Author Topic: Half angle formulae in further trig (Read 1646 times) Tweet Share

hinakamishiro · « **on:** July 24, 2016, 11:55:05 am »

Hi guys!
I was wondering if anyone could helpfully explain half angle theorems in further trig to me? Cause in having a lot of trouble with them and I really don't get it $:-\$ Thanks!

RuiAce · « **Reply #1 on:** July 24, 2016, 11:57:03 am »

What do you mean?
$\sin 2x = 2\sin x \cos x\\ \text{so }\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$
$\text{Similarly }\cos 2x = \cos^2x-\sin^2x\\ \text{so }\cos x = \cos^2\frac{x}{2}-\sin^2\frac{x}{2}$
Unless you mean the tangent half-angle substitution t=tan(x/2)

hinakamishiro · « **Reply #2 on:** July 24, 2016, 12:22:56 pm »

Quote from: RuiAce on July 24, 2016, 11:57:03 am

What do you mean?
$\sin 2x = 2\sin x \cos x\\ \text{so }\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$
$\text{Similarly }\cos 2x = \cos^2x-\sin^2x\\ \text{so }\cos x = \cos^2\frac{x}{2}-\sin^2\frac{x}{2}$
Unless you mean the tangent half-angle substitution t=tan(x/2)

Yes I mean't the tan one as well, sorry fot the confusion. But I was also wondering how to derive exact values for half angles? Thanks I hope that's clearer.

RuiAce · « **Reply #3 on:** July 24, 2016, 12:31:16 pm »

$\text{Something like }\sin 15^\circ\text{ is best done using }\sin(45^\circ - 30^\circ)\\ \\ \sin 22.5^\circ\text{ is a bit awkward but }\\ \tan 22.5^\circ\text{ can be done using }\tan 45^\circ = 1 = \frac{2\tan 22.5^\circ}{1-\tan^2 22.5^\circ}\\ \text{As can }\cos 22.5^\circ\text{ using }\cos 45^\circ = \frac{1}{\sqrt2}=2\cos^2 22.5^\circ - 1$
$\text{The idea with a tangent half-angle substitution is that }\\ \textbf{IF }\text{you let }t=\tan \frac{x}{2}\\ \text{The following become consequences:}\\ \begin{align*}\sin x &= \frac{2t}{1+t^2}\\ \cos x &= \frac{1-t^2}{1+t^2}\\ \tan x &= \frac{2t}{1-t^2}\end{align*}$
$\text{The proof for the tan one is immediate; it's just a double angle formula.}\\ \text{One way to prove it for sine:}\\ \begin{align*}LHS&=\sin x\\ &= 2\sin \frac{x}{2}\cos \frac{x}{2}\\ &= \frac{2\sin \frac{x}{2}}{\cos \frac{x}{2}} \times \cos^2 \frac{x}{2}\\ &= 2\tan \frac{x}{2} \times \frac{1}{\sec^2\frac{x}{2}}\\ &= 2t \times \frac{1}{1+\tan^2\frac{x}{2}}\\ &= \frac{2t}{1+t^2}\\ &= RHS\end{align*}$
Use of them is rare, but viable. They're mostly an alternative to the auxiliary angle method in MX1, but fact is you should always prefer auxiliary angle over these.

hinakamishiro · « **Reply #4 on:** July 24, 2016, 12:40:22 pm »

Quote from: RuiAce on July 24, 2016, 12:31:16 pm

$\text{Something like }\sin 15^\circ\text{ is best done using }\sin(45^\circ - 30^\circ)\\ \\ \sin 22.5^\circ\text{ is a bit awkward but }\\ \tan 22.5^\circ\text{ can be done using }\tan 45^\circ = 1 = \frac{2\tan 22.5^\circ}{1-\tan^2 22.5^\circ}\\ \text{As can }\cos 22.5^\circ\text{ using }\cos 45^\circ = \frac{1}{\sqrt2}=2\cos^2 22.5^\circ - 1$
$\text{The idea with a tangent half-angle substitution is that }\\ \textbf{IF }\text{you let }t=\tan \frac{x}{2}\\ \text{The following become consequences:}\\ \begin{align*}\sin x &= \frac{2t}{1+t^2}\\ \cos x &= \frac{1-t^2}{1+t^2}\\ \tan x &= \frac{2t}{1-t^2}\end{align*}$
$\text{The proof for the tan one is immediate; it's just a double angle formula.}\\ \text{One way to prove it for sine:}\\ \begin{align*}LHS&=\sin x\\ &= 2\sin \frac{x}{2}\cos \frac{x}{2}\\ &= \frac{2\sin \frac{x}{2}}{\cos \frac{x}{2}} \times \cos^2 \frac{x}{2}\\ &= 2\tan \frac{x}{2} \times \frac{1}{\sec^2\frac{x}{2}}\\ &= 2t \times \frac{1}{1+\tan^2\frac{x}{2}}\\ &= \frac{2t}{1+t^2}\\ &= RHS\end{align*}$
Use of them is rare, but viable. They're mostly an alternative to the auxiliary angle method in MX1, but fact is you should always prefer auxiliary angle over these.

I see this makes it all much clearer, thanks ^^

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Author Topic: Half angle formulae in further trig (Read 1646 times) Tweet Share

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