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
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$\begingroup$ 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
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$\begingroup$ 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.
answered Nov 16 '13 at 12:43
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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
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
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:
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:
