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Trigonometric ratios provide relationship between the sides and angles of a right angled triangle.
The plane is basically divided into four quadrants by the rectangular co-ordinate axis. An angle is said to be positive if it is measured in anti-clockwise direction from the positive x- axis and similarly, it is considered to be negative if it is measured in the clockwise direction.
In Fig. 1, note that
∠xoy = π/2, ∠xox’ = π, ∠xoy’ = 3π/2.
Here PiQi denote the y coordinate and hence are positive whenever above the x axis, while it is negative when below the x- axis. Though OPi is always taken as positive, OQi is positive only in the positive direction of x- axis and negative when along the negative direction.
The trigonometric formulas are easy to memorize and fetch some direct questions. Various trigonometric ratios have different signs depending on the quadrant they lie in.
Two angles whose sum is 90o are called complementary angles while those whose sum equals 180o are called supplementary angles.
Since the trigonometric ratios are ratio between two sides of a right angled triangle with respect to an angle, so they are real numbers.
The domain of sine and cos is R, i.e. the real numbers while their value varies from -1 to 1.
The domain of tan x is R – {(2n + 1) π/2 , n ∈ I} while its range is R.
The domain of cosec x is R – {nπ, n ∈ I} while its range is R – {x: -1 < x < 1}.
The domain of cot x is R – {nπ, n ∈ I} while its range is R.
The domain of sec x is R – {(2n + 1) π/2 , n ∈ I} while its range is R except {x: -1 < x < 1}.
sin (nπ + (–1)n) θ = sin θ, n ∈ I
cos (2nπ + π) = cos π, n ∈ I.
tan (nπ + π) = tan π, n ∈ I
Watch this Video for more reference
Given below is a trigonometric ratios chart which depicts the values of sin and cosine functions at different angles:
Illustration:
Given A = sin2θ + cos4θ, then for all real values of
(a) 1 ≤ A ≤ 2
(b) 3/4 ≤ A ≤ 1
(c) 13/16 ≤ A ≤ 1
(d) 3/4 ≤ A ≤ 13/16
Solution: Given A = sin2θ + cos4θ
= sin2θ + (1 - sin2θ)2
Hence, A = sin4θ - sin2θ +1.
This gives A = (sin2θ – 1/2)2 + 3/4
Since 0 ≤ sin2θ ≤ 1
Hence, 0 ≤ (sin2θ – 1/2)2 ≤ 1/4
Therefore, 3/4 ≤ A ≤ 1.
The expression
3[sin4 (3π/2 – α) + sin4 (3π + α)] – 2[sin6 (π/2 + α) + sin6 (5π – α)] is equal to
(a) 0
(b) 1
(c) 3
(d) sin 4α + cos 6α
Solution:
The given expression is
3[sin4 (3π/2 – α) + sin4 (3π + α)] – 2[sin6 (π/2 + α) + sin6 (5π – α)]
= 3(cos4 α + sin4 α) -2 (cos6 α + sin6 α)
= 3(1-2 cos2 α sin2 α) -2(1-3 cos2 α sin2 α)
= 3-6 cos2 α sin2 α – 2 + 6 cos2 α sin2 α
= 1.
Look into the Revision Notes on Trigonometry for a quick revision.
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