Question

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

Answer 1

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.