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Nataly_w [17]
3 years ago
11

Tan(x+pi) + 2sin(x+pi)=0can you help prove this and make the left side equal the right?

Mathematics
1 answer:
vladimir1956 [14]3 years ago
5 0

This is not an identity but an equation. You're supposed to find the values of x for which this is true. It is *not* true for all values of x.

To see why:

\tan x has period \pi, which means \tan(x+\pi)=\tan x.

The angle sum identity for \sin x says that

\sin(x+\pi)=\sin x\cos \pi+\cos x\sin\pi=-\sin x

So

\tan(x+\pi)+2\sin(x+\pi)=\tan x-2\sin x

By definition of \tan x,

\tan x=\dfrac{\sin x}{\cos x}

and so

\tan x-2\sin x=\sin x\left(\dfrac1{\cos x}-2\right)

which is only 0 if either \sin x=0 (which only happens for certain values of x) or \dfrac1{\cos x}-2=0\implies\cos x=\dfrac12 (which also only happens for certain values of x). It's these values of x you want to find.

\sin x=0 whenever x is a multiple of \pi, i.e. x=n\pi for any integer n.

If 0\le x, then \cos x=\dfrac12 is true for x=\dfrac\pi3 and x=\dfrac{5\pi}3. Then to account for all other possible values, we add a multiple of 2\pi, so that x=\dfrac\pi3+2n\pi or x=\dfrac{5\pi}3+2n\pi for integers n.

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Not 100% but all real number so D
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3 years ago
(3^n*3^2)^4=3^28<br> Enter the value of n for the equation<br> N= ?
maksim [4K]
Hello! And thank you for your question!

Use Pemdas to get

3^(n+2)*4=3^28

Rewrite the equation:

3^4(n+2) = 3^28

Cancel the base of 3:

4(n + 2) = 28

Then divide 4 on both sides:

2 + n = 28/4

Simplify 28/4:

2 + n = 7

Subtract 2 on both sides:

n = 7 - 2

Finally simplify 7 - 2:

n = 5

Final Answer:

n = 5
8 0
3 years ago
Find the approximate area between the curve f(x) = -4x² + 32x and on the x-axis on the interval [0,8] using 4 rectangles. Use th
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Split up the interval [0, 8] into 4 equally spaced subintervals:

[0, 2], [2, 4], [4, 6], [6, 8]

Take the right endpoints, which form the arithmetic sequence

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where 1 ≤ <em>i</em> ≤ 4.

Find the values of the function at these endpoints:

f(r_i)=-4{r_i}^2+32r_i=-16i^2+64i

The area is given approximately by the Riemann sum,

\displaystyle\int_0^8f(x)\,\mathrm dx\approx\sum_{i=1}^4f(r_i)\Delta x_i

where \Delta x_i=\frac{8-0}4=2; so the area is approximately

\displaystyle2\sum_{i=1}^4(-16i^2+64i)=-32\sum_{i=1}^4i^2+128\sum_{i=1}^4i=-32\cdot\frac{4\cdot5\cdot9}6+128\cdot\frac{4\cdot5}2=\boxed{320}

where we use the formulas,

\displaystyle\sum_{i=1}^ni=\frac{n(n+1)}2

\displaystyle\sum_{i=1}^ni^2=\frac{n(n+1)(2n+1)}6

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3 years ago
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3 years ago
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6 0
3 years ago
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