Sum up this trigonometric function, if you can!!

Algebra Level 5

If the value of ∑ r = 1 100 tan ⁡ 2 r − 1 cos ⁡ 2 r = tan ⁡ 4 n − tan ⁡ 1 \sum _{ r=1 }^{ 100 }{ \frac { \tan\quad { 2 }^{ r-1 } }{ \cos\quad { 2 }^{ r } } } =\quad \tan\quad { 4 }^{ n }\quad -\quad \tan\quad 1 , find n, where 'n' is a natural number.


The answer is 50.

Ninad Akolekar by Ninad Akolekar , 23, India

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1 solution

U Z
Nov 9, 2014

t a n ( 2 r 2 ) c o s ( 2 r ) \frac{tan(\frac{2^{r}}{2})}{cos(2^{r})}

Let 2 r = x 2^{r} = x

s i n ( x 2 ) c o s ( x 2 ) c o s x = s i n ( x − x 2 ) c o s ( x 2 ) c o s x \frac{sin(\frac{x}{2})}{cos(\frac{x}{2})cosx} =\frac{sin(x - \frac{x}{2})}{cos(\frac{x}{2})cosx}

s i n x c o s ( x 2 ) − c o s x s i n ( x 2 ) c o s ( x 2 ) c o s x \frac{sinxcos(\frac{x}{2}) - cosxsin(\frac{x}{2})}{cos(\frac{x}{2})cosx}

= t a n x − t a n x 2 = tanx - tan\frac{x}{2}

= ∑ r = 1 100 t a n 2 r − t a n 2 r − 1 = \sum_{r = 1}^{100} tan2^{r} - tan2^{r - 1}

= t a n 2 − t a n 1 + t a n 4 − t a n 2 + . . . . . . . . . . . . . . . . . . + t a n 2 100 − t a n 2 99 = tan2 - tan1 + tan4 - tan2 + .................. + tan2^{100} - tan2^{99}

= t a n 2 100 − t a n 1 = tan2^{100} - tan1

= t a n 4 50 − t a n 1 = tan4^{50} - tan1

16 Helpful 0 Interesting 0 Brilliant 0 Confused

well done,Megh

Ayush Verma - 6 years, 7 months ago

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Thank you sir

U Z - 6 years, 7 months ago

Also splved by Taylor and Maclaurin series But these method is very easy

harsh soni - 6 years, 7 months ago

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