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

Last updated date: 13th Jun 2024
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For answering this question we will simplify the given expression using the trigonometric identities ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ and we will also use the formulae ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ and then verify the given condition. From the basic concept, we know that the square of any number can never be negative in the real numbers set and we also know that the value of sine and cosine functions lie between -1 and 1.
Now considering from the question we have the trigonometric expression given as $1-2\sin \theta \cos \theta \ge 0$ for all $\theta \in R$ .
From the basic concept we know the trigonometric identity given as ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ now we will replace the $1$ in the expression with value then we will have ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta -2\sin \theta \cos \theta$ . We can observe that it is in the form of ${{a}^{2}}+{{b}^{2}}-2ab$ so this can be simply written as ${{\left( \sin \theta -\cos \theta \right)}^{2}}$ .
From the basic concept, we know that the square of any number can never be negative in the real numbers set. That is it will be always positive that is greater than zero or equals to zero. So we can say that, $1-2\sin \theta \cos \theta \ge 0$ for all $\theta \in R$