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

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Given $\sin α = \frac{5}{8},$ with $θ$ in quadrant I, find $\cos (2 α) .$

$\cos (2 α) = \frac{7}{32}$

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Using the double-angle formula for cosine without exact values

Use the double-angle formula for cosine to write $\cos (6 x)$ in terms of $\cos (3 x) .$

\begin{array}{l} \cos (6 x) = \cos (3 x + 3 x) \\ = \cos 3 x \cos 3 x - \sin 3 x \sin 3 x \\ = \cos^{2} 3 x - \sin^{2} 3 x \end{array}

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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 + \sin (2 θ) = {(\sin θ + \cos θ)}^{2}

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

\begin{array}{l} {(\sin θ + \cos θ)}^{2} = \sin^{2} θ + 2 \sin θ \cos θ + \cos^{2} θ \\ = (\sin^{2} θ + \cos^{2} θ) + 2 \sin θ \cos θ \\ = 1 + 2 \sin θ \cos θ \\ = 1 + \sin (2 θ) \end{array}

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Try It

Establish the identity: $\cos^{4} θ - \sin^{4} θ = \cos (2 θ) .$

$\cos^{4} θ - \sin^{4} θ = (\cos^{2} θ + \sin^{2} θ) (\cos^{2} θ - \sin^{2} θ) = \cos (2 θ)$

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Verifying a double-angle identity for tangent

Verify the identity:

\tan (2 θ) = \frac{2}{\cot θ - \tan θ}

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.

\begin{array}{l} \tan (2 θ) = \frac{2 \tan θ}{1 - \tan^{2} θ} & Double-angle formula \\ = \frac{2 \tan θ (\frac{1}{\tan θ})}{(1 - \tan^{2} θ) (\frac{1}{\tan θ})} \begin{matrix}  \end{matrix} & Multiply by a term that results in desired numerator . \\ = \frac{2}{\frac{1}{\tan θ} - \frac{\tan^{2} θ}{\tan θ}} \\ = \frac{2}{\cot θ - \tan θ} & Use reciprocal identity for \frac{1}{\tan θ} . \end{array}

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Try It

Verify the identity: $\cos (2 θ) \cos θ = \cos^{3} θ - \cos θ \sin^{2} θ .$

$\cos (2 θ) \cos θ = (\cos^{2} θ - \sin^{2} θ) \cos θ = \cos^{3} θ - \cos θ \sin^{2} θ$

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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 $\cos (2 θ) = 1 - 2 \sin^{2} θ .$ Solve for $\sin^{2} θ :$

\begin{array}{l} \cos (2 θ) = 1 - 2 \sin^{2} θ \\ 2 \sin^{2} θ = 1 - \cos (2 θ) \\ \sin^{2} θ = \frac{1 - \cos (2 θ)}{2} \end{array}

Next, we use the formula $\cos (2 θ) = 2 \cos^{2} θ - 1.$ Solve for $\cos^{2} θ :$

\begin{array}{l} \cos (2 θ) = 2 \cos^{2} θ - 1 \\ 1 + \cos (2 θ) = 2 \cos^{2} θ \\ \frac{1 + \cos (2 θ)}{2} = \cos^{2} θ \end{array}

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

\begin{array}{l} \tan^{2} θ = \frac{\sin^{2} θ}{\cos^{2} θ} \\ = \frac{\frac{1 - \cos (2 θ)}{2}}{\frac{1 + \cos (2 θ)}{2}} & Substitute the reduction formulas. \\ = (\frac{1 - \cos (2 θ)}{2}) (\frac{2}{1 + \cos (2 θ)}) \begin{matrix}  \end{matrix} \\ = \frac{1 - \cos (2 θ)}{1 + \cos (2 θ)} \end{array}

A General Note

Reduction formulas

The reduction formulas are summarized as follows:

\sin^{2} θ = \frac{1 - \cos (2 θ)}{2}

\cos^{2} θ = \frac{1 + \cos (2 θ)}{2}

\tan^{2} θ = \frac{1 - \cos (2 θ)}{1 + \cos (2 θ)}

Writing an equivalent expression not containing powers greater than 1

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

We will apply the reduction formula for cosine twice.

