Question

How do you find the exact value of $\cos ({\text{theta}})$ if $\sin ({\text{theta}}) = \dfrac{{ - 2}}{3}?$

Answer 1

Hint:We know that$\cos x$ is the ratio of base to hypotenuse of a right angled triangle whereas $\sin x$ is the ratio of height to hypotenuse, given that $x$ is the base angle of the right angled triangle. Use this definition and Pythagoras theorem to find the value of $\cos ({\text{theta}})$.
Find the quadrants in which $\sin x$ is negative and then find whether $\cos x$ is negative or positive in the respective quadrants and accordingly put signs.

Complete step by step solution:
Firstly we will find the value of $\cos \theta $ in the first quadrant in which the value $\sin \theta = \dfrac{{ - 2}}{3}$ will become $\dfrac{2}{3}$ because all trigonometric functions have positive values in the first quadrant.
Since we know that $\sin \theta $ is a ratio of height to hypotenuse of a right angled triangle and in this problem \[{\text{height}}\;{\text{:}}\;{\text{hypotenuse}}\] is given $2:3$
so we will assume the values of height and hypotenuse to be $2x\;{\text{and}}\;3x$ respectively.
Now to find $\cos \theta = \dfrac{{{\text{base}}}}{{{\text{hypotenuse}}}}$ , first we have to find the value of base of the right angled triangle, which can be known by use of Pythagoras theorem as follows
\[{c^2} = {a^2} + {b^2},\;{\text{where}}\;a,\;b\;{\text{and}}\;c\] are height, base and hypotenuse of the right angled triangle respectively.
We can find base, $b = \sqrt {{c^2} - {a^2}} $
$
\Rightarrow b = \sqrt {{c^2} - {a^2}} \\
\Rightarrow b = \sqrt {{{(3x)}^2} - {{(2x)}^2}} \\
\Rightarrow b = \sqrt {9{x^2} - 4{x^2}} \\
\Rightarrow b = \sqrt {5{x^2}} \\
\Rightarrow b = \sqrt 5 x\, \\
\therefore \cos \theta = \dfrac{{{\text{base}}}}{{{\text{hypotenuse}}}} = \dfrac{{\sqrt 5 x}}{{3x}} =
\dfrac{{\sqrt 5 }}{3} \\
$
Now, we will find the sign of $\cos \theta $ , since we know that $\sin \theta $ becomes negative in
$2nd\;{\text{and}}\;3rd$ quadrants and we also $\cos \theta $ have positive values in $2nd$ and negative values in $3rd$ quadrant.
$\therefore $ we will have two values of $\cos \theta $ for $\sin \theta = \dfrac{{ - 2}}{3}$,
$ \Rightarrow \cos \theta = \dfrac{{\sqrt 5 }}{3}\;{\text{and}}\;\dfrac{{ - \sqrt 5
}}{3}\;{\text{in}}\;2nd\;{\text{and}}\;3rd$ quadrants respectively.
Note: “theta” is represented by $\theta $ We can solve this by use of a trigonometric identity as follows
We know that
$
{\sin ^2}\theta + {\cos ^2}\theta = 1 \\
\Rightarrow {\cos ^2}\theta = 1 - {\sin ^2}\theta \\
\Rightarrow \cos \theta = \pm \sqrt {1 - {{\sin }^2}\theta } \\
$
Putting given value of $\sin \theta = \dfrac{{ - 2}}{3}$
\[
\Rightarrow \cos \theta = \pm \sqrt {1 - {{\sin }^2}\theta } \\
\Rightarrow \cos \theta = \pm \sqrt {1 - {{\left( {\dfrac{{ - 2}}{3}} \right)}^2}} \\
\Rightarrow \cos \theta = \pm \sqrt {1 - \dfrac{4}{9}} \\
\Rightarrow \cos \theta = \pm \sqrt {\dfrac{{9 - 4}}{9}} = \pm \sqrt {\dfrac{5}{9}} = \dfrac{{ \pm
\sqrt 5 }}{3} \\
\]
Trigonometric identities are very useful for this type of problem.