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sign of trigonometry ratios in various quadrants
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sign of trigonometry ratios in various quadrants
 and the concept of trigonometry ratio
for [(pi/2) - x] , [(pi/2) + x], [pi - x],[pi + x],[(3pi/2) - x] , [(3pi/2) + x],[2pi - x],[- x] .

  note that 180 degrees is pi radians
90 degrees is (pi / 2)
radians

45 degrees is (pi / 4) radians

60 degrees is (pi / 3) radians
 

30 degrees is (pi / 6) radians


270 degrees is (3pi / 2) radians

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( function, cofunction) pair
 for [ (pi / 2) + x ] [ (pi / 2)- x ] , [ (3pi / 2) + x ], [ (3pi / 2) - x ]

(sin x , cos x), (tan x , cot x), (sec x , cosec x)

**for those involving odd multiples of (pi/2)
*************************************************

sign of trigonometry ratios in various
                  quadrants
To remember this use

A S T C      or

A
ll Silver Tea Cups

In the first quadrant
[ 0 to (pi / 2) ], all trigonometric ratios are positive

In the second quadrant [  (pi / 2) to pi  ] , sine and cosec are  positive and other
trigonometric ratios are negative

 
In the third quadrant [ pi to (3pi / 2) ] , tan and cot are  positive and other
trigonometric ratios are negative


 In the fourth quadrant [  (3pi / 2 ) ro 2pi ] , cos and sec are  positive and other
trigonometric ratios are negative


Assume x is an acute angle ( x lies in the I quadrant )

[ (pi / 2) -x ] will lie in the I quadrant
[ (pi / 2) +x ] will lie in the II quadrant


[ pi -x ] will lie in the II quadrant
[ pi + x ] will lie in the III quadrant


[ (3pi / 2) -x ] will lie in the III quadrant
[ (3pi / 2) +x ] will lie in the IV quadrant


[ (2pi) -x ] will lie in the IV quadrant
[-x ] will lie in the IV quadrant

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x lies in the I quadrant
so sin x, cos x , tan x, cosec x, sec x, cot x are all positive

now
[ (pi / 2) - x ] will lie in the II quadrant

so sin
[ (pi / 2) - x ] and cosec[ (pi / 2) - x ]
 
cos[ (pi / 2)  - x ], tan[ (pi / 2) - x ],sec[ (pi / 2) - x ],cot[ (pi / 2) - x ]
are all positive
but for [ (pi / 2) - x you have to use cofunction pair
(sin x , cos x), (tan x , cot x), (sec x , cosec x) with the correct sign
therefore
sin[ (pi / 2) - x ] = cos x
cos[ (pi / 2) - x ] =   sin x
tan[ (pi / 2) -  x ] = cot x
cosec[ (pi / 2) - x ] = sec x
sec[ (pi / 2) - x ] =   cosec x
cot[ (pi / 2) - x ] = tan x


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If x lies in the I quadrant
so sin x, cos x , tan x, cosec x, sec x, cot x are all positive

now
[ (pi / 2) +x ] will lie in the II quadrant

so sin
[ (pi / 2) +x ] and cosec[ (pi / 2) +x ] are positive
while
cos[ (pi / 2) +x ], tan[ (pi / 2) +x ],sec[ (pi / 2) +x ],cot[ (pi / 2) +x ]
are all negative
but for [ (pi / 2) + x you have to use cofunction pair
(sin x , cos x), (tan x , cot x), (sec x , cosec x) with the correct sign
therefore
sin[ (pi / 2) +x ] = cos x
cos[ (pi / 2) +x ] =  - sin x
tan[ (pi / 2) +x ] = -cot x
cosec[ (pi / 2) +x ] = sec x
sec[ (pi / 2) +x ] =  - cosec x
cot[ (pi / 2) +x ] = -tan x
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If x lies in the first quadrant
[pi - x] lies in the second quadrant
therefore sin[pi - x] and cosec[pi - x] are positive
while cos[pi - x], tan[pi - x],sec[pi - x],cot[pi - x] are negative
You do not need to use cofunction and only need to use the same function for [pi - x]

sin[pi - x] = sin x
cos[pi - x] = -cos x
tan[pi - x] = -tan x
cosec[pi - x] = cosec x
sec[pi - x] = -sec x
cot[pi - x] = -cot x
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If x lies in the first quadrant
[pi + x] lies in the third quadrant
therefore tan[pi + x] and cot[pi + x] are positive
while sin[pi + x], cos[pi +x], cosec[pi + x], sec[pi + x], are negative
You do not need to use cofunction and only need to use the same function for [pi + x]

sin[pi + x] = -sin x
cos[pi + x] = -cos x
tan[pi + x] = tan x
cosec[pi+ x] = -cosec x
sec[pi + x] = -sec x
cot[pi + x] = cot x
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If x lies in the first quadrant
[(3pi/2) - x] lies in the third quadrant
therefore tan[(3pi/2) - x] and cot[[(3pi/2) - x] are positive
while sin[[(3pi/2) - x], cos[(3pi/2) - x], cosec[(3pi/2) - x], sec[(3pi/2) - x], are negative
You t need to use cofunction pair (sin x , cos x), (tan x , cot x), (sec x , cosec x) with the correct sign for [(3pi/2) - x]
sin[(3pi/2) - x] = -cos x
cos[(3pi/2) - x] = -sin x
tan[(3pi/2) - x]= cot x
cosec[(3pi/2) - x] = -sec x
sec[(3pi/2) - x] = -cosec x
cot[(3pi/2) - x] = tan x

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If x lies in the first quadrant
[(3pi/2) + x] lies in the fourth quadrant
therefore cos[(3pi/2) + x] and sec[[(3pi/2) + x] are positive
while sin[[(3pi/2) + x], tan [(3pi/2) + x], cosec[(3pi/2) + x], cot[(3pi/2) + x], are negative
You t need to use cofunction pair (sin x , cos x), (tan x , cot x), (sec x , cosec x) with the correct sign for [(3pi/2) + x]
sin[(3pi/2) + x] = -cos x
cos[(3pi/2) + x] = sin x
tan[(3pi/2) + x]= -cot x
cosec[(3pi/2) + x] = sec x
sec[(3pi/2) + x] = cosec x
cot[(3pi/2) - x] = -tan x

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If x lies in the first quadrant
[2pi - x] lies in the fourth quadrant
therefore cos[2pi - x] and sec[2pi - x] are positive
while sin[2pi - x], tan[2pi - x],cosec[2pi - x],cot[2pi - x] are negative
You do not need to use cofunction and only need to use the same function for [2pi - x]

sin[2pi - x] = -sin x
cos[2pi - x] = cos x
tan[2pi - x] = -tan x
cosec[2pi - x] = -cosec x
sec[2pi - x] = sec x
cot[2pi - x] = -cot x

note that [2pi-x] can also be treated as [-x] so that
sin[ - x] = -sin x
cos[ - x] = cos x
tan[ - x] = -tan x
cosec[- x] = -cosec x
sec[ - x] = sec x
cot[- x] = -cot x
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trigonometric identity and ratio of  certain standard angles

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