Thre trig identity questions (1 Viewer)

sinophile · Apr 19, 2009

The first one i can do, but the proof is very long. I was hoping someone could provide a more succinct one. Second one, not sure if my answer is correct. Thir done I can't do at all.

gurmies · Apr 19, 2009

$Question\,\,2$

$\frac{\sin5\theta }{\sin\theta}-\frac{\cos5\theta}{\cos\theta} = \frac{\sin5\theta\cos\theta-\cos5\theta\sin\theta}{\sin\theta\cos\theta}$

$= \frac{\sin(5\theta-\theta)}{\sin\theta\cos\theta}$

$= \frac{\sin4\theta}{\sin\theta\cos\theta}$

$= \frac{2\sin2\theta\cos2\theta}{\frac{1}{2}\sin2\theta}$

$= 4\cos2\theta$

$Question\,\,5$

$\frac{\sin^{3}\theta+\cos^{3}\theta}{\sin\theta+\cos\theta}-1 = \frac{(\sin\theta+\cos\theta)(\sin^{2}\theta-\sin\theta\cos\theta+\cos^{2}\theta)}{\sin\theta+\cos\theta}-1$

$= \sin^{2}\theta - \sin\theta\cos\theta + \cos^{2}\theta - 1$

$= -\sin\theta\cos\theta$

$= -\frac{1}{2}\sin2\theta$

$Question \,\, 10$

$\sin8A\sin2A = \sin(5A+3A)\sin(5A-3A)$

$= (\sin5A\cos3A+\cos5A\sin3A)(\sin5A\cos3A-\cos5A\sin3A)$

$= \sin^{2}5A\cos^{2}3A - \cos^{2}5A\sin^{2}3A$

$= \sin^{2}5A(1-\sin^{2}3A)-\sin^{2}3A(1-\sin^{2}5A)$

$= \sin^{2}5A - \sin^{2}3A$

$= LHS$

Thre trig identity questions (1 Viewer)

sinophile

Well-Known Member

gurmies

Drover

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