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How do you solve $sin (cos x) = \frac{\sqrt{2}}{4}$ $sin (cosx) = sqrt2/4$ ?

3 Answers

Jun 11, 2015

Take the arcsin to determine cos(x); then take arccos to determine $x$ $x$ (= 1.201)

Explanation:

If $sin (cos (x)) = \frac{\sqrt{2}}{4}$ $sin(cos(x)) = sqrt(2)/4$
then
$XXXX$ $color(white)("XXXX")$ $cos (x)$ $cos(x)$
$XXXX$ $color(white)("XXXX")$ $= arcsin (sin (cos (x)))$ $= arcsin(sin(cos(x)))$

$XXXX$ $color(white)("XXXX")$ $= arcsin (\frac{\sqrt{2}}{4})$ $=arcsin(sqrt(2)/4)$
$XXXX$ $color(white)("XXXX")$ $XXXX$ $color(white)("XXXX")$ (using a calculator)
$XXXX$ $color(white)("XXXX")$ $= 0.361367$ $=0.361367$

If $cos (x) = 0.361367$ $cos(x) = 0.361367$
then
$XXXX$ $color(white)("XXXX")$ $x$ $x$
$XXXX$ $color(white)("XXXX")$ $= arccos (cos (x))$ $= arccos(cos(x))$

$XXXX$ $color(white)("XXXX")$ $= arccos (0.361367)$ $=arccos(0.361367)$
$XXXX$ $color(white)("XXXX")$ $XXXX$ $color(white)("XXXX")$ (again, using a calculato)
$XXXX$ $color(white)("XXXX")$ $= 1.201$ $= 1.201$

Note: I have assumed all values are in radians

Jun 11, 2015

Solve $sin (arccos (\frac{\sqrt{2}}{4}))$ $sin (arccos (sqrt2/4))$

Explanation:

On the trig unit circle,

$cos x = \frac{\sqrt{2}}{4} = 0.35$ $cos x= sqrt2/4 = 0.35$ --> $x = \pm 69.30$ $x = +- 69.30$ deg

a . sin y = sin (69.30) = 0.94-->

y = 69.30 and y = 180 - 69.30 = 110.70 deg

b. sin y = sin (-69.30) = -0.94 -> sin (360 - 69.30) = 290.70

and y = 180 + 69.30 = 249.30

Within period (0, 360), there are 4 answers;

69.30; 110.70; 249.30 and 290.70.

Check by calculator: cos 290.70 = cos 249.30 = cos 110.70 = cos 69.30 = 0.35

Jun 11, 2015

Alternate Answer The given equation can not be solved since $sin (cos (x))$ $sin(cos(x))$ is a meaningless expression.

Explanation:

$cos x$ $cos x$ can be interpreted to mean one of two things:
1. ${cos}_{r} (x)$ $cos_r(x)$ where $x$ $x$ is an angle measured in radians;
2. ${cos}_{d} (x)$ $cos_d(x)$ where $x$ $x$ is an angle measured in degrees.

Whichever interpretation is used
$cos x$ $cos x$ is a ratio of tow sides of a right-triangle (it is not an angle measurement).

Similarly, $sin (θ)$ $sin(theta)$ can be interpreted as
1. ${sin}_{r} (θ)$ $sin_r(theta)$ where $θ$ $theta$ is an angle measured in radians;
2. ${sin}_{d} (θ)$ $sin_d(theta)$ where $θ$ $theta$ is an angle measured in degrees.

Note that $sin (θ)$ $sin(theta)$ does not have any meaning unless $θ$ $theta$ is an angle.
In particular $sin (θ)$ $sin(theta)$ is meaningless if $θ$ $theta$ is not an angle.

Since the value of $cos x$ $cos x$ is not an angle,
$sin (cos x)$ $sin(cos x)$ is meaningless.

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