Question

The greatest and least value of \[\sin x\cos x\] are?
1) \[1, - 1\]
2) \[\dfrac{1}{2}, - \dfrac{1}{2}\]
3) \[\dfrac{1}{4}, - \dfrac{1}{4}\]
4) \[2, - 2\]

Answer 1

Hint: In this question, we have to find the greatest and least value of \[\sin x\cos x\] . We will use the identity/formula \[\sin 2x = 2\sin x\cos x\] to find the solution.
We will transform \[\sin x\cos x\] into \[\sin 2x\] by multiplying and dividing by \[2\]simultaneously.
We will also use the fact that the value of sine lies between \[ - 1\] and \[1\].

Complete step by step solution:
This problem is based on application of trigonometric identity. Trigonometry is the branch of mathematics that deals with triangles, their ratio of sides and angle. Trigonometric identity is the relationship between ratios of angles. For example, \[{\sin ^2}x + {\cos ^2}x = 1\] is a trigonometric identity.
Trigonometric identities help to solve the question based on t-ratios of angles. We can use these identities to find the value of any t-ratio.
Consider the given question,
We have to find the greatest and least value of \[\sin x\cos x\]
Let’s take the function, \[\sin x\cos x\]
Multiplying and dividing \[\sin x\cos x\] simultaneously by \[2\] we have,
 \[ = \dfrac{{2\sin x\cos x}}{2}\]
From trigonometric identity \[\sin 2x = 2\sin x\cos x\] , we have
Hence, \[\dfrac{{2\sin x\cos x}}{2} = \dfrac{{\sin 2x}}{2}\]
Now we know that the value of sine always lies between \[ - 1\] and \[1\].
i.e. \[ - 1 \leqslant \sin 2x \leqslant 1\]
0n dividing the above inequality by 2. we have,
Therefore, \[\dfrac{{ - 1}}{2} \leqslant \dfrac{{\sin 2x}}{2} \leqslant \dfrac{1}{2}\] or \[\dfrac{{ - 1}}{2} \leqslant \sin x\cos x \leqslant \dfrac{1}{2}\]
Hence, we see that \[\sin x\cos x\] lies between \[ - \dfrac{1}{2}\] and \[\dfrac{1}{2}\] .
Hence the greatest value is \[\dfrac{1}{2}\] and least value is \[ - \dfrac{1}{2}\].
Hence option (\[2\]) is correct
So, the correct answer is “Option 2”.

Note: There are many trigonometric formulas , Some important Trigonometric formulas are
 \[
  \sin 2x = 2\sin x\cos x \\
  \cos 2x = {\cos ^2}x - {\sin ^2}x \\
  \tan 2x = \dfrac{{2\tan x}}{{1 - {{\tan }^2}x}} \;
    \]
While solving an inequality, we must divide, multiply, add and subtract in all parts of inequality simultaneously.
The basic difference between inequality and equation is that equation is equal to zero while inequality has order relationship with greater than, less than etc.