Given #sectheta=sqrt2, sintheta=-sqrt2/2# to find the remaining trigonometric function?

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Don Share
Feb 17, 2017

Answer:

x = absolute value of y
r = #sqrt2# times x = #sqrt2# times absolute value of y.

Explanation:

one way to think about it is in terms of x, y, and r.
x may be positive or negative, y may be positive or negative.
r is Always positive.

secθ = #sqrt2# meaning #r/x# is #sqrt2#
remember r is positive, so we can conclude x is also positive because #r/x# is positive root two.

sinθ = #-sqrt(2)/2# meaning #y/r# is #-sqrt(2)/2#

remember r is always positive so we can conclude that y is negative because y over r is negative.

so we can say r is positive, x is positive, y is negative.

now lets try and find the values.

#r/x# is #sqrt2# ... and... #y/r# is #-sqrt(2)/2#

#sqrt2# is the same as saying #sqrt(2)/1#
so lets take a look one more time

#r/x# = #sqrt(2)/1# ... and... #y/r# = #-sqrt(2)/2#

r = #sqrt2# times larger than x

and if you flip the y and r around you can see

r = #2/sqrt(2)# times bigger than (the absolute value of) y.

and #2/sqrt(2)# is actually the same value as #sqrt(2)#

what's really confusing in this problem is the fact that

#-sqrt(2)/2# = #-1/sqrt(2)#

when x = 1, r = root 2 is the same as when x = root 2, r = 2.
when y = negative root 2, r = 2 is the same as when y = -1, r = root 2.

We could actually say the answer in one of two main ways:

when x= 1, y= -1, and r= #sqrt(2)#
or
when x = #sqrt2#, y = #-sqrt2#, and r = 2
take your pick, either line will work.

now just substitute the three numbers into the fast:

sinθ = #y/r# cosθ = #x/r# tanθ = #y/x#

secθ = #r/x# cscθ = #r/y# cotθ = #x/y#

now lets use when x= 1, y= -1, and r= #sqrt2#

sinθ = #-1/sqrt(2)# cosθ = #1/sqrt(2)# tanθ = #-1/1#

secθ = #sqrt(2)/1# cscθ = #sqrt(2)/y# cotθ = #1/-1#

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