7.3 Double-angle, half-angle, and reduction formulas  (Page 2/8)

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Given $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\alpha =\frac{5}{8},$ with $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ in quadrant I, find $\text{\hspace{0.17em}}\mathrm{cos}\left(2\alpha \right).$

$\mathrm{cos}\left(2\alpha \right)=\frac{7}{32}$

Using the double-angle formula for cosine without exact values

Use the double-angle formula for cosine to write $\text{\hspace{0.17em}}\mathrm{cos}\left(6x\right)\text{\hspace{0.17em}}$ in terms of $\text{\hspace{0.17em}}\mathrm{cos}\left(3x\right).$

Using double-angle formulas to verify identities

Establishing identities using the double-angle formulas is performed using the same steps we used to derive the sum and difference formulas. Choose the more complicated side of the equation and rewrite it until it matches the other side.

Using the double-angle formulas to establish an identity

Establish the following identity using double-angle formulas:

$1+\mathrm{sin}\left(2\theta \right)={\left(\mathrm{sin}\text{\hspace{0.17em}}\theta +\mathrm{cos}\text{\hspace{0.17em}}\theta \right)}^{2}$

We will work on the right side of the equal sign and rewrite the expression until it matches the left side.

Establish the identity: $\text{\hspace{0.17em}}{\mathrm{cos}}^{4}\theta -{\mathrm{sin}}^{4}\theta =\mathrm{cos}\left(2\theta \right).$

${\mathrm{cos}}^{4}\theta -{\mathrm{sin}}^{4}\theta =\left({\mathrm{cos}}^{2}\theta +{\mathrm{sin}}^{2}\theta \right)\left({\mathrm{cos}}^{2}\theta -{\mathrm{sin}}^{2}\theta \right)=\mathrm{cos}\left(2\theta \right)$

Verifying a double-angle identity for tangent

Verify the identity:

$\mathrm{tan}\left(2\theta \right)=\frac{2}{\mathrm{cot}\text{\hspace{0.17em}}\theta -\mathrm{tan}\text{\hspace{0.17em}}\theta }$

In this case, we will work with the left side of the equation and simplify or rewrite until it equals the right side of the equation.

Verify the identity: $\text{\hspace{0.17em}}\mathrm{cos}\left(2\theta \right)\mathrm{cos}\text{\hspace{0.17em}}\theta ={\mathrm{cos}}^{3}\theta -\mathrm{cos}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}{\mathrm{sin}}^{2}\theta .$

$\mathrm{cos}\left(2\theta \right)\mathrm{cos}\text{\hspace{0.17em}}\theta =\left({\mathrm{cos}}^{2}\theta -{\mathrm{sin}}^{2}\theta \right)\mathrm{cos}\text{\hspace{0.17em}}\theta ={\mathrm{cos}}^{3}\theta -\mathrm{cos}\text{\hspace{0.17em}}\theta {\mathrm{sin}}^{2}\theta$

Use reduction formulas to simplify an expression

The double-angle formulas can be used to derive the reduction formulas    , which are formulas we can use to reduce the power of a given expression involving even powers of sine or cosine. They allow us to rewrite the even powers of sine or cosine in terms of the first power of cosine. These formulas are especially important in higher-level math courses, calculus in particular. Also called the power-reducing formulas, three identities are included and are easily derived from the double-angle formulas.

We can use two of the three double-angle formulas for cosine to derive the reduction formulas for sine and cosine. Let’s begin with $\text{\hspace{0.17em}}\mathrm{cos}\left(2\theta \right)=1-2\text{\hspace{0.17em}}{\mathrm{sin}}^{2}\theta .\text{\hspace{0.17em}}$ Solve for $\text{\hspace{0.17em}}{\mathrm{sin}}^{2}\theta :$

Next, we use the formula $\text{\hspace{0.17em}}\mathrm{cos}\left(2\theta \right)=2\text{\hspace{0.17em}}{\mathrm{cos}}^{2}\theta -1.\text{\hspace{0.17em}}$ Solve for $\text{\hspace{0.17em}}{\mathrm{cos}}^{2}\theta :$

The last reduction formula is derived by writing tangent in terms of sine and cosine:

Reduction formulas

The reduction formulas    are summarized as follows:

${\mathrm{sin}}^{2}\theta =\frac{1-\mathrm{cos}\left(2\theta \right)}{2}$
${\mathrm{cos}}^{2}\theta =\frac{1+\mathrm{cos}\left(2\theta \right)}{2}$
${\mathrm{tan}}^{2}\theta =\frac{1-\mathrm{cos}\left(2\theta \right)}{1+\mathrm{cos}\left(2\theta \right)}$

Writing an equivalent expression not containing powers greater than 1

Write an equivalent expression for $\text{\hspace{0.17em}}{\mathrm{cos}}^{4}x\text{\hspace{0.17em}}$ that does not involve any powers of sine or cosine greater than 1.

We will apply the reduction formula for cosine twice.

can you not take the square root of a negative number
Suppose P= {-3,1,3} Q={-3,-2-1} and R= {-2,2,3}.what is the intersection
can I get some pretty basic questions
In what way does set notation relate to function notation
Ama
is precalculus needed to take caculus
It depends on what you already know. Just test yourself with some precalculus questions. If you find them easy, you're good to go.
Spiro
the solution doesn't seem right for this problem
what is the domain of f(x)=x-4/x^2-2x-15 then
x is different from -5&3
Seid
All real x except 5 and - 3
Spiro
how to prroved cos⁴x-sin⁴x= cos²x-sin²x are equal
Don't think that you can.
Elliott
how do you provided cos⁴x-sin⁴x = cos²x-sin²x are equal
What are the question marks for?
Elliott
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
Abena
find the equation of the line if m=3, and b=-2
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
Where do the rays point?
Spiro
x=-b+_Гb2-(4ac) ______________ 2a
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
so good
abdikarin
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
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
William
what is f(x)=
I don't understand
Joe
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
Darius
Thanks.
Thomas
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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
Thomas
Now it shows, go figure?
Thomas