Question

PLEASE HELP PRECALCULUS WILL MARK BRAINLIEST -SEE ATTACHMENT-

Answer 1

Answer:

$\cos(arcsin(\frac{1}{4}))=\frac{\sqrt{15}}{4}$.

Step-by-step explanation:

We want to evaluate cos(arcsin(1/4)) probably without a calculator.

If you did want a calculator answer, that would be 0.968245837.

Alright so anyways, this is the way I begin these trig(arctrig( )) types of problems when the trig parts are different.

Let u=arcsin(1/4).

If u=arcsin(1/4) then sin(u)=1/4.

So we want to find cos(u) given sin(u)=1/4.  (I just replace arcsin(1/4) in cos(arcsin(1/4)) with u.)

Let's use a Pythagorean Identity:

$\cos^2(u)+\sin^2(u)=1$.

Let's plug in 1/4 for sin(u):

$\cos^2(u)+(\frac{1}{4})^2=1$

Simplify a bit:

$\cos^2(u)+\frac{1}{16}=1$

Subtract 1/16 on both sides:

$\cos^2(u)=1-\frac{1}{16}$

Simplify the right hand side:

$\cos^2(u)=\frac{15}{16}$

Take the square root of both sides:

$\cos(u)=\pm \sqrt{\frac{15}{16}}$

Separate the square thing to the numerator and denominator:

$\cos(u)=\pm \frac{\sqrt{15}}{\sqrt{16}}$

Replace $\sqrt{16}$ with 4 since $4^2=16$:

$\cos(u)=\pm \frac{\sqrt{15}}{4}$

Now how do we determine if the cosine should be positive or negative.

arcsin(1/4) is an angle that is going to be between -pi/2 and pi/2 due to restrictions upon the sine curve to be one to one.

cosine of an angle between -pi/2 and pi/2 is going to be positive because these are the 1st and 4th quadrant where the x-coordinate is positive (the cosine value is positive)

$\cos(u)=\frac{\sqrt{15}}{4}$

So recall u=arcsin(1/4):

$\cos(arcsin(\frac{1}{4}))=\frac{\sqrt{15}}{4}$.

For fun, put $\frac{\sqrt{15}}{4}$.  If you don't get  0.968245837 then you made a mistake in the above reasoning. We do get that so the results of the calculator and our trigonometry/algebra confirm each other.