Trig: Ratios of Double Angles (1 Viewer)

Gigacube · Jun 19, 2011

I'm stuck on how to solve this question.

Find the exact value of:
$\cos22.5^{\circ} \sin22.5^{\circ}$

So far all I can think of is:
$1-\sin 2x^{\circ}$

The answer in the book is:
$\frac{1}{2\sqrt{2}} = \frac{\sqrt{2}}{4}$

funnytomato · Jun 19, 2011

sin(2a) = 2 sin(a)cos(a)?

funnytomato · Jun 19, 2011

funnytomato said:
sin(2a) = 2 sin(a)cos(a)?

where a=22.5°

SpiralFlex · Jun 19, 2011

$\sin 2\theta = 2\sin\theta\cos\theta$

Hence,

$\frac{\sin 2\theta}{2}=\sin\theta\cos\theta$

Now we can see that,

$\sin \theta \cos \theta$ is identical to our expression of $\sin22.5^\circ\cos22.5^\circ$

So,

$\theta =22.5^\circ$

$\frac{\sin45^\circ }{2}=\sin22.5^\circ \cos 22.5^\circ$

$=\frac{\frac{\sqrt{2}}{2}}{2}$

$=\frac{\sqrt{2}}{4}$

Gigacube · Jun 19, 2011

funnytomato said:
where a=22.5°

Thanks!

SpiralFlex said:
$\sin 2\theta = 2\sin\theta\cos\theta$

Hence,

$\frac{\sin 2\theta}{2}=\sin\theta\cos\theta$

Now we can see that,

$\sin \theta \cos \theta$ is identical to our expression of $\sin \22.5 \cos\ 22.5$

So,

$\frac{\sin \45}{2}=\sin\22.5\cos\22.5$

$=\frac{\frac{\sqrt{2}}{2}}{2}$

$=\frac{\sqrt{2}}{4}$

Thanks! Nice invalid eqn. button.

SpiralFlex · Jun 19, 2011

Gigacube said:
Thanks!

Thanks! Nice invalid eqn. button.

Fixed.

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Trig: Ratios of Double Angles (1 Viewer)

Gigacube

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funnytomato

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funnytomato

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SpiralFlex

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Gigacube

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SpiralFlex

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