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Phoenix [80]
3 years ago
15

1. What’s the exact value of cot[cot^-1 sqrt3/3] ?

Mathematics
2 answers:
vovikov84 [41]3 years ago
8 0
So the problems ask to find and calculate the exact value of the trigonometric equation in the following equations and the best answers would be the following:
#1. sqrt(3)/3
#2. Arcsine of zero is 0
#3. x/sqrt(4-x^2)
I hope you are satisfied with my answer and feel free to ask for more if you have questions and further clarifications. Have a nice day 

steposvetlana [31]3 years ago
6 0

Answer and Explanation :

1) Expression \cot[\cot^{-1}(\frac{\sqrt{3}}{3})]

We have to find the exact solution of the expression.

We know the inverse property,

\cot[\cot^{-1}x]=x

Applying the property,

\cot[\cot^{-1}(\frac{\sqrt{3}}{3})]=\frac{\sqrt{3}}{3}    

Therefore, The exact solution is \cot[\cot^{-1}(\frac{\sqrt{3}}{3})]=\frac{\sqrt{3}}{3}        

2) Expression \sin^{-1][\cos(\frac{\pi}{2})]

We have to find the exact solution of the expression.

We know that,

\cos(\frac{\pi}{2})=0

\sin(0)=0

Applying the property,

\sin^{-1][\cos(\frac{\pi}{2})]=\sin^{-1}[0]

\sin^{-1][\cos(\frac{\pi}{2})]=\sin^{-1}[\sin(0)]

Applying inverse property,

\sin^{-1}[\sin x]=x

\sin^{-1][\cos(\frac{\pi}{2})]=0

Therefore, The exact solution is \sin^{-1][\cos(\frac{\pi}{2})]=0

3) Expression \tan(\sin^{-1}(\frac{x}{2}))

We have to find the exact solution of the expression.

Let   \sin^{-1}(\frac{x}{2})=y

i.e.    \sin y=\frac{x}{2}

Squaring both side,  \sin^2 y=\frac{x^2}{4}

Now, The expression became \tan(y)

We can write tan in form of sin and cosine as

\tan y=\frac{\sin y}{\cos y}

We know, \cos y=\sqrt{1-\sin^2 y}

Substituting,

\tan y=\frac{\sin y}{\sqrt{1-\sin^2 y}}

Now, put the value of sin y

\tan y=\frac{\frac{x}{2}}{\sqrt{1-\frac{x^2}{4}}}

\tan y=\frac{\frac{x}{2}}{\sqrt{\frac{4-x^2}{4}}}

\tan y=\frac{\frac{x}{2}}{\frac{\sqrt{4-x^2}}{2}}

\tan y=\frac{x}{\sqrt{4-x^2}}

Therefore, The exact solution is \tan(\sin^{-1}(\frac{x}{2}))=\frac{x}{\sqrt{4-x^2}}

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Sin tita= 0.6892.find the value Of tita correct to two decimal places<br><br>​
Anvisha [2.4K]

Answer:

\theta \approx 6.28n + 2.38,  \quad  n \in \mathbb{Z}

or

\theta \approx 6.28n + 0.76, \quad n \in \mathbb{Z}

Considering \theta \in (0, 2\pi]

\theta \approx 2.38

or

\theta \approx 0.76

Step-by-step explanation:

\sin(\theta)=0.6892

We have:

\sin (x)=a \Longrightarrow x=\arcsin (a)+2\pi n \text{ or } x=\pi -\arcsin (a)+2\pi n \text{ as } n\in \mathbb{Z}

Therefore,

\theta= \arcsin (0.6892)+2\pi n, \quad n \in \mathbb{Z}

or

\theta = \pi -\arcsin (0.6892)+2\pi n, \quad  n\in \mathbb{Z}

---------------------------------

\theta \approx 6.28n + 2.38,  \quad  n \in \mathbb{Z}

or

\theta \approx 6.28n + 0.76, \quad n \in \mathbb{Z}

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3.1 times 10 ^3 in ordinary form
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Ten students share 8 pieces of poster board equally. What fracyion pf a poece of poster board does each student get?
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Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Draw a typical approximating
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Answer:

V=25088π vu

Step-by-step explanation:

Because the curves are a function of "y" it is decided to take the axis of rotation as y

, according to the graph 1 the cutoff points of f(y)₁ and f(y)₂ are ±2

f(y)₁ = 7y²-28;  f(y)₂=28-7y²

y=0;   x=28-0 ⇒ x=28

x=0;   0 = 7y²-28 ⇒ 7y²=28 ⇒ y²= 28/7 =4 ⇒ y=√4 =±2

Knowing that the volume of a solid of revolution  V=πR²h, where R²=(r₁-r₂) and h=dy then:

dV=π(7y²-28-(28-7y²))²dy ⇒dV=π(7y²-28-28+7y²)²dy = 4π(7y²-28)²dy

dV=4π(49y⁴-392y²+784)dy integrating on both sides

∫dV=4π∫(49y⁴-392y²+784)dy ⇒ solving  ∫(49y⁴-392y²+784)dy

49∫y⁴dy-392∫y²dy+784∫dy =

V=4π( 49\frac{y^{5} }{5}-392\frac{y^{3}}{3}+784y ) evaluated -2≤y≤2, or 2(0≤y≤2), also

V=8\pi(49\frac{2^{5} }{5}-392\frac{2^{3} }{3}+784.2)  ⇒ V=25088π vu

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What is 3/4 x 1/2 as a fraction
Rasek [7]

Answer:

=  \frac{3}{4}  \times  \frac{1}{2}

=  \frac{3}{8}

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