Answer:
[tex]\cos(arcsin(\frac{1}{4}))=\frac{\sqrt{15}}{4}[/tex].
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:
[tex]\cos^2(u)+\sin^2(u)=1[/tex].
Let's plug in 1/4 for sin(u):
[tex]\cos^2(u)+(\frac{1}{4})^2=1[/tex]
Simplify a bit:
[tex]\cos^2(u)+\frac{1}{16}=1[/tex]
Subtract 1/16 on both sides:
[tex]\cos^2(u)=1-\frac{1}{16}[/tex]
Simplify the right hand side:
[tex]\cos^2(u)=\frac{15}{16}[/tex]
Take the square root of both sides:
[tex]\cos(u)=\pm \sqrt{\frac{15}{16}}[/tex]
Separate the square thing to the numerator and denominator:
[tex]\cos(u)=\pm \frac{\sqrt{15}}{\sqrt{16}}[/tex]
Replace [tex]\sqrt{16}[/tex] with 4 since [tex]4^2=16[/tex]:
[tex]\cos(u)=\pm \frac{\sqrt{15}}{4}[/tex]
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)
[tex]\cos(u)=\frac{\sqrt{15}}{4}[/tex]
So recall u=arcsin(1/4):
[tex]\cos(arcsin(\frac{1}{4}))=\frac{\sqrt{15}}{4}[/tex].
For fun, put [tex]\frac{\sqrt{15}}{4}[/tex]. 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.
