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What is sin^-1(cos 5pi/6) ?

Ratnaker Mehta
Featured 3 weeks ago

$- \frac{\pi}{3}$.

Explanation:

Recall the definition of the ${\sin}^{-} 1$ function :

${\sin}^{-} 1 x = \theta , \left(- 1 \le x \le 1\right) \iff x = \sin \theta , - \frac{\pi}{2} \le \theta \le \frac{\pi}{2}$.

Now, $\cos \left(\frac{5}{6} \pi\right) = \cos \left(\pi - \frac{\pi}{6}\right) = - \cos \left(\frac{\pi}{6}\right) = - \frac{\sqrt{3}}{2}$.

$\therefore {\sin}^{-} 1 \left(\cos \left(\frac{5}{6} \pi\right)\right) = {\sin}^{-} 1 \left(- \frac{\sqrt{3}}{2}\right)$.

Since, #sin(-pi/3)=-sin(pi/3)=-sqrt3/2, &, -pi/3 in [-pi/2,pi/2]#,

${\sin}^{-} 1 \left(- \frac{\sqrt{3}}{2}\right) = - \frac{\pi}{3}$.

$\therefore {\sin}^{-} 1 \left(\cos \left(\frac{5}{6} \pi\right)\right) = {\sin}^{-} 1 \left(- \frac{\sqrt{3}}{2}\right) = - \frac{\pi}{3}$.

What is the amplitude, period and the phase shift of #y = 2 sin (1/4 x)#?

Jim G.
Featured 2 weeks ago

$2 , 8 \pi , 0$

Explanation:

$\text{the standard form of the sine function is}$

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{y = a \sin \left(b x + c\right) + d} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

$\text{amplitude "=|a|," period } = \frac{2 \pi}{b}$

$\text{phase shift "=-c/b" and vertical shift } = d$

$\text{here } a = 2 , b = \frac{1}{4} , c = d = 0$

$\text{amplitude "=|2|=2," period } = \frac{2 \pi}{\frac{1}{4}} = 8 \pi$

$\text{there is no phase shift}$

If #sinA + sin^2A=1# and #a cos^12A+ b cos^8A + c cos^6A-1 = 0# then find #(b + c)/(a + b) #?

George C.
Featured 2 weeks ago

If $a , b , c$ are integers, then for some integer $k$, we have:

$\left(a , b , c\right) = \left(1 - k , 2 k , 4 - k\right)$

and:

$\frac{b + c}{a + b} = \frac{k + 4}{k + 1}$

Explanation:

Given:

$\sin A + {\sin}^{2} A = 1$

We have:

$0 = 4 \left({\sin}^{2} A + \sin A - 1\right)$

$\textcolor{w h i t e}{0} = 4 {\sin}^{2} A + 4 \sin A + 1 - 5$

$\textcolor{w h i t e}{0} = {\left(2 \sin A + 1\right)}^{2} - {\left(\sqrt{5}\right)}^{2}$

$\textcolor{w h i t e}{0} = \left(2 \sin A + 1 - \sqrt{5}\right) \left(2 \sin A + 1 + \sqrt{5}\right)$

So:

$\sin A = \frac{- 1 + \sqrt{5}}{2} \text{ }$ or $\text{ } \sin A = \frac{- 1 - \sqrt{5}}{2}$

We can discount the latter since it gives a value out of the range $\left[- 1 , 1\right]$, so is not satisfied by any Real value of $A$.

So:

$\sin A = - \frac{1}{2} + \frac{\sqrt{5}}{2}$

Then:

${\sin}^{2} A = {\left(- \frac{1}{2} + \frac{\sqrt{5}}{2}\right)}^{2} = \frac{1}{4} - \frac{\sqrt{5}}{2} + \frac{5}{4} = \frac{3}{2} - \frac{\sqrt{5}}{2}$

So:

${\cos}^{2} A = 1 - {\sin}^{2} A = 1 - \left(\frac{3}{2} - \frac{\sqrt{5}}{2}\right) = - \frac{1}{2} + \frac{\sqrt{5}}{2}$

Also:

${\cos}^{4} A = {\left({\cos}^{2} A\right)}^{2} = {\left(- \frac{1}{2} + \frac{\sqrt{5}}{2}\right)}^{2} = \frac{3}{2} - \frac{\sqrt{5}}{2}$

${\cos}^{6} A = \left({\cos}^{2} A\right) \left({\cos}^{4} A\right)$

$\textcolor{w h i t e}{{\cos}^{6} A} = \left(- \frac{1}{2} + \frac{\sqrt{5}}{2}\right) \left(\frac{3}{2} - \frac{\sqrt{5}}{2}\right)$

$\textcolor{w h i t e}{{\cos}^{6} A} = - \frac{3}{4} + \frac{\sqrt{5}}{4} + 3 \frac{\sqrt{5}}{4} - \frac{5}{4}$

