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Problem We need to find the period of the following: $f(x)=(\sin(x))(\cos(x))$ using basic trigonometric identities which is as follows:

My steps disclaimer! I know the steps but I will pin point where I am confused and please explain so Steps:

  • 1) $f(x)=\sin(x)\cos(x)$
  • 2) $f(x)=\frac{1}{2}\sin(2x)$ <-- I do not understand the transition from line 1 to line 2
  • 3) therefore period is $\pi$
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  • $\begingroup$ Are you familiar with trig identities such as $\sin(x+y)=\sin(x)\cos(y)+\cos(x)\sin(y)$? $\endgroup$ – stewbasic Jul 11 '16 at 6:18
  • $\begingroup$ yes its just the basic trig identities $\endgroup$ – John Rawls Jul 11 '16 at 6:22
  • $\begingroup$ @JohnRawls: Then put $y=x$ and you get $\sin(2x)=2\sin(x)\cos(x)$. $\endgroup$ – André Nicolas Jul 11 '16 at 6:26
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Use

1) $2\sin \alpha \cdot \cos \alpha = \sin 2 \alpha$

2) If $y=a\sin(kx+b)+c$, then period is $T=\frac{2\pi}{k}$.

Hence,

$$f(x)=\sin x \cos x= \frac12 \cdot 2\sin x \cos x=\frac12 \sin 2x$$ $$T=\frac{2\pi}{2}=\pi$$

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One of the trig identities is the following:

$$\sin(2\theta) = 2\sin(\theta)\cos(\theta)$$

from which the result follows immediately.

Various proofs for this identity exist, but I prefer using complex analysis to prove the result:

Given that $\sin(z) = \frac{e^{iz}-e^{-iz}}{2i}$ and $\cos(z)=\frac{e^{iz}+e^{-iz}}{2}$ one has:

$$2\sin(z)\cos(z) = 2\cdot \frac{e^{iz}-e^{-iz}}{2i}\cdot \frac{e^{iz}+e^{-iz}}{2} = \frac{e^{2iz}+1-1-e^{-2iz}}{2i}=\frac{e^{i(2z)}-e^{-i(2z)}}{2i}=\sin(2z)$$

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