Given cot x=1/5, what is the value cos^2 x?
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We have cot x= 1/5. We need the value of (cos x)^2
(sin x)^2 + (cos x)^2 = 1
divide all terms by (sin x)^2
=> 1 + (cot...
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cot x= 1/tan x
therefore,
since 1/5= 1/tan x, tan x =5
In a right angled triangle, tan x= opposite side/ adjacent side = 5/1=5
so, hypo. side= square root 26
cos x = adjacent side / hypo. side= 1/ square root 26
therefore since cos 2x= 1- 2cos^2 x= 1- 2(1 / square root 26) ^2
If cot x = 1/5, then tan x = 1/cot x = 5.
But tan x = sin x/cos x => sin x/cos x = 5
sin x = 5*cos x
We'll raise to square both sides:
(sin x)^2 = 25*(cos x)^2
We'll add (cos x)^2 both sides:
(sin x)^2 + (cos x)^2 = 25*(cos x)^2 + (cos x)^2
But, from Pythagorean identity, we'll have:
(sin x)^2 + (cos x)^2 = 1
1 = 26*(cos x)^2
We'll divide by 26 and we'll use symmetric property:
(cos x)^2 = 1/26
The value of (cos x)^2 is (cos x)^2 = 1/26.
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