\begin{array}{l} \cos^{4} x = {(\cos^{2} x)}^{2} \\ = {(\frac{1 + \cos (2 x)}{2})}^{2} & {Substitute reduction formula for cos}^{2} x . \\ = \frac{1}{4} (1 + 2 \cos (2 x) + \cos^{2} (2 x)) \\ = \frac{1}{4} + \frac{1}{2} \cos (2 x) + \frac{1}{4} (\frac{1 + \cos 2 (2 x)}{2}) \begin{matrix}  \end{matrix} & {Substitute reduction formula for cos}^{2} x . \\ = \frac{1}{4} + \frac{1}{2} \cos (2 x) + \frac{1}{8} + \frac{1}{8} \cos (4 x) \\ = \frac{3}{8} + \frac{1}{2} \cos (2 x) + \frac{1}{8} \cos (4 x) \end{array}

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Questions & Answers

for the "hiking" mix, there are 1,000 pieces in the mix, containing 390.8 g of fat, and 165 g of protein. if there is the same amount of almonds as cashews, how many of each item is in the trail mix?

ADNAN Reply

linear speed of an object

Melissa Reply

an object is traveling around a circle with a radius of 13 meters .if in 20 seconds a central angle of 1/7 Radian is swept out what are the linear and angular speed of the object

Melissa

test

Matrix

how to find domain

Mohamed Reply

like this: (2)/(2-x) the aim is to see what will not be compatible with this rational expression. If x= 0 then the fraction is undefined since we cannot divide by zero. Therefore, the domain consist of all real numbers except 2.

Dan

define the term of domain

Moha

if a>0 then the graph is concave

Angel Reply

if a<0 then the graph is concave blank

Angel

what's a domain

Kamogelo Reply

The set of all values you can use as input into a function su h that the output each time will be defined, meaningful and real.

Spiro

how fast can i understand functions without much difficulty

Joe Reply

what is inequalities

Nathaniel

functions can be understood without a lot of difficulty. Observe the following: f(2) 2x - x 2(2)-2= 2 now observe this: (2,f(2)) ( 2, -2) 2(-x)+2 = -2 -4+2=-2

Dan

what is set?

Kelvin Reply

a colony of bacteria is growing exponentially doubling in size every 100 minutes. how much minutes will it take for the colony of bacteria to triple in size

Divya Reply

I got 300 minutes. is it right?

Patience

no. should be about 150 minutes.

Jason

It should be 158.5 minutes.

ok, thanks

Patience

100•3=300 300=50•2^x 6=2^x x=log_2(6) =2.5849625 so, 300=50•2^2.5849625 and, so, the # of bacteria will double every (100•2.5849625) = 258.49625 minutes

Thomas

158.5 This number can be developed by using algebra and logarithms. Begin by moving log(2) to the right hand side of the equation like this: t/100 log(2)= log(3) step 1: divide each side by log(2) t/100=1.58496250072 step 2: multiply each side by 100 to isolate t. t=158.49

Dan

what is the importance knowing the graph of circular functions?

Arabella Reply

can get some help basic precalculus

ismail Reply

What do you need help with?

Andrew

how to convert general to standard form with not perfect trinomial

Camalia Reply

can get some help inverse function

ismail

Rectangle coordinate

Asma Reply

how to find for x

Jhon Reply

it depends on the equation

Robert

yeah, it does. why do we attempt to gain all of them one side or the other?

Melissa

how to find x: 12x = 144 notice how 12 is being multiplied by x. Therefore division is needed to isolate x and whatever we do to one side of the equation we must do to the other. That develops this: x= 144/12 divide 144 by 12 to get x. addition: 12+x= 14 subtract 12 by each side. x =2

Dan

whats a domain

mike Reply

The domain of a function is the set of all input on which the function is defined. For example all real numbers are the Domain of any Polynomial function.

Spiro

Spiro; thanks for putting it out there like that, 😁

Melissa

foci (–7,–17) and (–7,17), the absolute value of the differenceof the distances of any point from the foci is 24.

Churlene Reply

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Source: OpenStax, Precalculus. OpenStax CNX. Jan 19, 2016 Download for free at https://legacy.cnx.org/content/col11667/1.6

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