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






11 and 12 Grade Math

From Trigonometric Functions of any Angles to HOME PAGE


New! Comments

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.



Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.



Share this page: What’s this?

Recent Articles

  1. Fraction in Lowest Terms |Reducing Fractions|Fraction in Simplest Form

    Feb 28, 24 04:07 PM

    Fraction 8/16
    There are two methods to reduce a given fraction to its simplest form, viz., H.C.F. Method and Prime Factorization Method. If numerator and denominator of a fraction have no common factor other than 1…

    Read More

  2. Equivalent Fractions | Fractions |Reduced to the Lowest Term |Examples

    Feb 28, 24 01:43 PM

    Equivalent Fractions
    The fractions having the same value are called equivalent fractions. Their numerator and denominator can be different but, they represent the same part of a whole. We can see the shade portion with re…

    Read More

  3. Fraction as a Part of Collection | Pictures of Fraction | Fractional

    Feb 27, 24 02:43 PM

    Pictures of Fraction
    How to find fraction as a part of collection? Let there be 14 rectangles forming a box or rectangle. Thus, it can be said that there is a collection of 14 rectangles, 2 rectangles in each row. If it i…

    Read More

  4. Fraction of a Whole Numbers | Fractional Number |Examples with Picture

    Feb 24, 24 04:11 PM

    A Collection of Apples
    Fraction of a whole numbers are explained here with 4 following examples. There are three shapes: (a) circle-shape (b) rectangle-shape and (c) square-shape. Each one is divided into 4 equal parts. One…

    Read More

  5. Identification of the Parts of a Fraction | Fractional Numbers | Parts

    Feb 24, 24 04:10 PM

    Fractional Parts
    We will discuss here about the identification of the parts of a fraction. We know fraction means part of something. Fraction tells us, into how many parts a whole has been

    Read More