promath The sin(π/2-x) formula is given as follows: sin(π/2-x)=cosx …(∗) The formula of sin(pi/2-θ) is given by sin(π/2-θ)=cosθ. In this post, we will learn how to compute sin(pi/2-x) or sin(θ-pi/2). Proof of sin(pi/2-x) Formula To establish the above formula (∗), that is, the formula of sin(π/2-x), we need to apply the below formula: sin(a-b) = sina … Read more
The general solutions of sinx=-1 are given by x=(4n-1)π/2 where n is any integer. In this post, we will learn how to find the general solutions of the trigonometric equation sinx=-1. sinx=-1 General Solution Question: Find the general solutions of sinx= -1. Answer: As sin(-π/2)= -1, the equation sinx= -1 can be written sinx = … Read more
The general solution of sinx=1 is given by x=(4n+1)π/2 where n is any integer. In this post, we will learn how to find the general solutions of the trigonometric equation sinx=1. sinx=1 General Solution Question: Find the general solutions of sinx=1. Answer: As sin(π/2)=1, the equation sinx=1 can be written sinx = sin(π/2). ∴ x=mπ+(-1)m … Read more
There is a lot of confusion on which is bigger between million and billion. In this post, we will find the answer. Let us now find both billion and million in numbers. 1 billion = 10,000 lakhs 1 million = 10 lakhs Since 1 lakh = 1,00,000 we have that 1 billion = 1,000,000,000 1 … Read more
The Maclaurin series expansion of ex or the Taylor series expansion of ex at x=0 is given by the following summation: ex = $\sum_{n=0}^\infty \dfrac{x^n}{n!}$ = $1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+\cdots$. In this post, we will learn how to find the series expansion of ex. Taylor Series Expansion of ex at x=0 The Maclaurin series expansion of a function … Read more
For any real number c, we have limx→ccosx = cosc and cosx is defined for all real numbers. Thus, the function cosx is continuous everywhere. Now, we will prove that cosx is continuous for all values of x by the epsilon-delta method. We will use the following two formulas: Prove cosx is continuous Let f(x)=cosx … Read more
The simplification or the formula of 1+cot2x is given as follows: 1+cot2x = cosec2x. In this post, we will learn how to establish the formula of 1+cot2x. Proof of 1+cot2x Formula Question: Prove the formula 1+cot2x = cosec2x. Solution: L.H.S. = 1+cot2x = $1+(\dfrac{\cos x}{\sin x})^2$ as cotx=cosx/sinx. = $1+\dfrac{\cos^2 x}{\sin^2 x}$ = $\dfrac{\sin^2 x … Read more
The general solution of the differential equation dy/dx=y/x is given by y=cx where c is any constant. To solve dy/dx = y/x, we will use the separation of variables method. Let us learn how to find the solution of dy/dx=y/x. General Solution of dy/dx=y/x Question: Find the general solution of $\dfrac{dy}{dx}=\dfrac{y}{x}$. Answer: $\dfrac{dy}{dx}=\dfrac{y}{x}$ We can … Read more
The integration of the square root of 1+sin2x is given by ∫√(1+sin2x) dx = cosx-sinx+C where C is an integration constant. In this post, we will learn how to integrate square root of 1+sin2x. Integral of Square Root of 1+sin2x Question: Find the integral of 1+sin2x, that is, find ∫√(1+sin2x) dx. Answer: We will use … Read more
The Maclaurin series expansion of cosx or the Taylor series expansion of cosx at x=0 is given as follows: cosx = $\sum_{n=0}^\infty \dfrac{(-1)^n}{(2n)!}x^{2n}$ = $1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+\cdots$ Taylor Series Expansion of Sinx at x=0 Note that the Maclaurin series expansion of f(x)=cosx or the Taylor series of a function f(x) at $x=0$ is given by the following … Read more