It-is-known-that-5pi-lt-lt-13pi-2-cos-1-4-calculate-sin2-

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 by kelly33 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 by MJS 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 by kelly33 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 by MJS last updated on 19/Mar/19

\arccos (- x) = \pm (\frac{π}{2} + \arcsin x)

this should be clear \dots

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 by kelly33 last updated on 19/Mar/19

t h a n k s r

Commented by kelly33 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 \dots

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 by kelly33 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 \dots

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 by MJS 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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