find the principal and general solution of cosec x = -2
Answers
Question :-- find the principal and general solution of cosec x = -2 ?
Concept used :--
→ -1≤y≤ 1, that is, the value of y is between -1 and 1, so there is a solution.
The set of all solutions to sin(x) = y is
x = sin^(-1)y + 2kπ and x = −sin^(-1)y + (2k+1)π,
where k can be any integer; that is, the solutions for x consist of sin^(-1)y plus all even multiples of π, together with minus sin-1(y) plus all odd multiples of π.
→ The principal value of sin^(−1) x is α, where - π/2 ≤ α ≤ π/2 and α satisfies the equation sin θ = x.
→ cosec x = 1/sinx
→ sin30° = 1/2
→ sin150° = -1/2
→ sin(π+x) = - sinx
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Solution :----
Given , cosec x = (-2)
→ 1/sinx = (-2)
Cross - Multiplying we get,
→ sinx = (-1/2)
→ sinx = - sin30°
→ sinx = - sin(π/6)
→ sinx = sin(π+π/6)
→ sinx = sin(7π/6)
Therefore in general, sin^(−1)(-1/2) = θ = nπ + (-1)^n π/6 and the angle nπ + (- 1)^n 7π/6 is called the general value of sin^(-1)(-1/2).
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Now , The positive or negative least numerical value of the angle is called the principal value ...
So,
==>> Principal value of sinx = sin(7π/6) = - π/2 ≤ (-π/6) ≤ π/2
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Step-by-step explanation:
Using this logic ,In right ΔABC
sine of ∠ A = P/H = BC/AC
cosine of ∠ A = B/H = AB/AC
tangent of ∠ A = P/B = BC/AB
cosecant of ∠ A = 1/ sine of A = AC/BC
secant of ∠ A = 1/cosine of A = AC/AB
cotangent of ∠ A = 1/tangent of A = AB/BC
The value of sin A or cos A never exceeds 1, whereas the value of sec A or cosec A is always greater than or equal to 1.
Trigonometric ratios of an acute angle in a right triangle express the relationship between the angle and the length of its sides.