Trigonometric ratios of 360 plus theta
 (360 + θ)

In this section we will learn trigonometric ratios of 360 plus theta (360 + θ).
We know that the terminal sides of co-terminal angles always coincide and θ and (360 + θ) are co-terminal angles.

Clearly, 360 + θ and θ are co-terminal angles.
sin(360 + θ) = Sin[360 -(- θ)]
But we know that -sin(- θ) = sinθ
$\underline{sin(360 + θ)= sin θ} $

Again,
cos(360 + θ) = cos[360-(- θ)]
But we know that cos(- θ) =cosθ
$\underline{cos(360 + θ)= cos θ} $

Now,
tan(360 + θ) = tan[360-(- θ)]
But we know that -tan(- θ) = tanθ
$\underline{tan(360 + θ)= tan θ} $

As we know that, csc θ = $\frac{1}{sinθ}$
∴ csc (360+θ) = $\frac{1}{sin(360+θ)}$
But sin(360+θ) = sin(θ)
∴ csc (360+θ)= $\frac{1}{sin θ}$
$\underline{csc(360 + θ)= csc θ} $

And,
sec θ = $\frac{1}{cosθ}$
∴ sec (360+θ) = $\frac{1}{cos(360+θ)}$
But cos(360+θ) = cos(θ)
∴ sec (360+θ)= $\frac{1}{cos θ}$
$\underline{sec(360 + θ)= cos θ} $

Again,
cot θ = $\frac{1}{tanθ}$
∴ cot (360+θ) = $\frac{1}{tan(360+θ)}$
But tan(360+θ) = tan(θ)
∴ cot (360+θ)= $\frac{1}{ tan θ}$
$\underline{cot(360 + θ)= cot θ} $

For any positive integer 'n', angle (360 x n + θ) is co-terminal to angle θ. Therefore, for any positive integer n, we have,
sin(360 x n + θ ) = sin θ
cos (360 x n + θ ) = cos θ
tan(360 x n + θ ) = tan θ
csc(360 x n + θ ) = csc θ
sec (360 x n + θ ) = sec θ
cot (360 x n + θ ) = cot θ

Examples on Trigonometric ratios of 360 plus theta
(360 + θ)

1) Find the values of the following trigonometric ratios :
(i) csc $(390^{0}$
Solution : csc $(390)^{0}$
390 = 360 + 30
∴ csc $(390)^{0}$ = csc (360 + 30) = csc 30
But csc 30 = 2
⇒ csc (390) = 2

(ii) cot $(570^{0}$
Solution : cot $(570^{0}$
We have, 570 = 90 x 6 + 30
Clearly, 570 is in the IIIrd quadrant and 6 is even integer so the multiple of 90 is even.
∴ cot $(570^{0}$ = cot $(90^{0} \times 6 + 30) $ = cot 30 = $\sqrt{3}$

(iii) tan $(480^{0}$
Solution : tan $(480^{0}$
We have, 480 = 90 x 5 + 30
Clearly, 480 is in the IInd quadrant and 5 is odd integer so the multiple of 90 is odd.
∴ tan $(480^{0}$ = tan $(90^{0} \times 5 + 30) $ = -cot 30 = -$\sqrt{3}$

2) Prove that : cos(510). cos(330). + sin (390). cos(120) = -1
Solution : Consider,
=cos(510). cos(330). + sin (390). cos(120)
= cos(90 x 5 + 60). cos(90 x 3 + 60). + sin (90 x 4 + 30). cos(90 x 1 + 60)
= (-sin 60).(sin 60) + (sin 30).(-sin 30)
= $\frac{-\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} + \frac{1}{2} \times \frac{-1}{2}$
= $\frac{-3}{4} - \frac{1}{4}$
= - 1


11th grade math

From Trigonometric ratios of 360 plus theta(360 +Θ) to Home