Question

How do you find the exact value of $\sec ({120^0})$ ?

Answer 1

We can solve this using two ways. First we can solve this using complementary angles. Second way is by using the addition formula for the cosine function.Sine, cosine, tangent, cosecant, secant and cotangent are the six types of trigonometric functions; sine, cosine and tangent are the main functions while cosecant, secant and cotangent are the reciprocal of sine, cosine and tangent respectively. First we find the value of $\cos ({120^0})$ then taking reciprocal of that we will have the desired result.

Complete step by step answer:
Given, $\sec ({120^0})$.
We know that reciprocal of cosine is secant. That is
$\sec ({120^0}) = \dfrac{1}{{\cos ({{120}^0})}}$
Now we find the value of $\cos ({120^0})$.
We can write 120 as a sum of 90 and 30.
That is ${120^0} = {90^0} + {30^0}$.
$\cos ({120^0}) = \cos ({90^0} + {30^0})$
$\Rightarrow\cos ({120^0}) = \cos ({90^0} + {30^0})$
But we know that $\cos ({90^0} + \theta ) = - \sin \theta$. The negative sign is because the cosine lies in the third quadrant and in the third quadrant cosine is negative.
$\cos ({120^0}) = - \sin ({30^0})$
$\Rightarrow\cos ({120^0}) = - \dfrac{1}{2}$.
Thus we have
$\cos ({120^0}) = - \dfrac{1}{2}$.
Now we have $\sec ({120^0}) = \dfrac{1}{{\cos ({{120}^0})}}$
$\sec ({120^0}) = \dfrac{1}{{\left( { - \dfrac{1}{2}} \right)}}$
$\Rightarrow \sec ({120^0}) = - 2$. This is the required answer.
We can solve this using the cosine addition formula. That is,
$\cos (a + b) = \cos (a).\cos (b) - \sin (a).\sin (b)$ . Here $a = {90^0}$ and $b = {30^0}$.
Substituting we have,
$\cos ({90^0} + {30^0}) = \cos ({90^0}).\cos ({30^0}) - \sin ({90^0}).\sin ({30^0})$
We know the values and substituting we have,
$\cos ({120^0}) = \left( {0 \times \dfrac{{\sqrt 3 }}{2}} \right) - \left( {1 \times \dfrac{1}{2}} \right)$
$\cos ({120^0}) = - \dfrac{1}{2}$.
To find secant value we have $\sec ({120^0}) = \dfrac{1}{{\cos ({{120}^0})}}$
$\sec ({120^0}) = \dfrac{1}{{\left( { - \dfrac{1}{2}} \right)}}$
$\therefore\sec ({120^0}) = - 2$.
In both the cases we have the same answer.

Note: Remember the formula sine and cosine addition and subtract formula well.The additional formula for cosine is $\cos (a + b) = \cos (a).\cos (b) - \sin (a).\sin (b)$. Similarly we have $\sin (a - b) = \sin (a).\cos (b) - \cos (a).\sin (b)$ and $\cos (a - b) = \cos (a).\cos (b) + \sin (a).\sin (b)$. Depending on the angle we use the required formulas. Also remember A graph is divided into four quadrants, all the trigonometric functions are positive in the first quadrant, all the trigonometric functions are negative in the second quadrant except sine and cosine functions, tangent and cotangent are positive in the third quadrant while all others are negative and similarly all the trigonometric functions are negative in the fourth quadrant except cosine and secant.