I know that if Tan A =Tan B then A = B + nπ
This is because Tan has a periodicity of π
What is the equivalent formula for Sin A = Sin B and Cos A = Cos B
Please explain also why these formulas are true
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$\begingroup$Using Prosthaphaeresis Formulas,
$$\sin A-\sin B=0\iff2\sin\frac{A-B}2\cos\frac{A+B}2=0$$
If $\sin\dfrac{A-B}2=0\implies\dfrac{A-B}2=m\pi\iff A=2m\pi+B$
If $\cos\dfrac{A+B}2=0\implies\dfrac{A+B}2=(2m+1)\dfrac\pi2\iff A=2m\pi+\pi-B$
Similarly, we can derive $\cos A-\cos B=0\implies A=2r\pi\pm B$
where $m,r$ are any integers
Can you take it home from here?
$\endgroup$ 0 $\begingroup$$ A = 2n\pi + B $ since $ Sin $ and $Cos$ has a period $2n\pi$
$\endgroup$ 2 $\begingroup$$\sin A=\sin B$ gives $$A=B+2n\pi \vee A=\pi-B+2n\pi$$
$\cos A=\cos B$ gives $$A=B+2n\pi \vee A=-B+2n\pi$$
(for $n\in\mathbb{Z}$)
Explanation
For $\sin A=\sin B$, think of two angles in the unit circle. Sine is the $y$-coordinate, so either the angles are the same (first case) or we seek the other angle that has the same $y$-coordinate (second case).
The same for $\cos A=\cos B$, think of two angles in the unit circle. Cosine is the $x$-coordinate, so either the angles are the same (first case) or we seek the other angle that has the same $x$-coordinate (second case).
$\endgroup$ $\begingroup$$A=B+2k\pi$ or $A=\pi−B+2k\pi$ with $k\in\mathbb{Z}$ for $\sin(A)=\sin(B)$.
$A=B+2k\pi$ or $A=−B+2k\pi$ with $k\in\mathbb{Z}$ for $\cos(A)=\cos(B)$.
As stated in the comments, this can easily be seen by drawing the function and a horizontal line intersecting the function.
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