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Elena-2011 [213]

3 years ago

11

Find cos(20) if 0=pi/3

1 answer:

mario62 [17]3 years ago

7 0

$\bf \textit{Double Angle Identities}
\\ \quad \\
cos(2\theta)=
\begin{cases}
cos^2(\theta)-sin^2(\theta)\\
1-2sin^2(\theta)\\
\boxed{2cos^2(\theta)-1}
\end{cases}\\\\
-------------------------------\\\\
cos(2\theta)\implies 2cos^2(\theta)-1\qquad \theta=\frac{\pi }{3}\implies 2\left[ cos\left( \frac{\pi }{3} \right)\right]^2-1
\\\\\\
2\left[ \cfrac{1}{2} \right]^2-1\implies 2\cdot \cfrac{1}{4}-1\implies \cfrac{1}{2}-1\implies -\cfrac{1}{2}$

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Find the particular solution that satisfies the differential equation and the initial condition.

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Answer:

1) $y =4x^2 +7$

2) $y =7s^2 -3s^4 +181$

Step-by-step explanation:

Assuming that our function is $y = f(x)$ for the first case and $y=f(s)$ for the second case.

Part 1

We can rewrite the expression like this:

$\frac{dy}{dx} =8x$

And we can reorder the terms like this:

$dy = 8 x dx$

Now if we apply integral in both sides we got:

$\int dy = 8 \int x dx$

And after do the integrals we got:

$y = 4x^2 +c$

Now we can use the initial condition $y(0) =7$

$7 = 4(0)^2 +c, c=7$

And the final solution would be:

$y =4x^2 +7$

Part 2

We can rewrite the expression like this:

$\frac{dy}{ds} =14s -12s^3$

And we can reorder the terms like this:

$dy = 14s -12s^3 dx$

Now if we apply integral in both sides we got:

$\int dy = \int 14s -12s^3 ds$

And after do the integrals we got:

$y = 7s^2 -3s^4 +c$

Now we can use the initial condition $y(3) =1$

$1 = 7(3)^2 -3(3)^4 +c, c=1-63+243=181$

And the final solution would be:

$y =7s^2 -3s^4 +181$

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