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




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.



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.




Share this page: What’s this?

Recent Articles

  1. Formation of Square and Rectangle | Construction of Square & Rectangle

    Jul 16, 25 02:45 AM

    Construction of a Square
    In formation of square and rectangle we will learn how to construct square and rectangle. Construction of a Square: We follow the method given below. Step I: We draw a line segment AB of the required…

    Read More

  2. Perimeter of a Figure | Perimeter of a Simple Closed Figure | Examples

    Jul 16, 25 02:33 AM

    Perimeter of a Figure
    Perimeter of a figure is explained here. Perimeter is the total length of the boundary of a closed figure. The perimeter of a simple closed figure is the sum of the measures of line-segments which hav…

    Read More

  3. Formation of Numbers | Smallest and Greatest Number| Number Formation

    Jul 15, 25 11:46 AM

    In formation of numbers we will learn the numbers having different numbers of digits. We know that: (i) Greatest number of one digit = 9,

    Read More

  4. 5th Grade Quadrilaterals | Square | Rectangle | Parallelogram |Rhombus

    Jul 15, 25 02:01 AM

    Square
    Quadrilaterals are known as four sided polygon.What is a quadrilateral? A closed figure made of our line segments is called a quadrilateral. For example:

    Read More

  5. 5th Grade Geometry Practice Test | Angle | Triangle | Circle |Free Ans

    Jul 14, 25 01:53 AM

    Name the Angles
    In 5th grade geometry practice test you will get different types of practice questions on lines, types of angle, triangles, properties of triangles, classification of triangles, construction of triang…

    Read More