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How can I graphically show/prove that cos(A+B) is equal to cosA cosB + sinA sinB?

user366312
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  • I think there pictures google – CoffeeArabica May 07 '21 at 20:26
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    The right-hand side of the equation should be a difference, not a sum. – Geoffrey Trang May 07 '21 at 20:28
  • Did you mean $\cos(A+B)=\cos A\cos B-\sin A\sin B$ (\cos(A+B)=\cos A\cos B-\sin A\sin B), $\cos(A-B)=\cos A\cos B+\sin A\sin B$ (\cos(A-B)=\cos A\cos B+\sin A\sin B) or $\cos(A\pm B)=\cos A\cos B\mp\sin A\sin B$ (\cos(A\pm B)=\cos A\cos B\mp\sin A\sin B)? Please use one of these between dollar signs when you edit your question to clarify. – J.G. May 07 '21 at 20:35
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    The question is not a duplicate of this question, but the pictorial answer https://math.stackexchange.com/a/1342/442 is definitely an answer to this question. – GEdgar May 07 '21 at 21:09

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Draw a unit circle. Label the origin $O$, and a point on the $+x$ axis $X$. Draw an angle $A$ that passes through point $P$ on the unit circle, and above it an angle $B$ that passes through point $Q$, so that the total angle from the $x$ axis is $A+B$. (My picture works most easily with two acute angles that sum to less than $\pi/2$.)

Now drop a perpendicular from $OQ$ onto $OP$ at point $R$, and drop a perpendicular from $OR$ to $OX$ at point $S$. Drop a perpendicular from $OQ$ all the way to $OX$ at $T$.

From there, you can work out the lengths of many sides of triangles, giving you products of trigonometric functions, so you can see how to add and subtract distances to get what you need.

As a bonus, the same diagram proves the formula for $\sin (A+B)$ as well.

RobertTheTutor
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