Im working on reflections and I am confused on what the reflection of A would be? I'm trying to find the reflection of (0,5) ove
r the Yaxis? The rule is this: (x,y)->(-x,y) but there is no such thing as -0......so... yea thats why im confused? Or is the reflection unchanging? is A'=0,5?
1 answer:
You might be interested in
<h3>Answer: 24/25</h3>
=================================================
Explanation:
Sine is given to be negative, and so is tangent. This only happens in quadrant Q4
Recall that y = sin(theta), so if sin(theta) < 0, then we're below the x axis.
If tan(theta) < 0, then this means cos(theta) > 0
So we have y < 0 and x > 0 which places the angle somewhere in Q4.
--------------------------
Draw a right triangle as shown below in the attached image. We have AC = 25 and BC = 7. Use the pythagorean theorem to find that AB = 24
So this is what your steps may look like
a^2+b^2 = c^2
7^2+b^2 = 25^2
b^2+49 = 625
b^2 = 625-49
b^2 = 576
b = sqrt(576)
b = 24
So because AB = 24, we know that the cosine of the angle is adjacent/hypotenuse = 24/25
---------------------------
As an alternative, you could use the trig identity
sin^2(x) + cos^2(x) = 1
and plug in the given value of sine to solve for cosine. The cosine value result will be positive since we're in Q4.
So,
sin^2(x) + cos^2(x) = 1
(-7/25)^2 + cos^2(x) = 1
(49/625) + cos^2(x) = 1
cos^2(x) = 1 - (49/625)
cos^2(x) = (625/625) - (49/625)
cos^2(x) = (625-49)/625
cos^2(x) = 576/625
cos(x) = sqrt(576/625)
cos(x) = sqrt(576)/sqrt(625)
cos(x) = 24/25
This is effectively a rephrasing of the previous section since the pythagorean trig identity is more or less the pythagorean theorem (just in a trig form)