Question

Show that $1-2\sin \theta \cos \theta \ge 0$ for all $\theta \in R$ .

Answer 1

Hint:When an inequation is given in terms of sine, cosine, tangent we must use any of the trigonometric identities to make the inequation solvable. There are many interrelations between sine, cosine, tan, secant these are interrelations called as identities. Whenever you see conditions such that $\theta \in R$, that means inequality is true for all angles. So, directly think of identity which will make your work easy.
${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ for all $\theta $ .

Complete step-by-step answer:
An inequality with sine, cosine or tangent in them is called trigonometric inequality. These are solved by some interrelations known beforehand.
All the interrelations which relate sine, cosine, tangent, secant, cotangent, cosecant are called trigonometric identities. These trigonometric identities solve the equation and make them simpler to understand for a proof. These are the main and crucial steps to take us nearer to result.
Given inequality in the question, which we need to prove:
$1-2\sin \theta \cos \theta \ge 0$ for all $\theta \in R$
Irrespective of angle, if some term is greater than or equal to zero then the first possibility must be it being a square.
So, to convert left hand side to square of a number, we need the identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$
By substituting this identity in place of 1, we get:
${{\sin }^{2}}\theta +{{\cos }^{2}}\theta -2\sin \theta \cos \theta $
By the algebraic identity of a, b being 2 numbers, we have
${{a}^{2}}+{{b}^{2}}-2ab={{\left( a-b \right)}^{2}}$
By using this identity left hand side will get a form of:
${{\left( \sin \theta -\cos \theta \right)}^{2}}$
As a square of numbers is always $\ge 0$ irrespective of number inside. So,
${{\left( \sin \theta -\cos \theta \right)}^{2}}\ge 0$
$1-2\sin \theta \cos \theta \ge 0$
Hence, we proved the required equation.

Note: Whenever you see $\ge 0$ in proof of inequality the maximum possible way is to convert L.H.S into a square of numbers. By this way you will have a solid proof for given $\ge 0$ inequality.We can also solve this question by replacing $2\sin \theta \cos \theta$ as $\sin{2} \theta$ as we know trigonometric identity $\sin2 \theta=2\sin \theta \cos \theta$ ,we get $1-\sin2 \theta$.We know the range of $\sin{2}\theta$ varies from $-1\le\theta\le1$ So we get positive value only i.e $\ge 0$ Hence verified. Students should remember the trigonometric identities and formulas to solve these types of questions..