Question

Why is cos(-x) = cos(x) but sin (-x) = -sin(x)?

Answer 1

Let $\rm{P(x,\ y)}$ be the coordinate on a plane and let the angle be $\theta$.

Consider reflecting on the x-axis.

Then the coordinate of the image and the angle are $\rm{P'(x,\ -y)}$ and $-\theta$.

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$\boxed{\rm{\sin \theta=\dfrac{y}{r}}}$

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$\rm{\sin (-\theta )=-\dfrac{y}{r}}$

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$\therefore \sin (-\theta )=-\sin \theta$

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$\boxed{\rm{\cos \theta=\dfrac{x}{r}}}$

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$\rm{\cos (-\theta )=\dfrac{x}{r}}$

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$\therefore \cos (-\theta )=\cos \theta$

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$\boxed{\rm{\tan \theta =\dfrac{y}{x}}}$

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$\rm{\tan (-\theta )=-\dfrac{y}{x}}$

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$\therefore \tan (-\theta )=-\tan \theta$

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Hence proven.

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$\Large\textrm{Learn More}$

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$\textbf{- Meaning of Trigonometric Functions}$

$\boxed{\rm{\sin \theta=\dfrac{y}{r}}}$

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$\boxed{\rm{\cos \theta=\dfrac{x}{r}}}$

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$\boxed{\rm{\tan \theta=\dfrac{y}{x}}}$

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$\textbf{- Periodicity}$

$\boxed{\rm{\sin(2\pi+\theta)=\sin\theta}}$

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$\boxed{\rm{\cos(2\pi+\theta)=\cos\theta}}$

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$\boxed{\rm{\tan(\pi+\theta)=\tan\theta}}$

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$\textbf{- Changing Signs on Quadrants (A.S.T.C.)}$

$\boxed{\rm{1st\ quadrant:\ x>0,\ y>0}}$

$\rm{\therefore \sin \theta >0,\ \cos \theta >0,\ \tan \theta >0\ \textbf{(All)}}$

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$\boxed{\rm{2nd\ quadrant:\ x<0,\ y>0}}$

$\rm{\therefore \sin \theta >0,\ \cos \theta <0,\ \tan \theta <0\ \textbf{(Sine)}}$

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$\boxed{\rm{3rd\ quadrant:\ x<0,\ y<0}}$

$\rm{\therefore \sin \theta <0,\ \cos \theta <0,\ \tan \theta >0\ \textbf{(Tangent)}}$

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$\boxed{\rm{4th\ quadrant:\ x>0,\ y<0}}$

$\rm{\therefore \sin \theta <0,\ \cos \theta >0,\ \tan \theta <0\ \textbf{(Cosine)}}$

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$\textbf{- Reciprocal Trigonometric Functions}$

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$\boxed{\rm{\sin\theta\csc\theta=1}}$

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$\boxed{\rm{\cos\theta\sec\theta=1}}$

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$\boxed{\rm{\tan\theta\cot\theta=1}}$

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$\textbf{- Consequencial Formula}$

$\rm{Angle:\ \dfrac{n}{2}\pi\pm\theta}$

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Consider $\theta$ as an acute angle.

The sign changes depend on the sign of the previous function.

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$\textrm{n is odd}$

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$\boxed{\rm{\sin\to\cos,\ \cos\to\sin,\ \tan\to\cot}}$

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$\textrm{n is even}$

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$\boxed{\rm{\sin\to\sin,\ \cos\to\cos,\ \tan\to\tan}}$