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Question: How do you write \(2\sin 3\cos 3\) as a single trigonometric function?...

How do you write 2sin3cos32\sin 3\cos 3 as a single trigonometric function?

Explanation

Solution

We know that sinθ\sin \theta is a periodic function with period 2π2\pi and also cosθ\cos \theta is periodic function with period 2π2\pi
The value of sinθ\sin \theta is maximum at π2\dfrac{\pi }{2} from 00 to 2π2\pi the value is 1.1.
The value of cosθ\cos \theta is maximum at 00{}^\circ and 2π2\pi from 00 to 2π2\pi the value is 1.1.
The sinθ\sin \theta is minimum at 0,π,2π0,\pi ,2\pi and the value is 00 from 00 to 2π2\pi
The sinθ\sin \theta is 1-1 an angle of 3π2\dfrac{3\pi }{2}
The cosθ\cos \theta is minimum at π2\dfrac{\pi }{2} and 3π2\dfrac{3\pi }{2} the value is 00 from 00 to 2π2\pi and cosθ\cos \theta is 1-1 and θ\theta is π\pi
When sinθ\sin \theta and cosθ\cos \theta are in the product of each other and twice of it. Then it is equal to sin of twice the angle.
2sinθcosθ=sin2θ2\sin \theta \cos \theta =\sin 2\theta

Complete step by step solution:
It is given that 2sin3cos32\sin 3\cos 3
Here 33 is the angle at sin and cos.
The angle of both are equal
Therefore, we can use the formula.
2sinθcosθ=sin2θ2\sin \theta \cos \theta =\sin 2\theta
We can put sin3\sin 3 in place at sinθ\sin \theta and cos3\cos 3 in place of cosθ\cos \theta
Therefore,
2sin3cos3=sin2×32\sin 3\cos 3=\sin 2\times 3
The product of 22 and 33 is 66
2sin3cos3=sin62\sin 3\cos 3=\sin 6

The value of 2sin3cos32\sin 3\cos 3 as a single trigonometric function is sin6.\sin 6.

Additional Information:
This question can be asked in the other way also,
For example
Split sin240\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.
240240 can be split as,
120+120120+120 we can write it as 2(120)2\left( 120 \right)
So, the sin240\sin 240 can be written as sin2(120)\sin 2\left( 120 \right)
And we know that,
sin2θ=2sinθcosθ\sin 2\theta =2\sin \theta \cos \theta
Here, 2θ=2(120)2\theta =2\left( 120 \right)
So, θ\theta will be 120120
sin240=2sin120cos120\sin 240=2\sin 120\cos 120
The sin240\sin 240 in two trigonometric terms in 2sin120cos120.2\sin 120\cos 120.

Note: In the question the θ\theta is 3.3. and the formula is only applicable if the angle of sin and cos are equal.
The maximum value of sin2θ\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 2sinθ2\sin \theta