Question

Find the value of $\cos 120{}^{\circ }$.

Answer 1

Hint: In a right-angled triangle with the length of the side opposite to angle θ as perpendicular (P), base (B) and hypotenuse (H):
 $\sin \theta ={\displaystyle \frac{P}{H}},\cos \theta ={\displaystyle \frac{B}{H}},\tan \theta ={\displaystyle \frac{P}{B}}$
 $P^2+B^2=H^2$ (Pythagoras' Theorem)
If a point P is at a distance $r$ from the origin, then the coordinates of the point are $(r\cos \theta ,r\sin \theta )$, where $\theta$ is angle made by line OP with the positive direction of the x-axis.
Find the coordinates of a point which is 1 unit far from the origin and makes an angle of $120{}^{\circ }$ with the x-axis. The x-co-ordinate of the point is the value of $\cos 120{}^{\circ }$.

Complete step by step answer:
Let us mark a point P on the graph paper, at distance of 1 unit from the origin, such that OP makes an angle of $120{}^{\circ }$ with the positive direction of the x-axis.
From the definition of trigonometric ratios, we know that the x-co-ordinate of the point P is the value of $\cos 120{}^{\circ }$ and the y-coordinate is the value of $\sin 120{}^{\circ }$.

The $\mathrm{\Delta }PSO$ in the above diagram is an equilateral triangle.
∴ $OP=OS=1$ ... (1)
Also, $\mathrm{\Delta }PTO\cong \mathrm{\Delta }PTS$ ... (Using ASA congruence)
∴ $ST=OT={\displaystyle \frac{1}{2}}OS$
⇒ $OT={\displaystyle \frac{1}{2}}\times 1={\displaystyle \frac{1}{2}}$ ... [Using equation (1)]
It means that the coordinates of the point T are $(-{\displaystyle \frac{1}{2}},0)$.
Since the point P is on the perpendicular line at point T, its x-co-ordinate must also be the same as that of T, i.e. $-{\displaystyle \frac{1}{2}}$.
It follows that $\cos 120{}^{\circ }=-{\displaystyle \frac{1}{2}}$.

Note: Now that we know the lengths of OP and OT, we can use Pythagoras' theorem and find PT as well, which is the y-coordinate of P, i.e. $\sin 120{}^{\circ }$.
Rule of CAST: In the IV, I, II and III quadrants, $\cos \theta$, All trigonometric ratios, $\sin \theta$ and $\tan \theta$ are positive, respectively.
Trigonometric Ratios for Allied Angles:
 $\sin (-\theta )=-\sin \theta$ $\cos (-\theta )=\cos \theta$
 $\sin (2n\pi +\theta )=\sin \theta$ $\cos (2n\pi +\theta )=\cos \theta$
 $\sin (n\pi +\theta )={(-1)}^n\sin \theta$ $\cos (n\pi +\theta )={(-1)}^n\cos \theta$
 $\sin [(2n+1){\displaystyle \frac{\pi }{2}}+\theta ]={(-1)}^n\cos \theta$ $\cos [(2n+1){\displaystyle \frac{\pi }{2}}+\theta ]={(-1)}^n(-\sin \theta )$
If one trigonometric ratio is known, we can use Pythagoras' Theorem and calculate the values of all other trigonometric ratios.