$\textcolor{w h i t e}{{\cos}^{6} A} = - 2 + \sqrt{5}$

${\cos}^{8} A = {\left({\cos}^{4} A\right)}^{2} = {\left(\frac{3}{2} - \frac{\sqrt{5}}{2}\right)}^{2} = \frac{9}{4} - \frac{3 \sqrt{5}}{2} + \frac{5}{4} = \frac{7}{2} - \frac{3 \sqrt{5}}{2}$

${\cos}^{12} A = {\left({\cos}^{6} A\right)}^{2} = {\left(- 2 + \sqrt{5}\right)}^{2} = 4 - 4 \sqrt{5} + 5 = 9 - 4 \sqrt{5}$

Then:

$0 = 2 \left(a {\cos}^{12} A + b {\cos}^{8} A + c {\cos}^{6} A - 1\right)$

$\textcolor{w h i t e}{0} = 2 \left(a \left(9 - 4 \sqrt{5}\right) + b \left(\frac{7}{2} - \frac{3}{2} \sqrt{5}\right) + c \left(- 2 + \sqrt{5}\right) - 1\right)$

$\textcolor{w h i t e}{0} = \left(18 a + 7 b - 4 c - 2\right) + \left(- 8 a - 3 b + 2 c\right) \sqrt{5}$

This defines a plane of irrational slope in $a , b , c$ coordinate space.

It is insufficient to determine a unique value for $\frac{b + c}{a + b}$

Perhaps the question is missing some specification, e.g. that $a , b , c$ are integers.

What integer lattice points lie on the plane?

They must satisfy:

$\left\{\begin{matrix}18 a + 7 b - 4 c - 2 = 0 \\ - 8 a - 3 b + 2 c = 0\end{matrix}\right.$

Adding twice the second equation to the first, we find:

$2 a + b - 2 = 0$

So in order that $a$ be an integer, we need $b$ to be even, i.e. $b = 2 k$ for some integer $k$

Then:

$a = \frac{1}{2} \left(2 - b\right) = \frac{1}{2} \left(2 - 2 k\right) = 1 - k$

and we find:

$c = \frac{1}{2} \left(8 a + 3 b\right) = \frac{1}{2} \left(8 - 8 k + 6 k\right) = 4 - k$

Then:

$\frac{b + c}{a + b} = \frac{2 k + \left(4 - k\right)}{\left(1 - k\right) + 2 k} = \frac{k + 4}{k + 1}$

For example, with $k = 3$ we find:

$a = - 2$, $b = 6$, $c = 1$ and $\frac{b + c}{a + b} = \frac{k + 4}{k + 1} = \frac{7}{4}$

How do you find trigonometric ratios of 30, 45, and 60 degrees?

Dean R.
Featured 2 weeks ago

The other answer is fine. I'll just point out that once the student has internalized these Two Tired Triangles of Trig (30/60/90 and 45/45/90) they're done. If you review the trig questions here you'll find the vast majority use just these triangles.

How do you graph #1 + 2sin(2x - pi)#?

sankarankalyanam
Featured 5 days ago

As detailed.

Explanation:

Standard form of sinusoidal function is $y = A \sin \left(B x - C\right) + D$

Given $y = 2 \sin \left(2 x - \pi\right) + 1$

$A = 2 , B = 2 , C = \pi , D = 1$

$A m p l i t u \mathrm{de} = | A | = 2$

$\text{Period } = \frac{2 \pi}{|} B | = \frac{2 \pi}{2} = \pi$

$\text{Phase Shift } = - \frac{C}{B} = \frac{\pi}{2}$, #color(red)(pi/2 " to the right"#

$\text{Vertical Shift } = D = 1$

graph{2 sin(2x -pi) + 1 [-10, 10, -5, 5]}

How do youFind all the solutions of the equation in the interval [0,2pi).? -5sinx = -2cos^2x + 4

Sean
Featured 1 week ago

$x = \frac{7 \pi}{6} , \frac{11 \pi}{6}$

Explanation:

.

$- 5 \sin x = - 2 {\cos}^{2} x + 4$

$- 5 \sin x = - 2 \left(1 - {\sin}^{2} x\right) + 4$

$- 5 \sin x = - 2 + 2 {\sin}^{2} x + 4$

$2 {\sin}^{2} x + 5 \sin x + 2 = 0$

$\sin x = \frac{- 5 \pm \sqrt{25 - 4 \left(2\right) \left(2\right)}}{4} = \frac{- 5 \pm 3}{4} = - 2 , - \frac{1}{2}$

$\sin x = - 2$ is invalid because $\sin$ values are always between $1$ and $- 1$.

$\sin x = - \frac{1}{2} , \therefore x = \frac{7 \pi}{6} , \frac{11 \pi}{6}$

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