Find $sin x$ if $cos x=tan y$, $cos y=tan z$, $cos z=tan x$
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If $cos x=tan y$, $cos y=tan z$, $cos z=tan x$, then find the value of $sin x$
My reference says $sin x=2sin 18^circ$, but how do I approach the problem ?
My Attempt
$$
sin x=tan x.cos x=tan x.tan y
$$
$$
tan^2z=cos^2y=frac{1}{1+tan^2y}=frac{1}{1+cos^2x}\
tan^2y=cos^2x=frac{1}{1+tan^2x}=frac{1}{1+cos^2z}
$$
I have no clue of whats the trick to solve this.
trigonometry
asked Nov 14 at 15:24
ss1729
1,640722
add a comment |
up vote
1
down vote
favorite
If $cos x=tan y$, $cos y=tan z$, $cos z=tan x$, then find the value of $sin x$
My reference says $sin x=2sin 18^circ$, but how do I approach the problem ?
My Attempt
$$
sin x=tan x.cos x=tan x.tan y
$$
$$
tan^2z=cos^2y=frac{1}{1+tan^2y}=frac{1}{1+cos^2x}\
tan^2y=cos^2x=frac{1}{1+tan^2x}=frac{1}{1+cos^2z}
$$
I have no clue of whats the trick to solve this.
trigonometry
asked Nov 14 at 15:24
ss1729
1,640722
1
answers.yahoo.com/question/…
– lab bhattacharjee
Nov 14 at 15:33
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
If $cos x=tan y$, $cos y=tan z$, $cos z=tan x$, then find the value of $sin x$
My reference says $sin x=2sin 18^circ$, but how do I approach the problem ?
My Attempt
$$
sin x=tan x.cos x=tan x.tan y
$$
$$
tan^2z=cos^2y=frac{1}{1+tan^2y}=frac{1}{1+cos^2x}\
tan^2y=cos^2x=frac{1}{1+tan^2x}=frac{1}{1+cos^2z}
$$
I have no clue of whats the trick to solve this.
trigonometry
asked Nov 14 at 15:24
ss1729
1,640722
If $cos x=tan y$, $cos y=tan z$, $cos z=tan x$, then find the value of $sin x$
My reference says $sin x=2sin 18^circ$, but how do I approach the problem ?
My Attempt
$$
sin x=tan x.cos x=tan x.tan y
$$
$$
tan^2z=cos^2y=frac{1}{1+tan^2y}=frac{1}{1+cos^2x}\
tan^2y=cos^2x=frac{1}{1+tan^2x}=frac{1}{1+cos^2z}
$$
I have no clue of whats the trick to solve this.
trigonometry
trigonometry
asked Nov 14 at 15:24
ss1729
1,640722
asked Nov 14 at 15:24
ss1729
1,640722
asked Nov 14 at 15:24
ss1729
1,640722
asked Nov 14 at 15:24
ss1729
1,640722
asked Nov 14 at 15:24
ss1729
1,640722
1,640722
1
answers.yahoo.com/question/…
– lab bhattacharjee
Nov 14 at 15:33
add a comment |
1
answers.yahoo.com/question/…
– lab bhattacharjee
Nov 14 at 15:33
1
1
answers.yahoo.com/question/…
– lab bhattacharjee
Nov 14 at 15:33
answers.yahoo.com/question/…
– lab bhattacharjee
Nov 14 at 15:33
add a comment |
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answers.yahoo.com/question/…
– lab bhattacharjee
Nov 14 at 15:33