Question 6cb94

Oct 10, 2016

By using Demoivre's theorem multiply the trig expression out and extract the imaginary part.

Explanation:

we shall use ${\left(\cos \theta + i \sin \theta\right)}^{n} = \cos n \theta = i \sin n \theta$

Since we want $\sin 4 \theta$ we shall let $n = 4$

so $\sin 4 \theta = I m \left({\left(\cos \theta + i \sin \theta\right)}^{4}\right)$
since ${\left(\cos \theta + i \sin \theta\right)}^{4} = \cos 4 \theta + i \sin 4 \theta$
expand ${\left(\cos \theta + i \sin \theta\right)}^{4}$

 (costheta+isintheta)^4= cos^4theta +4(cos^3theta(isintheta)) +6(cos^2theta(isintheta)^2)+4(costheta(isintheta)^3)+(isintheta)^4 

 (costheta+isintheta)^4= cos^4theta +4i(cos^3thetasintheta) -6(cos^2thetasin^2theta)-4i(costhetasin^3theta))+sin^4theta #
now extract the imaginary part

$\sin 4 \theta = I m \left({\left(\cos \theta + i \sin \theta\right)}^{4}\right)$

$\sin 4 \theta = 4 {\cos}^{3} \theta \sin \theta - 4 \cos \theta {\sin}^{3} \theta$

$\sin 4 \theta = 4 \left({\cos}^{3} \theta \sin \theta - \cos \theta {\sin}^{3} \theta\right)$

as required

Oct 11, 2016

Prove sin 4t

Explanation:

Use trig identity: sin (a + b) = sin a.cos b + sin b.cos a
sin 4t = sin (2t + 2t) = sin 2t. cos 2t + cos 2t.sin 2t = 2sin 2t.cos 2t =
Since,
sin 2t = 2sin t.cos t
$\cos 2 t = {\cos}^{2} t - {\sin}^{2} t$,
There for:
$\sin 4 t = 4 \sin t . \cos t \left({\cos}^{2} t - {\sin}^{2} t\right)$
$\sin 4 t = 4 \left(\sin t . {\cos}^{3} t - \cos t . {\sin}^{3} t\right)$