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I cannot seem to understand why this is true. Same for $\cos(x) = -\cos(180^{\circ}-x)$ and $\tan(x) = -\tan(180^{\circ}-x)$. Without the use of the compound angle formulas.

Thanks

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  • $\begingroup$ Remember (or study) the extension of the basic trigonometric functions from a straight-angled triangle to the unit circle in the plane: it's immediate from this and the only way I can see without the compound angle formulae, which are based on the unit circle thing, indeed. $\endgroup$ – DonAntonio Nov 16 '13 at 11:34
  • $\begingroup$ You can always verify this using addition formula for $\sin(180^{\circ}-x)$. But this has stronger geometric explanation using unit circle. $\endgroup$ – Cortizol Nov 16 '13 at 11:55
  • $\begingroup$ Any answer to this question will have to depend very heavily on your definition of $\sin$ as a function defined on all of ${\mathbb R}$. $\endgroup$ – Christian Blatter Nov 24 '13 at 19:41
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Hint Rebember that $ \tan( 180^o \pm x ) = \frac{\sin( 180^o \pm x)}{\cos(180^o\pm x)}. $ Now draw the trigonometric cycle and observes the measures of sine and cosine.

enter image description here

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There are geometric reasons for the relations $\sin(\pi-x)=\sin x$ and $\cos(\pi-x)= -\cos x$ (I prefer not using degrees, change $\pi$ into degrees, if you want).

The historic definition of sine and cosine are by means of rectangle triangles. If $ABC$ is a triangle with a right angle in $B$ and $\alpha$ is the angle with vertex in $A$, then $$ \sin\alpha=\frac{BC}{AC},\quad \cos\alpha=\frac{AB}{AC}, $$ so that the relation $\sin^2\alpha+\cos^2\alpha=1$ follows from Pythagoras' theorem. It follows also, by the very definition, that $$ \sin\left(\frac{\pi}{2}-\alpha\right)=\cos\alpha,\quad \cos\left(\frac{\pi}{2}-\alpha\right)=\sin\alpha. $$

This defines the sine and the cosine for all acute angles. What can we do for obtuse angles?

Consider the triangle in the following figure

enter image description here

If we call $a=BC$, $b=AC$, $c=AB$ (I omit the bar to denote the length for simplicity) and $d=AH$ (where $CH$ is the perpendicular to $AB$), we can easily see by a double application of Pythagoras' theorem that $$ a^2=b^2+c^2-2cd $$ or, introducing the cosine, $$ a^2=b^2+c^2-2bc\cos\alpha. $$

Suppose now we have an obtusangle triangle

enter image description here

With the same notation as before, we have $$ a^2=b^2+c^2+2cd $$ or, introducing the cosine $$ a^2=b^2+c^2+2bc\cos(\pi-\alpha) $$ because now $d=b\cos(\pi-\alpha)$. By defining $$ \cos\alpha=-\cos(\pi-\alpha) $$ for an obtuse angle $\alpha$, the relation $$ a^2=b^2+c^2-2bc\cos\alpha $$ (al-Kashi theorem) holds for all triangles. It only remains to supplement the definition of $\cos(\pi/2)$, but by Pythagoras' theorem we need to set $$ \cos\frac{\pi}{2}=0 $$

For the sine, the reasoning can start from the sine's theorem:

enter image description here

The figure, with some simple reasoning, gives $$ a\sin\alpha=2R $$ where $R=OB=OC=OA$. If you take the diametral opposite point $A'$ and of $A$, you immediately get that the sine of the supplementary angle $\alpha'$ of $\alpha$ must obey, if its sine were defined, $$ a\sin\alpha'=2R $$ that means $$ \sin(\pi-\alpha)=\sin\alpha. $$ Again, this forces us to set $\sin(\pi/2)=1$, which agrees with $\cos(\pi/2)=0$. For the zero angle, we can use the complementary angle rules to conclude we have to define $\sin0=0$ and $\cos0=1$.

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Draw a circle in the Cartesian plane with the centre at $0$ and see that $\sin x$ is the $y$ axis and figure out that $\sin x$ is positive in the the upper part of the unit disk.

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Because, if you take $\cos(180°-x)$ you are effectively getting the opposite of $\cos(x)$. Due to $180°$ being halfway over the circle diameter. So if you do take the opposite of that $-\cos(180°-x)$ you get $x$ back, because you switch the sign.

Easy to see with this image:

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