# need description on how to do this

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• Jun 3rd 2010, 11:53 PM
mamann
need description on how to do this
Consider: y=sin(x+pi/9)

(a) the graph intersects the x-axis at point A. Find the x co-ordinate of A where 0 < x < pi

(b) Solve the equation sin(x+pi/9)=-1/2 for 0 < x < 2pi

the "<" are less than or equal to in both cases.

Thank you
• Jun 4th 2010, 02:37 AM
Mush
Quote:

Originally Posted by mamann
Consider: y=sin(x+pi/9)

(a) the graph intersects the x-axis at point A. Find the x co-ordinate of A where 0 < x < pi

(b) Solve the equation sin(x+pi/9)=-1/2 for 0 < x < 2pi

the "<" are less than or equal to in both cases.

Thank you

For part a), when an equation crosses the x axis, the y coordinate is zero, hence the equation becomes:

$0 = sin(x+\frac{\pi}{9}$

Now, when you take the inverse sine of 0 you get an infinite number of solutions, but only one of them lies in the interval $[0, \pi]$. Find that solution, then it's just a case of solving algebraically for x.

For part b it is the exact same process except that you need to find take the inverse sine of $\frac{-1}{2}$ and find the solution to it which lies in the given interval.
• Jun 4th 2010, 03:01 AM
mamann
yea but how do i solve for x?
Apparently you cant just apply sin to x and pi/9.
you have to do something where sin(a+b) = (sin a)(cos b) + (sin b)(cos a)
and that gives me:
(sin x)(cos pi/9) + (sin pi/9)(cos x) = 0
Note: pi/9 is in radians which turns out to be 20 degrees.

and i got to here but i dont know what to do after this:

0.94sinx + 0.34cosx = 0
• Jun 4th 2010, 03:11 AM
Mush
Quote:

Originally Posted by mamann
yea but how do i solve for x?
Apparently you cant just apply sin to x and pi/9.
you have to do something where sin(a+b) = (sin a)(cos b) + (sin b)(cos a)
and that gives me:
(sin x)(cos pi/9) + (sin pi/9)(cos x) = 0
Note: pi/9 is in radians which turns out to be 20 degrees.

and i got to here but i dont know what to do after this:

0.94sinx + 0.34cosx = 0

Ah, yes, you could do that.

From there, you divide the equation by $\cos(x)$ which gives:

$0.94 \frac{\sin(x)}{\cos(x)} + 0.34 = 0$

You should know that $\frac{\sin(x)}{\cos(x)} = \tan(x)$, and you can solve from there.
• Jun 4th 2010, 04:47 AM
skeeter
Quote:

Originally Posted by mamann
Consider: y=sin(x+pi/9)

(a) the graph intersects the x-axis at point A. Find the x co-ordinate of A where 0 < x < pi

(b) Solve the equation sin(x+pi/9)=-1/2 for 0 < x < 2pi

the "<" are less than or equal to in both cases.

Thank you

(a) the graph of $y = \sin\left(x + \frac{\pi}{9}\right)$ is the graph of $y = \sin{x}$ shifted left $\frac{\pi}{9}$ units.

since $y = \sin{x}$ crosses the x-axis at $x = \pi$ , the graph of $y = \sin\left(x + \frac{\pi}{9}\right)$ will cross the x-axis at $x = \pi - \frac{\pi}{9} = \frac{8\pi}{9}$

(b) from the unit circle ...

$\sin\left(\frac{7\pi}{6}\right) = \sin\left(\frac{11\pi}{6}\right) = -\frac{1}{2}$

so, solve the equations to get exact solutions ...

$
x + \frac{\pi}{9} = \frac{7\pi}{6}
$

$
x + \frac{\pi}{9} = \frac{11\pi}{6}
$