2
$\begingroup$

Prove $$\tan\left( \frac {a+b}{2}\right) = \frac{\sin a+\sin b} {\cos a + \cos b } $$

Can someone help? I separated $\tan$ and did double-angle, but I just went into a circle and couldn't get the trig functions without the halves.

$\endgroup$
1
6
$\begingroup$

Use the formulas $$\sin a+\sin b=2\sin\left(\frac{a+b}{2}\right)\cos\left(\frac{a-b}{2}\right),$$ $$\cos a+\cos b=2\cos\left(\frac{a+b}{2}\right)\cos\left(\frac{a-b}{2}\right).$$

Therefore, we have

$$\frac{\sin a+\sin b}{\cos a+\cos b}=\frac{\sin\displaystyle\left(\frac{a+b}{2}\right)}{\cos\displaystyle\left(\frac{a+b}{2}\right)}=\tan\left(\frac{a+b}{2}\right).$$

$\endgroup$
0
$\begingroup$

If $a,b$ are acute angles one can proceed as follows:

Draw a parallelogram with all four sides of length $1,$ as follows. The first side has one endpoint at $(0,0)$ and the angle between it and the $x$-axis is $a$. Its other endpoint is therefore at $(\cos a,\sin a).$ The second side begins at that latter point and its angle with the $x$-axis is $b.$ Its other endpoint is therefore $(\sin a + \sin b, \cos a+\cos b).$ The tangent of the angle from the origin to this last point is therefore $(\sin a+\sin b)/(\cos a + \cos b).$ But the geometric symmetry of the parallelogram shows that that angle is $(a+b)/2.$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.