It is known that 5π<α<((13π)/2). cosα=−(1/4) calculate sin2α


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Question Number 56498 by kelly33 last updated on 17/Mar/19

$I t i s k n o w n t h a t 5 π < α < \frac{13 π}{2} . c o s α = - \frac{1}{4}$ $c a l c u l a t e s i n 2 α$
Commented bykelly33 last updated on 17/Mar/19

$p l e a s e h e l p m e$
	Answered by tanmay.chaudhury50@gmail.com last updated on 17/Mar/19

	$s i n 2 α = 2 s i n α c o s α$ $i g n o r i n g s i g n s i n α = \frac{\sqrt{15}}{4}$ $n o w 5 π < α < \frac{13 π}{2}$ $s o 10 π < 2 α < 13 π$ $(5 \times 2 π) < 2 α < (6 \times 2 π + π)$ $s o 2 α l i e s i n s e c o n d q u a d r a n t$ $s o s i n 2 α = 2 \times \frac{\sqrt{15}}{4} \times \frac{1}{4} = \frac{\sqrt{15}}{8}$ $i t h i n k q u e s t i o n i t s e l f h a s s o m e e r r o r . .$ $i t s h o u d b e 6 π < α < \frac{13 π}{2}$
	Commented byMJS last updated on 17/Mar/19

	$\cos α = - \frac{1}{4} \Rightarrow α = 2 π n \pm (\frac{π}{2} + \arcsin \frac{1}{4})$ $5 π < α < \frac{13 π}{2} \Rightarrow α = \frac{11 π}{2} - \arcsin \frac{1}{4}$ $\sin 2 α = \frac{\sqrt{15}}{8}$
	Commented bykelly33 last updated on 18/Mar/19

	$α = 2 π n \pm (\frac{π}{2} + a r c s i n \frac{1}{4})$ $w h e r e d i d t h i s f o r m u l a c o m e f r o m ?$
	Commented byMJS last updated on 19/Mar/19

	$\arccos (- x) = \pm (\frac{π}{2} + \arcsin x)$ $this should be clear . . .$ $but \cos α has a period of 2 π$ $\cos α = - \frac{1}{4}$ $\cos (α + 2 π n) = - \frac{1}{4}; n \in Z$ $α + 2 π n = \arccos (- \frac{1}{4})$ $α + 2 π n = \pm (\frac{π}{2} + \arcsin \frac{1}{4})$ $α = - 2 π n \pm (\frac{π}{2} + \arcsin \frac{1}{4}) but n \in Z \Rightarrow$ $\Rightarrow we can write$ $α = 2 π n \pm (\frac{π}{2} + \arcsin \frac{1}{4})$
	Commented bykelly33 last updated on 19/Mar/19

	$t h a n k s r$
	Commented bykelly33 last updated on 19/Mar/19

	$S o r r y, y o u c a n e x p l a i n h o w {i t}^{'} s h e r e . . .$ $5 π < α < \frac{13 π}{2} \Rightarrow α = \frac{11 π}{2} - \arcsin \frac{1}{4}$ $c o m e t o t h e f i n a l r e s u l t ?$ $A n d w h e r e t h i s v a l u e d h a s e m e r g e d ?$ $α = \frac{11 π}{2} - \arcsin \frac{1}{4}$
	Commented bykelly33 last updated on 19/Mar/19

	$S o r r y, y o u c a n e x p l a i n h o w {i t}^{'} s h e r e . . .$ $5 π < α < \frac{13 π}{2} \Rightarrow α = \frac{11 π}{2} - \arcsin \frac{1}{4}$ $c o m e t o t h e f i n a l r e s u l t ?$ $A n d w h e r e t h i s v a l u e d h a s e m e r g e d ?$ $α = \frac{11 π}{2} - \arcsin \frac{1}{4}$
	Commented byMJS last updated on 19/Mar/19

	$general result :$ $α = 2 π n \pm (\frac{π}{2} + \arcsin \frac{1}{4})$ $now we must test which n \in Z makes α fit$ $into the interval .$ $α_{n}^{-} = 2 π n - (\frac{π}{2} + \arcsin \frac{1}{4})$ $α_{n}^{+} = 2 π n + (\frac{π}{2} + \arcsin \frac{1}{4})$ $] 5 π; \frac{13}{2} π [\approx] 15.708; 20.420 [$ $α_{3}^{-} \approx 17.026 is the only one to fit$ $\Rightarrow α = 6 π - (\frac{π}{2} + \arcsin \frac{1}{4}) = \frac{11 π}{2} - \arcsin \frac{1}{4}$ $\sin 2 α = \sin (11 π - 2 \arcsin \frac{1}{4}) =$ $[because of period 2 π : \sin (11 π - θ) = \sin θ]$ $= \sin (2 \arcsin \frac{1}{4}) =$ $[\sin 2 θ = 2 \sin θ \cos θ]$ $= 2 (\sin \arcsin \frac{1}{4}) (\cos \arcsin \frac{1}{4}) =$ $= 2 \times \frac{1}{4} \times \cos \arcsin \frac{1}{4} =$ $[\cos \arcsin t = \sin \arcsin t = \sqrt{1 - t^{2}}]$ $= \frac{1}{2} \sqrt{1 - \frac{1}{16}} = \frac{\sqrt{15}}{8}$
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