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Trigonometry Functions and Inverse Functions
1. Sin∅ = 1 / cosec∅
sin∅ = P / H
if ∅ = 30౦ and P = 4
sin30 = 4/ H
from table we can find sin 30 = 1/2
1/2 = 4/ H
by cross multiplying
H = 8
now we have
P = 4, H = 8 and we can find base side
i.e. P² + B² = H²
4² + B² = 8²
B² = 64 – 16
B² = 48
B = √ 48
B = 6.9
now we can find all the ratios from conversions
- Behaviour with trig functions with signs of angles:-
if in place of ∅ there is -∅ then
- sin(-∅) = – sin∅
- cos (-∅) = cos∅
- tan (-∅) = – tan ∅
- cot -(∅) = – cot∅
- sec (- ∅) = sec ∅
- cosec (-∅) = – cosec ∅
- Some of the Restrictions of Trigonometrical Ratios are:
● If A and B two acute angles and sin A = sin B, then we get A = B. If we cancel sign from both side to get the result, that is wrong.
For Example: If Sin θ = 60°, then θ = 60°. But we should not cancel sign from both sides to get the result.
● Cos 2θ ≠ 2 Cos θ
● Sin A/Sin B ≠ A/B
● Sin A ± Sin B ≠ Sin (A ≠ B)
● Cos θ does not imply cos × θ; in fact, it represents the ratio of perpendicular and hypotenuse with respect to the angle θ of a right-angled triangle.
● Cos2 θ means (cos θ)2 or cos θ × cos θ; do not write (cos θ)2 = cos θ2 since cos θ2 implies cos (θ × θ).
Similarly we can write,
sin3 θ for (sin θ)3;
tan5 θ for (tan θ)5;
sec7 θ for (sec θ)7; etc,.
These are the restrictions of trigonometrical ratios need to be followed in case of learning the trigonometric ratios.
these above mentioned rules are to be carried out for each solution of questions of Trigonometry.
- INVERSE TRIGONOMETRIC FUNCTIONS:-
inverse functions behaviour and solutions can be understands from this :-
if sin 30° = 1/2
30° = (sin)౼¹ 1/2
generally we can conclude that
if sin∅ = x
them ∅ = sin౼¹ x
and this process will work similarly for other trig ratios.
following are some examples for better understanding of this concept:-
1. Find the general values of sin−1−1 (- √3/2)
Let, sin−1−1 (- √3/2) = θ
Therefore, sin θ = – √3/2
⇒ sin θ = – sin (π/3)
⇒ sin θ = (- π/3)
Therefore, the general value of sin−1−1 (- √3/2) = θ = nπ – (- 1)nn π/3, where, n = 0 or any integer.
2. Find the general values of cot−1−1 (- 1)
Let, cot−1−1 (- 1) = θ
Therefore, cot θ = – 1
⇒ cot θ = cot (- π/4)
Therefore, the general value of cot−1−1 (- 1) = θ = nπ – π/4, where, n = 0 or any integer.
Hope you’ve found this tutorial about trig identities and inverse trig functions, useful. Please let me know how do you feel with this topic by commenting
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