Answer:
[tex]\displaystyle \cos ( \theta _{1} ) = - \frac{ \sqrt{15}}{ 4 }[/tex]
Step-by-step explanation:
on a unit circle there're 4 Quadrant. on Q:I sin and cos both are positive,on Q:II cos is negative and sin positive, on Q:III both sin and cos are negative and on Q:IV cos is positive and sin negative.
actually a unit circle is a coordinate plane but
- there's (cos,sin) instead of (x,y)
it'll be required later
well to solve the problem we can consider Pythagorean theorem which states that the square of sin and cos is equal to 1. therefore,
[tex] \displaystyle \sin ^{2} ( \theta) + \cos ^{2} ( \theta) = 1[/tex]
[tex] \rm \displaystyle \implies \boxed{\cos ^{} ( \theta) = \sqrt{1 - \sin ^{2} ( \theta) }}[/tex]
in this case,
- [tex] \rm \theta \: \: is \: \: \theta_{1}[/tex]
- sin[tex]\theta[/tex] is ¼
Thus substitute:
[tex]\displaystyle \cos ( \theta _{1} ) = \sqrt{1 - \left( \frac{1}{4} \right) ^{2} }[/tex]
simplify square:
[tex]\displaystyle \cos ( \theta _{1} ) = \sqrt{1 - \frac{1}{16} }[/tex]
simplify substraction:
[tex]\displaystyle \cos ( \theta _{1} ) = \sqrt{ \frac{16 - 1}{16} }[/tex]
simplify numerator:
[tex]\displaystyle \cos ( \theta _{1} ) = \sqrt{ \frac{15}{16} }[/tex]
recall redical rule:
[tex]\displaystyle \cos ( \theta _{1} ) = \frac{ \sqrt{15}}{ \sqrt {16} }[/tex]
simplify square root:
[tex]\displaystyle \cos ( \theta _{1} ) = \frac{ \sqrt{15}}{ 4 }[/tex]
since cos is negative on Q:II hence,
[tex]\displaystyle \cos ( \theta _{1} ) = \boxed{- \frac{ \sqrt{15}}{ 4 }}[/tex]
and we're done!
