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2 years ago

Angle θ is in standard position. if sin(θ) = − 1/3, and π < θ < 3π/2 , find cos(θ).

Mathematics

1 answer:

Step2247 [10]2 years ago

4 0

Answer:

Step-by-step explanation:

$sin^2({ \theta})+cos^2({ \theta})=1\\cos({ \theta})=-/+\sqrt{1-sin^2({ \theta})} \\cos({ \theta})=-\sqrt{1-sin^2({ \theta})} \\ --- > sin({ \theta})=\frac{1}{3} (known)\\cos({ \theta})=-\sqrt{1-sin^2({ \theta})} =-\sqrt{1-(-\frac{1}{3})} =-\sqrt{\frac{9-1}{9}} =-\frac{2x^{2} }{3}$

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Find T5(x) : Taylor polynomial of degree 5 of the function f(x)=cos(x) at a=0 . (You need to enter function.) T5(x)= Find all va

Burka [1]

Answer:

$\bf cos(x)\approx1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{4!}=\\\\=1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{24}$

The polynomial is an approximation with an error less than or equals to 0.002652 for x in the interval

[-1.113826815, 1.113826815]

Step-by-step explanation:

According to Taylor's theorem

$\bf f(x)=f(0)+f'(0)x+f''(0)\displaystyle\frac{x^2}{2}+f^{(3)}(0)\displaystyle\frac{x^3}{3!}+f^{(4)}(0)\displaystyle\frac{x^4}{4!}+f^{(5)}(0)\displaystyle\frac{x^5}{5!}+R_6(x)$

with

$\bf R_6(x)=f^{(6)}(c)\displaystyle\frac{x^6}{6!}$

for some c in the interval (-x, x)

In the particular case f

f(x)=cos(x)

we have

$\bf f'(x)=-sin(x)\\f''(x)=-cos(x)\\f^{(3)}(x)=sin(x)\\f^{(4)}(x)=cos(x)\\f^{(5)}(x)=-sin(x)\\f^{(6)}(x)=-cos(x)$

therefore

$\bf f'(x)=-sin(0)=0\\f''(0)=-cos(0)=-1\\f^{(3)}(0)=sin(0)=0\\f^{(4)}(0)=cos(0)=1\\f^{(5)}(0)=-sin(0)=0$

and the polynomial approximation of T5(x) of cos(x) would be

$\bf cos(x)\approx1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{4!}=\\\\=1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{24}$

In order to find all the values of x for which this approximation is within 0.002652 of the right answer, we notice that

$\bf R_6(x)=-cos(c)\displaystyle\frac{x^6}{6!}$

for some c in (-x,x). So

$\bf |R_6(x)|\leq|\displaystyle\frac{x^6}{6!}|=\displaystyle\frac{|x|^6}{6!}$

and we must find the values of x for which

$\bf \displaystyle\frac{|x|^6}{6!}\leq0.002652$

Working this inequality out, we find

$\bf \displaystyle\frac{|x|^6}{6!}\leq0.002652\Rightarrow |x|^6\leq1.90944\Rightarrow\\\\\Rightarrow |x|\leq\sqrt[6]{1.90944}\Rightarrow |x|\leq1.113826815$

Therefore the polynomial is an approximation with an error less than or equals to 0.002652 for x in the interval

[-1.113826815, 1.113826815]

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3 years ago

What's 95% of 85 do not round your answer

TEA [102]

80.75 my dude

(85/100)*95

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3 years ago

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What is the value of x? x20=58 A. 0.625 B. 2.5 C. 12.5 D. 32.0

OLga [1]

Answer:

2.9 is the correct information

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3 years ago

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2+2=??? PLEASE ANSWER

Dvinal [7]

Answer 4

Step-by-step explanation:

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1 year ago

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A group of friends will host a party and split the cost of food and supplies evenly. The cost that each person will pay is inver

timofeeve [1]

Answer:

C=60/h

Step-by-step explanation:

Cost is inversely Proportional to the number of friends who decides to host.

If five (5) friends host the party,

And each friend pays $12 for food and supplies

The equation that describes the relationship is

C=k ×1/h

Where,

C= cost that each host pays

K=constant

h=number of host

Therefore,

$12=k×1/5

$12=k/5

Make k the subject of the formula

$12=k*1/5

Multiply both sides by 5

12×5=k*1/5×5

60=k*5/5

60=k*1

60=k

Sub the value of k into

C=k ×1/h

C=60*1/h

C=60/h

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3 years ago