Question #3d967

Question

Question #3d967

1 Answer

Impact of this question

1147 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-24T12:01:06

Please see below.

Explanation:

Let's set the two functions equal to each other to solve for the $x$ $x$ -coordinate of the intersection point:

$cos 2 x = 2 sin 3 x$ $cos2x=2sin3x$

But we know:

$sin (a + b) = sin a cos b + cos a sin b$ $sin(a+b)=sinacosb+cosasinb$

We can re-write our equation and use the above formula:

$cos 2 x = 2 sin (2 x + x)$ $cos2x=2sin(2x+x)$

$cos 2 x = 2 (sin 2 x cos x + cos 2 x sin x)$ $cos2x=2(sin2xcosx+cos2xsinx)$

$cos 2 x = 2 sin 2 x cos x + 2 cos 2 x sin x$ $cos2x=2sin2xcosx+2cos2xsinx$

But we know:

$sin 2 x = 2 sin x cos x$ $sin2x=2sinxcosx$ and $cos 2 x = 2 {cos}^{2} x - 1$ $cos2x=2cos^2x-1$ . Let's plug them in:

$2 {cos}^{2} x - 1 = 2 \cdot 2 sin x cos x cos x + 2 (2 {cos}^{2} x - 1) sin x$ $2cos^2x-1=2*2sinxcosxcosx+2(2cos^2x-1)sinx$

$2 {cos}^{2} x - 1 = 4 sin x {cos}^{2} x + 4 sin x {cos}^{2} x - 2 sin x$ $2cos^2x-1=4sinxcos^2x+4sinxcos^2x-2sinx$

$2 {cos}^{2} x - 1 = 8 sin x {cos}^{2} x - 2 sin x$ $2cos^2x-1=8sinxcos^2x-2sinx$

Let's use ${sin}^{2} x + {cos}^{2} x = 1$ $sin^2x+cos^2x=1$ and plug in $1 - {sin}^{2} x$ $1-sin^2x$ for ${cos}^{2} x$ $cos^2x$ to turn the entire equation in terms of $sin x$ $sinx$ :

$2 (1 - {sin}^{2} x) - 1 = 8 sin x (1 - {sin}^{2} x) - 2 sin x$ $2(1-sin^2x)-1=8sinx(1-sin^2x)-2sinx$

$2 - 2 {sin}^{2} x - 1 = 8 sin x - 8 {sin}^{3} x - 2 sin x$ $2-2sin^2x-1=8sinx-8sin^3x-2sinx$

$1 - 2 {sin}^{2} x = 6 sin x - 8 {sin}^{3} x$ $1-2sin^2x=6sinx-8sin^3x$

$8 {sin}^{3} x - 2 {sin}^{2} x - 6 sin x + 1 = 0$ $8sin^3x-2sin^2x-6sinx+1=0$

Here is the link for an online calculator for cubic roots:

http://www.1728.org/cubic.htm

You will get the following three answers for $sin x$ $sinx$ :

Answer 1: $sin x = (0.91842444565049)$ $sinx=(0.91842444565049)$ which means

$x = arcsin (0.91842444565049)$ $x=arcsin(0.91842444565049)$ or $x = 66.7 D e g r e e s$ $x=66.7 Degrees$

Answer 2: $sin x = (- 0.8320078223739)$ $sinx=(-0.8320078223739)$ which means

$x = arcsin (- 0.8320078223739)$ $x=arcsin(-0.8320078223739)$ or $x = - 56.3 D e g r e e s$ $x=-56.3 Degrees$

Answer 3: $sin x = (0.16358337672342)$ $sinx=(0.16358337672342)$ which means

$x = arcsin (0.16358337672342)$ $x=arcsin(0.16358337672342)$ or $x = 9.4 D e g r e e s$ $x=9.4 Degrees$

Only answers $1$ $1$ and $3$ $3$ are acceptable because they fall in the specified interval. As such the two curves intersect at two points whose $x$ $x$ -values are $66.7$ $66.7$ and $9.4 D e g r e e s$ $9.4 Degrees$

To find the angles between the curves at these points you need to find the slopes of the tangent lines to the curves at these points and then find the angle between the two tangents.

To find the slope of a tangent to a curve you take the derivative of the functions and evaluate it at the $x$ $x$ value of the tangent point.

For $y = cos 2 x$ $y=cos2x$

$\frac{d y}{d x} = - 2 sin 2 x$ $dy/dx=-2sin2x$

For $y = 2 sin 3 x$ $y=2sin3x$

$\frac{d y}{d x} = 6 cos 3 x$ $dy/dx=6cos3x$

At $x = 9.4 D e g r e e s$ $x=9.4 Degrees$ slope of tangent to $y = cos 2 x$ $y=cos2x$ is

$m_{1} = - 2 sin (2 \cdot 9.4) = - 2 sin (18.8) = - 2 \cdot 0.32$ $m_1=-2sin(2*9.4)=-2sin(18.8)=-2*0.32$ or $m_{1} = - 0.64$ $m_1=-0.64$

At this point, the slope of tangent to $y = 2 sin 3 x$ $y=2sin3x$ is

$m_{2} = 6 cos (3 \cdot 9.4) = 6 cos (28.2) = 6 \cdot 0.88$ $m_2=6cos(3*9.4)=6cos(28.2)=6*0.88$ or $m_{2} = 5.28$ $m_2=5.28$

Angle between tangent one and $x$ $x$ -axis is $arctan (m_{1}) = arctan (- 0.64) = - 32.62 D e g r e e s$ $arctan(m_1)=arctan(-0.64)=-32.62 Degrees$

Angle between tangent two and $x$ $x$ -axis is $arctan (m_{2}) = 79.27 D e g r e e s$ $arctan(m_2)=79.27 Degrees$

Then the angle between the two curves is $79.27 - (- 32.62) = 111.89 D e g r e e s$ $79.27-(-32.62)=111.89 Degrees$

At $x = 66.7 D e g r e e s$ $x=66.7 Degrees$

$m_{1} = - 2 sin (2 \cdot 66.7) = - 2 sin (133.4) = - 2 \cdot 0.73 = - 1.46$ $m_1=-2sin(2*66.7)=-2sin(133.4)=-2*0.73=-1.46$

$m_{2} = 6 cos (3 \cdot 66.7) = 6 cos (200.1) = 6 \cdot - 0.94 = - 5.64$ $m_2=6cos(3*66.7)=6cos(200.1)=6*-0.94=-5.64$

Angle between tangent one and $x$ $x$ -axis is $arctan (- 1.46) = - 55.59 D e g r e e s$ $arctan(-1.46)=-55.59 Degrees$

Angle between tangent two and $x$ $x$ -axis is $arctan (- 5.64) = - 79.94 D e g r e e s$ $arctan(-5.64)=-79.94 Degrees$

Angle between the two curves is $- 55.59 - (- 79.94) = 24.35 D e g r e e s$ $-55.59-(-79.94)=24.35 Degrees$

Science

Math

Humanities

... and beyond

Question #3d967

1 Answer

Explanation:

Related questions

Impact of this question