Let $f(0)=\sin \theta(\sin \theta+\sin 3 \theta)$ then $f(\theta)$ is:
A.$\geq 0$ only when $\theta \geq 0$
B.$\le 0$only when $\theta \leq 0$
C.$\le 0$ for all real $\theta$
D.$\ge 0$ for all real $\theta$
Answer
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Hint: In mathematics, a real number is a value of a continuous quantity that can represent a distance along a line (or alternatively, a quantity that can be represented as an infinite decimal expansion). Real numbers include all rational numbers (numbers that can be written like fractions) and all irrational numbers (numbers that cannot be written like fractions). We won't go into all the details here, but imaginary numbers are all multiples of something called the imaginary unit, which we write with the letter i.
Complete step-by-step answer:
Real numbers consist of zero $(0),$ the positive and negative integers $(-3,-1,2,4),$ and all the fractional and decimal values in between $(0.4,3.1415927,1 / 2)$. Real numbers are divided into rational and irrational numbers. A real number is any positive or negative number. This includes all integers and all rational and irrational numbers. For example, a program may limit all real numbers to a fixed number of decimal places.
In mathematics, the irrational numbers are all the real numbers which are not rational numbers. That is, irrational numbers cannot be expressed as the ratio of two integers. Irrational numbers can also be expressed as non-terminating continued fractions and many other ways.
$f(\theta)=\sin \theta(\sin \theta+\sin 3 \theta)$
Now it can be said that:
Applying identity $: \sin \mathrm{A}+\sin \mathrm{B}=2 \sin \dfrac{\mathrm{A}+\mathrm{B}}{2} \cos \dfrac{\mathrm{A} - \mathrm{B}}{2}$
$\Rightarrow f(\theta )=\sin \theta \left( 2\sin \left( \dfrac{\theta +3\theta }{2} \right)\cos \left( \dfrac{\theta -3\theta }{2} \right) \right)$
$\Rightarrow \mathrm{f}(\theta)=\sin \theta(2 \sin 2 \theta \cos \theta)[\because \cos (-\theta)=\cos \theta]$
$\Rightarrow \mathrm{f}(\theta)=2 \sin \theta \cos \theta \sin 2 \theta$
Now we have to apply identity to get: $\sin 2\theta =2\sin \theta \cos \theta $
$\Rightarrow \mathrm{f}(\theta)=\sin ^{2} 2 \theta[\because$ $\sin 2\theta =2\sin \theta \cos \theta $
$\therefore \mathrm{f}(\theta) \geq 0$ for all real $\theta$ and lies in region [0,1]
Hence, $f(\theta) \geq 0$ for all real $\theta$
So, the correct answer is option D.
Note: Non-real numbers are numbers that contain a square root of a negative number. Typically, the square root of -1 is denoted as "i", and imaginary numbers are expressed as a multiple of i. Real numbers are all rational and irrational numbers which include whole numbers, repeating decimals and non-repeating decimals.
Negative numbers don't have real square roots since a square is either positive or 0. The square roots of numbers that are not a perfect square are members of the irrational numbers. This means that they can't be written as the quotient of two integers. The set of real numbers includes every number, negative and decimal included, that exists on the number line. The set of real numbers is represented by the symbol $\mathrm{R}$. The set of integers includes all whole numbers (positive and negative), including 0. The set of integers is represented by the symbol $Z$.
Complete step-by-step answer:
Real numbers consist of zero $(0),$ the positive and negative integers $(-3,-1,2,4),$ and all the fractional and decimal values in between $(0.4,3.1415927,1 / 2)$. Real numbers are divided into rational and irrational numbers. A real number is any positive or negative number. This includes all integers and all rational and irrational numbers. For example, a program may limit all real numbers to a fixed number of decimal places.
In mathematics, the irrational numbers are all the real numbers which are not rational numbers. That is, irrational numbers cannot be expressed as the ratio of two integers. Irrational numbers can also be expressed as non-terminating continued fractions and many other ways.
$f(\theta)=\sin \theta(\sin \theta+\sin 3 \theta)$
Now it can be said that:
Applying identity $: \sin \mathrm{A}+\sin \mathrm{B}=2 \sin \dfrac{\mathrm{A}+\mathrm{B}}{2} \cos \dfrac{\mathrm{A} - \mathrm{B}}{2}$
$\Rightarrow f(\theta )=\sin \theta \left( 2\sin \left( \dfrac{\theta +3\theta }{2} \right)\cos \left( \dfrac{\theta -3\theta }{2} \right) \right)$
$\Rightarrow \mathrm{f}(\theta)=\sin \theta(2 \sin 2 \theta \cos \theta)[\because \cos (-\theta)=\cos \theta]$
$\Rightarrow \mathrm{f}(\theta)=2 \sin \theta \cos \theta \sin 2 \theta$
Now we have to apply identity to get: $\sin 2\theta =2\sin \theta \cos \theta $
$\Rightarrow \mathrm{f}(\theta)=\sin ^{2} 2 \theta[\because$ $\sin 2\theta =2\sin \theta \cos \theta $
$\therefore \mathrm{f}(\theta) \geq 0$ for all real $\theta$ and lies in region [0,1]
Hence, $f(\theta) \geq 0$ for all real $\theta$
So, the correct answer is option D.
Note: Non-real numbers are numbers that contain a square root of a negative number. Typically, the square root of -1 is denoted as "i", and imaginary numbers are expressed as a multiple of i. Real numbers are all rational and irrational numbers which include whole numbers, repeating decimals and non-repeating decimals.
Negative numbers don't have real square roots since a square is either positive or 0. The square roots of numbers that are not a perfect square are members of the irrational numbers. This means that they can't be written as the quotient of two integers. The set of real numbers includes every number, negative and decimal included, that exists on the number line. The set of real numbers is represented by the symbol $\mathrm{R}$. The set of integers includes all whole numbers (positive and negative), including 0. The set of integers is represented by the symbol $Z$.
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