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

##### 1 Answer

#### Answer

#### Answer:

#### Explanation

#### Explanation:

#### Answer:

x = absolute value of y

r =

#### 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 =

remember r is positive, so we can conclude x is also positive because

sinθ =

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.

so lets take a look one more time

r =

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

r =

and

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

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=

or

when x =

take your pick, either line will work.

now just substitute the three numbers into the fast:

sinθ =

secθ =

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

sinθ =

secθ =

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