Let f(x) be a function.
In general, to find whether f(x) is an even or odd function, we will plugin -x for x into f(x) and do the following check.
f(-x) = f(x) ----> f(x) is even function
f(-x) = -f(x) ----> f(x) is odd function
In trigonometry, we have the following six ratios
sin θ
csc θ
cos θ
sec θ
tan θ
cot θ
To find each of the above trigonometric ratios is even or odd, we have replace θ by -θ.
For example, in sin θ, if we replace θ by -θ, it becomes
sin (-θ)
In sin (-θ), the angle is negative. In a trigonometric ratio, if the angle is negative, we will consider quadrant IV.
In quadrant IV, the trigonometric ratios cosine and secant are positive and the remaning four trigonometric ratios are negative.
Then, we have
cos (-θ) = cos θ
sec (-θ) = sec θ
sin (-θ) = -sin θ
csc (-θ) = -csc θ
tan (-θ) = -tan θ
cot (-θ) = -cot θ
Since
cos (-θ) = cos θ
sec (-θ) = sec θ,
cos θ and sec θ are even trigonometric ratios and the rest are odd trigonometric ratios.
cos (-θ) = cos θ
sec (-θ) = sec θ
sin (-θ) = -sin θ
csc (-θ) = -csc θ
tan (-θ) = -tan θ
cot (-θ) = -cot θ
Problem 1 :
If the value of cos θ = 0.5, fins the value of cos (-θ).
Solution :
Since cos θ is even, we have
cos (-θ) = cos θ
= 0.5
Problem 2 :
If the value of tan θ = 1.2, find the value of tan (-θ).
Solution :
Since tan θ is odd, we have
tan (-θ) = -tan θ
= -1.2
Problem 3 :
If the value of csc θ = -2, find the value of csc (-θ).
Solution :
Since csc θ is odd, we have
csc (-θ) = -csc θ
= -(-2)
= 2
Problem 4 :
Solution :
Since cos θ is even, we have
Since tan θ is odd, we have
Evaluating sin θ :
Problem 5 :
Solution :
Since cos θ is even, we have
Since sin θ is odd, we have
Evaluating tan θ :
Problem 6 :
If the value of cos 60° = 0.5, find the value of cos (-480°).
Solution :
Since cos θ is even, we have
cos (-480°) = cos 480°
= cos (360° + 120°)
= cos 120°
120° is in quadrant II and its reference angle is 60°.
And also, cosine is negative in quadrant II.
= -cos 60°
= -0.5
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