Question-39529

Question Number 39529 by Rio Mike last updated on 07/Jul/18

Commented by Rio Mike last updated on 07/Jul/18

i n e e d h e l p g u y s . c a n s o m e o n e p l e a s d

e x p l a i n t o m e w h e n t o u s e e a c h

o f t h e a b o v e q u a d r a n t s a n d h o w ?

i n t r i g o n o m e t r y i c a n s o l v e t h e e q u a t i o n

b u t {w o n}^{'} t k n o w t h e q u a d r a n t t o u s e a n d h o w .

f o r e x a m p l e : s o l v e f o r θ t h e e q u a t i o n

2 {s i n}^{2} θ = s i n θ

2 {s i n}^{2} θ - s i n θ = 0

s i n θ (2 s i n θ - 1) = 0

E i t h e r s i n θ = 0 o r θ = {s i n}^{- 1} \frac{1}{2}

w h a t d o i d o n o w ?

Commented by Rio Mike last updated on 07/Jul/18

t h a n k s g u y s

Commented by tanmay.chaudhury50@gmail.com last updated on 07/Jul/18

s i n θ (2 s i n θ - 1) = 0

s i n θ = 0 s o θ = 0

2 s i n θ - 1 = 0

s i n θ = \frac{1}{2} = s i n \frac{Π}{6}

θ = \frac{Π}{6}

n o w t h i s t w o v a l u e l i e s i n f i r s t q u a a d r a n t

s i n θ = + v e w h e n θ 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 0 a l s o c a n b e w r i t t e n a s s i n (Π - 0)

t h a t i s s i n Π

n e x t s i n \frac{Π}{6} a l s o c a n b e w r i t t e n a s

s i n (Π - \frac{Π}{6}) \dots s i n \frac{5 Π}{6}

t h e r e i s f o r m u l a p l s w a i t i s h a l l p o s t

s i n θ = s i n α

θ = n Π + {(- 1)}^{n} α

Commented by tanmay.chaudhury50@gmail.com last updated on 07/Jul/18

Commented by MJS last updated on 07/Jul/18

in geometric examples you have to think

which solution fits . i . e . if you calculate an

angle of a triangle it must fit into this triangle

(α + β + γ = 180 °)

in “ free from any {praxis}^{″} examples {you}^{'} ll

have to deal with more than one possible

solutions, same as in equations of 2^{nd} or

higher degree .

it might help to think of the coordinates

of the points of the circle

x^{2} + y^{2} = 1 \Rightarrow

\Rightarrow \cos^{2} α + \sin^{2} α = 1 \Rightarrow

\Rightarrow P = (\begin{matrix} \cos α \\ \sin α \end{matrix})

0 ° < α < 90 ° \Rightarrow P in 1^{st} quadrant

90 ° < α < 180 ° \Rightarrow P in 2^{nd} quadrant

180 ° < α < 270 ° \Rightarrow P in 3^{rd} quadrant

180 ° < α < 360 ° \Rightarrow P in 4^{th} quadrant

α = 0 ° = 360 ° \Rightarrow P on x^{+}

α = 90 ° \Rightarrow P on y^{+}

α = 180 ° \Rightarrow P on x^{-}

α = 270 ° \Rightarrow P on y^{-}

electronic calculators handle angles in

] - 180 °; 180 °] instead of [0 °; 360 ° [or in

r adiant] - π; π] instead of [0; 2 π [

you have to be careful with arc - functions

(always remember what was the question,

or / and try the other possibilities to find out

which answer fits into the given example)

0 ° 45 ° 90 ° 135 ° 180 ° 225 ° 270 ° 315 ° 360 °

0 ° 45 ° 90 ° 135 ° 180 ° - 135 ° - 90 ° - 45 ° 0 °

0 \frac{π}{4} \frac{π}{2} \frac{3 π}{4} π \frac{5 π}{4} \frac{3 π}{2} \frac{7 π}{4} 2 π

0 \frac{π}{4} \frac{π}{2} \frac{3 π}{4} π - \frac{3 π}{4} - \frac{π}{2} - \frac{π}{4} 0

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