Trigonometric Functions of any Angles

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

• Basic Trigonometric Ratios and Their Names
• Restrictions of Trigonometrical Ratios
• Reciprocal Relations of Trigonometric Ratios
• Quotient Relations of Trigonometric Ratios
  
• Limit of Trigonometric Ratios
  
• Trigonometrical Identity
• Problems on Trigonometric Identities
• Elimination of Trigonometric Ratios
• Eliminate Theta between the equations
• Problems on Eliminate Theta
• Trig Ratio Problems
• Proving Trigonometric Ratios
• Trig Ratios Proving Problems
• Verify Trigonometric Identities
• Trigonometrical Ratios of 0°
• Trigonometrical Ratios of 30°
• Trigonometrical Ratios of 45°
• Trigonometrical Ratios of 60°
• Trigonometrical Ratios of 90°
• Trigonometrical Ratios Table
• Problems on Trigonometric Ratio of Standard Angle
  
• Trigonometrical Ratios of Complementary Angles
• Rules of Trigonometric Signs
• Signs of Trigonometrical Ratios
  
• All Sin Tan Cos Rule
• Trigonometrical Ratios of (- θ)
• Trigonometrical Ratios of (90° + θ)
• Trigonometrical Ratios of (90° - θ)
• Trigonometrical Ratios of (180° + θ)
• Trigonometrical Ratios of (180° - θ)
• Trigonometrical Ratios of (270° + θ)
• Trigonometrical Ratios of (270° - θ)
• Trigonometrical Ratios of (360° + θ)
• Trigonometrical Ratios of (360° - θ)
• Trigonometrical Ratios of any Angle
• Trigonometrical Ratios of some Particular Angles
• Trigonometric Ratios of an Angle
• Trigonometric Functions of any Angles
• Problems on Trigonometric Ratios of an Angle
• Problems on Signs of Trigonometrical Ratios

11 and 12 Grade Math

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