sin ⁡ ( 2 θ ) = n sin ⁡ ( θ ) cos ⁡ ( θ ) \sin(2θ) = n\sin(θ)\cos(θ)

Algebra Level 2

sin ⁡ ( 2 θ ) = n sin ⁡ ( θ ) cos ⁡ ( θ ) \large \sin(2θ) = n \sin(θ)\cos(θ)

Determine the value of n n in the equation above.


The answer is 2.

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2 solutions

James Watson
Aug 2, 2020

sin ⁡ ( 2 θ ) = sin ⁡ ( θ + θ ) = sin ⁡ ( θ ) cos ⁡ ( θ ) + cos ⁡ ( θ ) sin ⁡ ( θ ) = 2 sin ⁡ ( θ ) cos ⁡ ( θ ) \begin{aligned} \sin(2\theta) &= \sin(\theta + \theta) \\ &= \sin(\theta)\cos(\theta) + \cos(\theta)\sin(\theta) \\ &= 2\sin(\theta)\cos(\theta) \end{aligned}

Therefore n = 2 n=\boxed{2}

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This is an identity: s i n ( 2 θ ) = n s i n ( θ ) c o s ( θ ) sin(2θ) = nsin(θ)cos(θ) .

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