We will learn how to solve various type of problems on trigonometric functions of any angles.
1. Is the equation 2 sin\(^{2}\) θ - cos θ + 4 = 0 possible?
Solution:
2 sin\(^{2}\) θ – cos θ + 4 = 0
⇒ 2(1 - cos\(^{2}\) θ) - cos θ + 4 = 0
⇒ 2 - 2 cos\(^{2}\) θ - cos θ + 4 = 0
⇒ - 2 cos\(^{2}\) θ - cos θ + 6 = 0
⇒ 2 cos\(^{2}\) θ + cos θ - 6 = 0
⇒ 2 cos\(^{2}\) θ + 4 cos θ - 3 cos θ - 6 = 0
⇒ 2 cos θ (cos θ + 2) - 3 (cos θ + 2) = 0
⇒ (cos θ + 2) (2 cos θ - 3) = 0
⇒ (cos θ + 2) = 0 or (2 cos θ - 3) = 0
⇒ cos θ = - 2 or cos θ = 3/2, both of which are impossible as -1 ≤ cos θ ≤ 1.
Hence, the equation 2sin\(^{2}\) θ - cos θ + 4 = 0 is not possible.
2. Simplify the expression: \(\frac{sec (270° - θ) sec (90° - θ) - tan (270° - θ) tan (90° + θ)}{cot θ + tan (180° + θ) + tan (90° + θ) + tan (360° - θ) + cos 180°}\)
Solution:
First we will simplify the numerator {sec (270° - θ) sec (90° - θ) - tan (270° - θ) tan (90° + θ)};
= sec (3 ∙ 90° - θ) sec (90° - θ) - tan (3 ∙ 90° - θ) tan (90° + θ)
= - csc θ ∙ csc θ - cot θ (- cot θ)
= - csc\(^{2}\) θ + cot\(^{2}\) θ
= - (csc\(^{2}\) θ - cot\(^{2}\) θ)
= - 1
And, now we will simplify the denominator {cot θ + tan (180°
+ θ) +
tan (90° + θ) + tan (360° - θ) + cos 180°};
= cot θ
+ tan (2 ∙ 90° + θ) + tan (90° + θ) + tan (4 ∙ 90° - θ) + cos (2 ∙ 90° - 0°)
= cot θ + tan θ - cot θ - tan θ - cos 0°
= - cos 0°
= 1
Therefore, the given expression = (-1)/(-1) = 1
3. If tan α = -4/3, find the value of (sin α + cos α).
Solution:
We know that, sec\(^{2}\) α = 1 + tan\(^{2}\) α and tan α = - 4/3
Therefore, sec\(^{2}\) α = 1 + (-4/3)\(^{2}\)
sec\(^{2}\) α = 1 + 16/9
sec\(^{2}\) α = 25/9
Therefore, sec α = ± 5/3
Therefore, cos α = ± 3/5
Again, sin\(^{2}\) α = 1 - cos\(^{2}\) α
sin\(^{2}\) α = 1 - (± 3/5)\(^{2}\); since, cos α = ± 3/5
sin\(^{2}\) α = 1 - (9/25)
sin\(^{2}\) α = 16/25
Therefore, sin α = ± 4/5
Now, tan α is negative; hence, α lies either in the second or in the fourth quadrant.
If α lies in the second quadrant then sin α is positive and cos α is negative.
Hence, we take, sin α = 4/5 and cos α = - 3/5
Therefore, sin α + cos
α = 4/5 - 3/5 = 1/5
Again, if α lies in the fourth quadrant then sin α is negative and cos α is positive.
Hence, we take, sin α = -4/5 and cos α = 3/5
Therefore, sin α + cos
α = - 4/5 + 3/5 = -1/5
Therefore, the required values of (sin α + cos α) = ± 1/5.
● Trigonometric Functions
11 and 12 Grade Math
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