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How do you write $2\sin 3\cos 3$ as a single trigonometric function?

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Last updated date: 26th Jul 2024
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Answer
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Hint: We know that $\sin \theta $ is a periodic function with period $2\pi $ and also $\cos \theta $ is periodic function with period $2\pi $
The value of $\sin \theta $ is maximum at $\dfrac{\pi }{2}$ from $0$ to $2\pi $ the value is $1.$
The value of $\cos \theta $ is maximum at $0{}^\circ $ and $2\pi $ from $0$ to $2\pi $ the value is $1.$
The $\sin \theta $ is minimum at $0,\pi ,2\pi $ and the value is $0$ from $0$ to $2\pi $
The $\sin \theta $ is $-1$ an angle of $\dfrac{3\pi }{2}$
The $\cos \theta $ is minimum at $\dfrac{\pi }{2}$ and $\dfrac{3\pi }{2}$ the value is $0$ from $0$ to $2\pi $ and $\cos \theta $ is $-1$ and $\theta $ is $\pi $
When $\sin \theta $ and $\cos \theta $ are in the product of each other and twice of it. Then it is equal to sin of twice the angle.
$2\sin \theta \cos \theta =\sin 2\theta $

Complete step by step solution:
It is given that $2\sin 3\cos 3$
Here $3$ is the angle at sin and cos.
The angle of both are equal
Therefore, we can use the formula.
$2\sin \theta \cos \theta =\sin 2\theta $
We can put $\sin 3$ in place at $\sin \theta $ and $\cos 3$ in place of $\cos \theta $
Therefore,
$2\sin 3\cos 3=\sin 2\times 3$
The product of $2$ and $3$ is $6$
$2\sin 3\cos 3=\sin 6$

The value of $2\sin 3\cos 3$ as a single trigonometric function is $\sin 6.$

Additional Information:
This question can be asked in the other way also,
For example
Split $\sin 240$ in two trigonometric terms.
So, in this case you can do it as,
First let's split the angle which is present in the sin.
$240$ can be split as,
$120+120$ we can write it as $2\left( 120 \right)$
So, the $\sin 240$ can be written as $\sin 2\left( 120 \right)$
And we know that,
$\sin 2\theta =2\sin \theta \cos \theta $
Here, $2\theta =2\left( 120 \right)$
So, $\theta $ will be $120$
$\sin 240=2\sin 120\cos 120$
The $\sin 240$ in two trigonometric terms in $2\sin 120\cos 120.$

Note: In the question the $\theta $ is $3.$ and the formula is only applicable if the angle of sin and cos are equal.
The maximum value of $\sin 2\theta $ because both sin in common and only change is in the angle of both.
The maximum values will be different is there is term $2\sin \theta $