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

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#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

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Don Share
Feb 17, 2017

x = absolute value of y
r = $\sqrt{2}$ times x = $\sqrt{2}$ 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 = $\sqrt{2}$ meaning $\frac{r}{x}$ is $\sqrt{2}$
remember r is positive, so we can conclude x is also positive because $\frac{r}{x}$ is positive root two.

sinθ = $- \frac{\sqrt{2}}{2}$ meaning $\frac{y}{r}$ is $- \frac{\sqrt{2}}{2}$

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.

$\frac{r}{x}$ is $\sqrt{2}$ ... and... $\frac{y}{r}$ is $- \frac{\sqrt{2}}{2}$

$\sqrt{2}$ is the same as saying $\frac{\sqrt{2}}{1}$
so lets take a look one more time

$\frac{r}{x}$ = $\frac{\sqrt{2}}{1}$ ... and... $\frac{y}{r}$ = $- \frac{\sqrt{2}}{2}$

r = $\sqrt{2}$ times larger than x

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

r = $\frac{2}{\sqrt{2}}$ times bigger than (the absolute value of) y.

and $\frac{2}{\sqrt{2}}$ is actually the same value as $\sqrt{2}$

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

$- \frac{\sqrt{2}}{2}$ = $- \frac{1}{\sqrt{2}}$

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 = $\sqrt{2}$, y = $- \sqrt{2}$, and r = 2
take your pick, either line will work.

now just substitute the three numbers into the fast:

sinθ = $\frac{y}{r}$ cosθ = $\frac{x}{r}$ tanθ = $\frac{y}{x}$

secθ = $\frac{r}{x}$ cscθ = $\frac{r}{y}$ cotθ = $\frac{x}{y}$

now lets use when x= 1, y= -1, and r= $\sqrt{2}$

sinθ = $- \frac{1}{\sqrt{2}}$ cosθ = $\frac{1}{\sqrt{2}}$ tanθ = $- \frac{1}{1}$

secθ = $\frac{\sqrt{2}}{1}$ cscθ = $\frac{\sqrt{2}}{y}$ cotθ = $\frac{1}{-} 1$

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