To find the coordinates of a point being reflected across the x-axis you need to use the rule (x, -y) on the point. Example: If the point was (4, 7) you would make the y value negative since it is positive, which would give you (4, -7) which is the point after being reflected across the x-axis. I know this one is for reflecting on the x-axis but for future problems if you want to reflect across the y-axis you do the same thing but with the x value, so (-x, y)

7 0

2 years ago

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Question (c)! How do I know that t^5-10t^3+5t=0?<br> Thanks!

astra-53 [7]

(a) By DeMoivre's theorem, we have

$(\cos\theta+i\sin\theta)^5=\cos5\theta+i\sin5\theta$

On the LHS, expanding yields

$\cos^5\theta+5i\cos^4\theta\sin\theta-10\cos^3\theta\sin^2\theta-10i\cos^2\theta\sin^3\theta+5\cos\theta\sin^4\theta+i\sin^4\theta$

Matching up real and imaginary parts, we have for (i) and (ii),

$\cos5\theta=\cos^5\theta-10\cos^3\theta\sin^2\theta+5\cos\theta\sin^4\theta$
$\sin5\theta=5\cos^4\theta\sin\theta-10\cos^2\theta\sin^3\theta+\sin^5\theta$

(b) By the definition of the tangent function,

$\tan5\theta=\dfrac{\sin5\theta}{\cos5\theta}$
$=\dfrac{5\cos^4\theta\sin\theta-10\cos^2\theta\sin^3\theta+\sin^5\theta}{\cos^5\theta-10\cos^3\theta\sin^2\theta+5\cos\theta\sin^4\theta}$

$=\dfrac{5\tan\theta-10\tan^3\theta+\tan^5\theta}{1-10\tan^2\theta+5\tan^4\theta}$
$=\dfrac{t^5-10t^3+5t}{5t^4-10t^2+1}$

(c) Setting $\theta=\dfrac\pi5$ , we have $t=\tan\dfrac\pi5$ and $\tan5\left(\dfrac\pi5\right)=\tan\pi=0$ . So

$0=\dfrac{t^5-10t^3+5t}{5t^4-10t^2+1}$

At the given value of $t$ , the denominator is a non-zero number, so only the numerator can contribute to this reducing to 0.

$0=t^5-10t^3+5t\implies0=t^4-10t^2+5$

Remember, this is saying that

$0=\tan^4\dfrac\pi5-10\tan^2\dfrac\pi5+5$

If we replace $\tan^2\dfrac\pi5$ with a variable $x$ , then the above means $\tan^2\dfrac\pi5$ is a root to the quadratic equation,

$x^2-10x+5=0$

Also, if $\theta=\dfrac{2\pi}5$ , then $t=\tan\dfrac{2\pi}5$ and $\tan5\left(\dfrac{2\pi}5\right)=\tan2\pi=0$ . So by a similar argument as above, we deduce that $\tan^2\dfrac{2\pi}5$ is also a root to the quadratic equation above.

(d) We know both roots to the quadratic above. The fundamental theorem of algebra lets us write

$x^2-10x+5=\left(x-\tan^2\dfrac\pi5\right)\left(x-\tan^2\dfrac{2\pi}5\right)$

Expand the RHS and match up terms of the same power. In particular, the constant terms satisfy

$5=\tan^2\dfrac\pi5\tan^2\dfrac{2\pi}5\implies\tan\dfrac\pi5\tan\dfrac{2\pi}5=\pm\sqrt5$

But $\tanx>0$ for all $0$ , as is the case for $x=\dfrac\pi5$ and $x=\dfrac{2\pi}5$ , so we choose the positive root.

3 0

3 years ago

Write an explicit formula for the sequence -13,-6,1,8, ...

dimulka [17.4K]

Answer:

7n-20 I believe

Step-by-step explanation:

hope this helps