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Home / Trigonometry / TRIGONOMETRICAL EQUATIONS AND GENERAL VALUES| LECTURE – 1

TRIGONOMETRICAL EQUATIONS AND GENERAL VALUES| LECTURE – 1

General expression of all angles, one of whose trigonometrical ratio’s is zero

If the sign of an angle be zero, from definition, the length of the perpendicular from any point of one of its arms upon another is zero, so that the two arms must be in the same straight line. Evidently, therefore, such angle must be zero, or some multiple of π, odd or even.

Thus, if sin θ = 0, then θ = nπ,

n being zero, or any integer, positive or negative.

When the cosine of an angle is zero, the projection of any length along one arm upon the other is zero, and so the two arms must be at two right angles to one another. The angle must, therefore, be evidently either π/2 or 3π/ 2 or differ from these by complete revolutions; in other words, the angle may be any odd multiple of π/2.

Thus if cos θ = 0, then

(2n + 1)π/2,

n being zero, or any integer, positive or negative.

Again, if tan θ = 0, then its numerator sin θ is also zero;

And so θ = nπ.

Similarly if cot θ = 0, then cos θ = 0;

and so(2n + 1)π/2 ,

General expression of angles having the same sine (or cosecant)

Let α be any angle, positive or negative, such that its sine is equal to a given quantity k; for fixing up the idea, and for the sake of convenience , in particle the smallest positive angle having its sine for the given quantity k is taken as α. Let θ be any other angle whose sine is equal to k.

Then sin θ = sin α

⟹ sin θ – sin α = 0

⟹ 2 sin ½ (θ – α) cos ½ (θ + α) = 0

Therefore either sin ½ (θ – α) = 0

⟹ ½ (θ – α) = any multiple if π = mπ

⟹ ½ (θ – α) = mπ ————————————- (1)

Or, else cos ½ (θ + α) = 0

⟹ ½ (θ + α) = any odd multiple of π/2 = (2m +  1)π/2

⟹ ½ (θ + α) = (2m +  1)π/2 ————————- (2)

From (1) and (2), we get respectively

θ – α = 2mπ  ⟹  θ = α+2mπ  ———————– (3)

θ + α = (2m +  1)π  ⟹ θ = – α + (2m +  1)π  ———- (4)

Combining (3) and (4), we get

θ = (– 1 )nα + nπ ——————————————– (5)

n is zero, or any integer, positive or negative, odd or even.

If cosec θ = cosec α , then sin θ = sin α; hence all angles having the same cosecant as that of α are also given by the expression (5).

Thus all angles having the same sine or cosecant as that of α are given by nπ + α or nπ – α , depending on whether n is even or odd, that is by

nπ + (– 1 )nα,

n being zero, or any integer, positive or negative, odd or even.