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algol [13]
3 years ago
6

The point (1, −1) is on the terminal side of angle θ, in standard position. What are the values of sine, cosine, and tangent of

θ? Make sure to show all work.

Mathematics
1 answer:
mixer [17]3 years ago
6 0
Notice the picture below

x = 1 and y = -1

now... recall your SOH CAH TOA   \bf sin(\theta)=\cfrac{y}{r}
\qquad 
% cosine
cos(\theta)=\cfrac{x}{r}
\qquad 
% tangent
tan(\theta)=\cfrac{y}{x}

now.. what's the value for "r" or the radius? well, using the pythagorean theorem \bf c^2=a^2+b^2\implies c=\sqrt{a^2+b^2}\qquad 
\begin{cases}
a=x\\
b=y\\
c=r
\end{cases}

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vichka [17]
We can write the equation in point-form in this case. The equation should be:

y-y1 = m(x-x1)

y and x can be any point in the line, so we leave them just like that in the equation, we don’t have to add any numbers to them. m is the slope of the line, and (x1, y1) is the given point in the line, which is (-1, 1) in this case.

Now, just substitute the number in,

y - 1 = -2 [x - (-1)]
y - 1 = -2 (x+1)
y - 1 = -2x - 2
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So, 2x + y + 1 = 0 is your answer.


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Step-by-step explanation:

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