Bicycle ramps made for competition (see
[link] ) must vary in height depending on the skill level of the competitors. For advanced competitors, the angle formed by the ramp and the ground should be
$\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ such that
$\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\theta =\frac{5}{3}.\text{\hspace{0.17em}}$ The angle is divided in half for novices. What is the steepness of the ramp for novices? In this section, we will investigate three additional categories of identities that we can use to answer questions such as this one.
Using double-angle formulas to find exact values
In the previous section, we used addition and subtraction formulas for trigonometric functions. Now, we take another look at those same formulas. The
double-angle formulas are a special case of the sum formulas, where
$\text{\hspace{0.17em}}\alpha =\beta .\text{\hspace{0.17em}}$ Deriving the double-angle formula for sine begins with the sum formula,
Deriving the double-angle for cosine gives us three options. First, starting from the sum formula,
$\text{\hspace{0.17em}}\mathrm{cos}\left(\alpha +\beta \right)=\mathrm{cos}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\beta -\mathrm{sin}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\beta ,$ and letting
$\text{\hspace{0.17em}}\alpha =\beta =\theta ,$ we have
Similarly, to derive the double-angle formula for tangent, replacing
$\text{\hspace{0.17em}}\alpha =\beta =\theta \text{\hspace{0.17em}}$ in the sum formula gives
Given the tangent of an angle and the quadrant in which it is located, use the double-angle formulas to find the exact value.
Draw a triangle to reflect the given information.
Determine the correct double-angle formula.
Substitute values into the formula based on the triangle.
Simplify.
Using a double-angle formula to find the exact value involving tangent
Given that
$\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\theta =-\frac{3}{4}\text{\hspace{0.17em}}$ and
$\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ is in quadrant II, find the following:
$\mathrm{sin}\left(2\theta \right)$
$\mathrm{cos}\left(2\theta \right)$
$\mathrm{tan}\left(2\theta \right)$
If we draw a triangle to reflect the information given, we can find the values needed to solve the problems on the image. We are given
$\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\theta =-\frac{3}{4},$ such that
$\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ is in quadrant II. The tangent of an angle is equal to the opposite side over the adjacent side, and because
$\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ is in the second quadrant, the adjacent side is on the
x -axis and is negative. Use the
Pythagorean Theorem to find the length of the hypotenuse:
We see that we to need to find
$\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ and
$\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\theta .\text{\hspace{0.17em}}$ Based on
[link] , we see that the hypotenuse equals 5, so
$\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\theta =\frac{3}{5},$ and
$\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\theta =-\frac{4}{5}.\text{\hspace{0.17em}}$ Substitute these values into the equation, and simplify.
In this formula, we need the tangent, which we were given as
$\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\theta =-\frac{3}{4}.\text{\hspace{0.17em}}$ Substitute this value into the equation, and simplify.
No because a negative times a negative is a positive. No matter what you do you can never multiply the same number by itself and end with a negative
lurverkitten
Actually you can. you get what's called an Imaginary number denoted by i which is represented on the complex plane. The reply above would be correct if we were still confined to the "real" number line.
Liam
Suppose P= {-3,1,3} Q={-3,-2-1} and R= {-2,2,3}.what is the intersection
Someone should please solve it for me
Add 2over ×+3 +y-4 over 5
simplify (×+a)with square root of two -×root 2 all over a
multiply 1over ×-y{(×-y)(×+y)} over ×y
For the first question, I got (3y-2)/15
Second one, I got Root 2
Third one, I got 1/(y to the fourth power)
I dont if it's right cause I can barely understand the question.
Is under distribute property, inverse function, algebra and addition and multiplication function; so is a combined question
graph the following linear equation using intercepts method.
2x+y=4
Ashley
how
Wargod
what?
John
ok, one moment
UriEl
how do I post your graph for you?
UriEl
it won't let me send an image?
UriEl
also for the first one... y=mx+b so.... y=3x-2
UriEl
y=mx+b
you were already given the 'm' and 'b'.
so..
y=3x-2
Tommy
Please were did you get y=mx+b from
Abena
y=mx+b is the formula of a straight line.
where m = the slope & b = where the line crosses the y-axis. In this case, being that the "m" and "b", are given, all you have to do is plug them into the formula to complete the equation.
Tommy
thanks Tommy
Nimo
0=3x-2
2=3x
x=3/2
then .
y=3/2X-2
I think
Given
co ordinates for x
x=0,(-2,0)
x=1,(1,1)
x=2,(2,4)
neil
"7"has an open circle and "10"has a filled in circle who can I have a set builder notation
I've run into this:
x = r*cos(angle1 + angle2)
Which expands to:
x = r(cos(angle1)*cos(angle2) - sin(angle1)*sin(angle2))
The r value confuses me here, because distributing it makes:
(r*cos(angle2))(cos(angle1) - (r*sin(angle2))(sin(angle1))
How does this make sense? Why does the r distribute once
this is an identity when 2 adding two angles within a cosine. it's called the cosine sum formula. there is also a different formula when cosine has an angle minus another angle it's called the sum and difference formulas and they are under any list of trig identities
Brad
strategies to form the general term
carlmark
consider r(a+b) = ra + rb. The a and b are the trig identity.
Mike
How can you tell what type of parent function a graph is ?
generally by how the graph looks and understanding what the base parent functions look like and perform on a graph
William
if you have a graphed line, you can have an idea by how the directions of the line turns, i.e. negative, positive, zero
William
y=x will obviously be a straight line with a zero slope
William
y=x^2 will have a parabolic line opening to positive infinity on both sides of the y axis
vice versa with y=-x^2 you'll have both ends of the parabolic line pointing downward heading to negative infinity on both sides of the y axis
William
y=x will be a straight line, but it will have a slope of one. Remember, if y=1 then x=1, so for every unit you rise you move over positively one unit. To get a straight line with a slope of 0, set y=1 or any integer.
Aaron
yes, correction on my end, I meant slope of 1 instead of slope of 0
Typically a function 'f' will take 'x' as input, and produce 'y' as output. As
'f(x)=y'.
According to Google,
"The range of a function is the complete set of all possible resulting values of the dependent variable (y, usually), after we have substituted the domain."
Thomas
Sorry, I don't know where the "Â"s came from. They shouldn't be there. Just ignore them. :-)
Thomas
GREAT ANSWER THOUGH!!!
Darius
Thanks.
Thomas
Â
Thomas
It is the Â that should not be there. It doesn't seem to show if encloses in quotation marks.
"Â" or 'Â' ... Â