Mathematics

Mathematics, 31.12.2019 01:31, jesus3426

Find the sum of the three smallest positive values of θ such that $4\cos^2(2\theta-\pi) =3$ .
(give your answer in radians.)

answer

Answers: 2

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Answers

Answer

answered: waltersamy30

∅= 13π/12

Step-by-step explanation:

Solve until you get cos(___) on one side and a value on the other side.

I got cos(2∅-π) = ±√(3)/2

I know that the only cos() value that gives me ±√(3)/2 are -5π/6, -π/6, and π/6 from looking at the unit circle; those 3 are the smallest value that would give me ±√(3)/2.

So i solve for what value of 2∅-π would give me -5π/6, -π/6, and π/6

by solving for

2∅₁-π = -5π/6, 2∅₂-π = -π/6, and 2∅₃-π = π/6

I got ∅₁=π/12, ∅₂=5π/12, ∅₃=7π/12.

The question wanted the sum of the three smallest value so i added it up and got 13π/12.

And thanks for peer review. Good catch on my mistakes.

Find the sum of the three smallest positive values of θ such that <img src=

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Answer

answered: breniljakenotro

$\frac{13}{12}\pi$

Step-by-step explanation:

$4\cos^2(2\theta-\pi)=3$

We want to first isolate the trig expression.

We will do this by first dividing both sides by 4 and then taking the square root of both sides.

$\cos^2(2\theta-\pi)=\frac{3}{4}$

$\cos(2\theta-\pi)=\pm \sqrt{\frac{3}{4}}$

Let's clean up the right hand side a little:

$\cos(2\theta-\pi)=\pm \frac{\sqrt{3}}{2}$

Break time to think about something to help us solve the above:

So $\cos(u)=\frac{\sqrt{3}}{2}$ when $u=\pm \frac{\pi}{6}$ in the one cycle of $\cos(u)$ .

So $\cos(u)=-\frac{\sqrt{3}}{2}$ when $u=\pm \frac{5\pi}{6}$ in the one cycle of $\cos(u)$ .

Back to it:

$\cos(2\theta-\pi)=\pm \frac{\sqrt{3}}{2}$

Solving one of those equations:

$\cos(2\theta-\pi)=\frac{\sqrt{3}}{2}$ when

$2\theta-\pi=\pm \frac{\pi}{6}+2\pi k$

Let's solve for $\theta$ .

Add $\pi$ on both sides:

$2\theta=\pm \frac{\pi}{6}+2\pi k+\pi$

Divide both sides by 2:

$\theta=\pm \frac{\pi}{12}+\pi k+\frac{\pi}{2}$

$\theta=\pm \frac{\pi}{12}+\frac{\pi}{2}+\pi k$

Now we also have to solve the other:

$\cos(2\theta-\pi)=-\frac{\sqrt{3}}{2}$ when

$2\theta-\pi=\pm \frac{5\pi}{6}+2\pi k$

Let's solve for $\theta$ .

Add $\pi$ on both sides:

$2\theta=\pm \frac{5\pi}{6}+2\pi k+\pi$

Divide both sides by 2:

$\theta=\pm \frac{5\pi}{12}+\pi k+\frac{\pi}{2}$

$\theta=\pm \frac{5\pi}{12}+\frac{\pi}{2}+\pi k$

So the full set of solutions is:

$\theta=\pm \frac{\pi}{12}+\frac{\pi}{2}+\pi k$

$\theta=\pm \frac{5\pi}{12}+\frac{\pi}{2}+\pi k$

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

I'm going to evaluate these expressions for some integer :

k=-2 we have:

$\theta=\pm \frac{\pi}{12}+\frac{\pi}{2}+\pi (-2)$

$\theta=\frac{-17}{12}\pi \text{ or } \frac{-19}{12}\pi$

$\theta=\pm \frac{5\pi}{12}+\frac{\pi}{2}+\pi (-2)$

$\theta=\frac{-13}{12}\pi \text{ or } \frac{-23}{12}\pi$

k=-1 we have:

$\theta=\pm \frac{\pi}{12}+\frac{\pi}{2}+\pi (-1)$

$\theta=\frac{-5}{12}\pi \text{ or } \frac{-7}{12}\pi$

$\theta=\pm \frac{5\pi}{12}+\frac{\pi}{2}+\pi (-1)$

$\theta=\frac{-1}{12}\pi \text{ or } \frac{-11}{12}\pi$

k=0 we have:

$\theta=\pm \frac{\pi}{12}+\frac{\pi}{2}+\pi (0)$

$\theta=\frac{7}{12}\pi \text{ or } \frac{5}{12}\pi$

$\theta=\pm \frac{5\pi}{12}+\frac{\pi}{2}+\pi (0)$

$\theta=\frac{11}{12}\pi \text{ or } \frac{1}{12}\pi$

k=1 we have:

$\theta=\pm \frac{\pi}{12}+\frac{\pi}{2}+\pi (1)$

$\theta=\frac{19}{12}\pi \text{ or } \frac{17}{12}\pi$

$\theta=\pm \frac{5\pi}{12}+\frac{\pi}{2}+\pi (1)$

$\theta=\frac{23}{12}\pi \text{ or } \frac{13}{12}\pi$

Anyways the values will keep getting larger from here.

We can see that the smallest positive values happen when k=0 .

So we have the first smallest is: \frac{1}{12}\pi[/tex].

The second smallest is $\frac{5}{12}\pi$ .

The third smallest is $\frac{7}{12}\pi$ .

So the sum of these numbers are:

$\frac{1}{12}\pi+\frac{5}{12}\pi+\frac{7}{12}\pi$

$\frac{1+5+7}{12}\pi$

$\frac{13}{12}\pi$

Answer

answered: Guest

The operation is suck a c@ck? !

Answer

answered: Guest

the length of o'p' is $1.5\times op$ .

step-by-step explanation:

it is given that the quadrilateral m'n'o'p' is the image of quadrilateral mnop after dilation.

the center of dilation is origin and the scale factor is 1.5.

in dilation the image and preimge are similar figures. it means the sides of image and preimage are proportional.

the scale factor is defined as

$k=\frac{o'p'}{op}$

the scale factor is 1.5.

$1.5=\frac{o'p'}{op}$

multiply both sides by op.

$1.5\times op=o'p'$

note: if the value of op is given, then multiply the value of op with 1.5 to get the length of side o'p'.

therefore length of o'p' is $1.5\times op$ .

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