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Integrating using half angle formula. Ask Question Asked 9 years, 4 months ago. Modified 9 years, 4 months ...
If this is not satisfying, you can try an alternative derivation that doesn't fuss with plus and minus signs: $$ \tan\frac\theta2=\frac{\sin\frac\theta2}{\cos\frac\theta2}=\frac{2\sin\frac\theta2\sin\frac\theta2}{2\sin\frac\theta2\cos\frac\theta2}=\frac{1-\cos\theta}{\sin\theta}, $$ using the double-angle identities $$ \cos2t=1-2\sin^2t $$ and ...
Only the half-angle approach with the conjugate works for both, so that is the correct formulation. Read on for the long answer. Consider the two cases of the input vector v being parallel and perpendicular to the rotation axis u, in the below formula.
I can derive the sin, cos and tan half angle formulas from the cosine double angle formula. But I'm having trouble deriving the sine half angle formula from the sine double angle formula. Below is my attempt at deriving sine half angle formula from sine double angle formula. And I could go no further. Could someone provide me with a hint? Edit 1:
Half angle formulas. The Half-Angle formulas for $\sin$ and $\cos$ are then obtained from the Double Angle formula for $\cos$ by writing, for example, $\cos\theta=\cos(2\cdot{\theta\over2})$ The $\tan$ formula here can easily be obtained from the other two. (Note the forms for the $\cos$ and $\sin$ formulas. These aren't to hard to memorize)
The formulas \eqref{1}, \eqref{2}, \eqref{3} explain the Heron–Brahmagupta–Bretschneider development better than I have seen anywhere else. This made me wonder what would happen if I analogously applied the half-angle formulas to formulas where half-angles explicitly appeared, such as Mollweide's (rather Newton's) formula or the law of ...
Another easy way with your upper triangle: By the whole triangle $\;\Delta APB\;$ and then a little algebra + trigonometry,
T. Bongers is right: When an angle is on the interval $0 < \theta < 2\pi$, the cosine of the angle is positive.
So the cosine half-angle formula says: Now, we know that co-terminal angles have equal cosines. Consider that $\cos (7\pi/4)$ = $\cos(-\pi/4)$. However, if you apply the half angle formula to $(7\pi/4)$ you get a different answer than if you apply the half angle formula to $(-\pi/4)$.
As "trigonographs" go, this one seems unsatisfying. Maybe I'm missing a cool geometric trick, but I don't see a way to tease-out the half-angle formulas from the elements given, without invoking a double-angle formula. Ideally, trigonographs are more self-evident.