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True [87]
2 years ago
14

use the formulas for lowering powers to rewrite the expression in terms of the first power of cosine cos^4

Mathematics
1 answer:
Firdavs [7]2 years ago
5 0

The expression cos⁴ θ in terms of the first power of cosine is <u>[ 3 + 2cos 2θ + cos 4θ]/8.</u>

The power-reducing formula, for cosine, is,

cos² θ = (1/2)[1 + cos 2θ].

In the question, we are asked to use the formulas for lowering powers to rewrite the expression in terms of the first power of cosine cos⁴ θ.

We can do it as follows:

cos⁴ θ

= (cos² θ)²

= {(1/2)[1 + cos 2θ]}²

= (1/4)[1 + cos 2θ]²

= (1/4)(1 + 2cos 2θ + cos² 2θ] {Using (a + b)² = a² + 2ab + b²}

= 1/4 + (1/2)cos 2θ + (1/4)(cos ² 2θ)

= 1/4 + (1/2)cos 2θ + (1/4)(1/2)[1 + cos 4θ]

= 1/4 + cos 2θ/4 + 1/8 + cos 4θ/8

= 3/8 + cos 2θ/4 + cos 4θ/8

= [ 3 + 2cos 2θ + cos 4θ]/8.

Thus, the expression cos⁴ θ in terms of the first power of cosine is <u>[ 3 + 2cos 2θ + cos 4θ]/8</u>.

Learn more about reducing trigonometric powers at

brainly.com/question/15202536

#SPJ4

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Which number is a solution of the inequality x less-than negative 4? Use the number line to help answer the question.
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Find the measure of an angle whose supplement measures seventeen times its measure
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In the diagram below, assume that all points are given in rectangular coordinates. Determine the polar coordinates for each poin
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Step-by-step explanation:

We have cartisean points. We are trying to find polar points.

We can find r by applying the pythagorean theorem to the x value and y values.

r  {}^{2} =  {x}^{2}  +  {y}^{2}

And to find theta, notice how a right triangle is created if we draw the base(the x value) and the height(y value). We also just found our r( hypotenuse) so ignore that. We know the opposite side and the adjacent side originally. so we can use the tangent function.

\tan(x)  =  \frac{y}{x}

Remeber since we are trying to find the angle measure, use inverse tan function

\tan {}^{ - 1} ( \frac{y}{x} )  =

Answers For 2,5

{2}^{2}  +  {5}^{2}  =  \sqrt{29}  = 5.4

So r=sqr root of 29

\tan {}^{ - 1} ( \frac{5}{2} )  = 68

So the answer is (sqr root of 29,68).

For -3,3

{ -3 }^{2}  +  {3}^{2}  =  \sqrt{18}  = 3 \sqrt{2}

\tan {}^{ - 1} ( \frac{3}{ - 3} )  =  - 45

Use the identity

\tan(x)  =  \tan(x + \pi)

So that means

\tan(x)  = 135

So our points are

(3 times sqr root of 2, 135)

For 5,-3.5

{5}^{2}  +  {3.5}^{2}  =  \sqrt{37.25}

\tan {}^{ - 1} ( \frac{ - 3.5}{ - 5} )  = 35

So our points are (sqr root of 37.25, 35)

For (0,-5.4)

{0}^{2}  +  { - 5.4}^{2}  = \sqrt{}  29.16 = 5.4

So r=5.4

\tan {}^{ - 1}  (0)  = undefined

So our points are (5.4, undefined)

4 0
3 years ago
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