Let S = 1 3  + 2 3  + 3 3  + .......... + (n-1) 3  + n 3 (m+1) 4  - m 4  = (m 4  + 4m 3  + 6m 2 + 4m + 1) - m 4  = 4m 3  + 6m 2 + 4m + 1 ------------ (i) put m = 1,2,3, ....... , (n-1), n in (i) Then, 2 4  - 1 4   = 4(1) 3  + 6(1) 2  + 4(1) + 1 ------------ (1) 3 4  - 2 4   = 4(2) 3  + 6(2) 2  + 4(2) + 1 ------------ (2) 4 4  - 3 4   = 4(3) 3  + 6(3) 2  + 4(3) + 1 ------------ (3) ................... n 4  - (n-1) 4   = 4(n-1) 3  + 6(n-1) 2  + 4(n-1) + 1 ------------ (n-1) (n+1) 4  - n 4   = 4(n) 3  + 6(n) 2  + 4(n) + 1 ------------ (n) Now, by adding equations (1),(2),(3), ...... ,(n-1) and (n), we get the following equation. (n+1) 4...
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Proof of the projection rules

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In triangle ABC, let us denote the magnitudes of the sides AB, BC, CA by c, a, b respectively and the angle BAC, angle CBA, angle ACB by A, B, C respectively. Proof :  From the cosine rules, we have cos(B) = ((c 2 + a 2 - b 2 )/2ca), cos(C) = ((a 2 + b 2 - c 2 )/2ab) Now, let us consider the right hand side expression in the equation to be proved. Therefore, b cos(C) + c cos(B) = b ( (a 2  + b 2  - c 2 )/2ab ) + c ( (c 2  + a 2  - b 2 )/2ca ) ⇒ b cos(C) + c cos(B) = ( (a 2  + b 2  - c 2 ) + (c 2  + a 2  - b 2 ) )/2a = (2a 2 )/2a = a. Therefore, a = b cos(C) + c cos(B) Similarly, we can prove that b = c cos(A) + a cos(C) and c = a cos(B) + b cos(A)
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