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Question Number 1830 by 112358 last updated on 10/Oct/15

S h o w t h a t g i v e n

4 c o s \frac{π}{3} + 2 \sqrt{3} s i n \frac{π}{3} = 5

t h e n

π = 3 {c o s}^{- 1} (\frac{5}{\sqrt{28}}) + 3 {t a n}^{- 1} (\frac{\sqrt{3}}{2}) .

Answered by Rasheed Soomro last updated on 10/Oct/15

^{″} 4 c o s \frac{π}{3} + 2 \sqrt{3} s i n \frac{π}{3}^{″} i s a c o n s t a n t e q u a l t o 5

s o t o s a y^{″} i f 4 c o s \frac{π}{3} + 2 \sqrt{3} s i n \frac{π}{3} = 5^{″}

i s m e a n i n g l e s s!

S i m i l a r l y^{″} 3 {c o s}^{- 1} (\frac{5}{\sqrt{28}}) + 3 {t a n}^{- 1} (\frac{\sqrt{3}}{2})^{″} i s

a l s o a c o n s t a n t a n d i t i s n o t d e p e n d a n t o n

^{″} 4 c o s \frac{π}{3} + 2 \sqrt{3} s i n \frac{π}{3} = 5^{″}

I t h i n k t h a t t h e Q u e s t i o n i s n o t m e a n i n g f u l!

Commented by 112358 last updated on 10/Oct/15

T h e o r i g i n a l q u e s t i o n i s t h i s .

W r i t e d o w n a v a l u e o f θ i n t h e

i n t e r v a l \frac{π}{4} < θ < \frac{π}{2} t h a t s a t i s f i e s

t h e e q u a t i o n

4 c o s θ + 2 \sqrt{3} s i n θ = 5 .

H e n c e, o r o t h e r w i s e, s h o w t h a t

π = 3 {c o s}^{- 1} (\frac{5}{\sqrt{28}}) + 3 {t a n}^{- 1} (\frac{\sqrt{3}}{2}) .

S h o w t h a t

π = 4 {s i n}^{- 1} (\frac{7 \sqrt{2}}{10}) - 4 {t a n}^{- 1} (\frac{3}{4}) .

I j u s t t h o u g h t t h a t f r o m s e t t i n g

θ = \frac{π}{3} t h e q u e s t i o n r e q u i r e d (t h o u g h

n o t s t r i c t l y) t h a t y o u w o r k f r o m

4 c o s \frac{π}{3} + 2 \sqrt{3} s i n \frac{π}{3} = 5 .

Commented by 112358 last updated on 10/Oct/15

I^{'} v e c o r r e c t e d t h e s t a t e m e n t a n d

I t h i n k I^{'} v e s o l v e d i t .

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