## Mathematics tutors

**Our IB Maths Tutors say that Trigonometry is a very important and interesting branch of Maths**

Trigonometry is one of the branches of Mathematics that fascinates Mathematicians. It tells about the relationships between the angles and sides of a triangle. Word trigonometry was started from the Greek word, where,

‘**TRI**‘ = *Three*

**‘GON**‘= sides and the **‘METRON’ =** to **measure. **It’s an ancient and probably most widely applicable branch in Mathematics.

### Parts of Trigonometry-

For basic learning, I am dividing trigonometry in two-part:-

1 Trigonometry based on right triangles

1 Trigonometry based on non-right triangles.

In this post, I will only discuss Trigonometry based on right triangles.

In a right triangle, there are three sides hypotenuse (the longest side), adjacent side(base) and the opposite side(perpendicular).

### Pythagoras theorem

If we know any two sides of the right triangle, we can find the third side by using Pythagoras theorem:-

##### ** P²+B²=H²**

In a right triangle, the ratios of sides can be written in many ways like P/H, B/H, P/B, H/P, H/B, B/P. each ratio is called a trigonometric ratio and these are represented by Latin words Sin, Cos, Tan, Cosec, Sec and Cot.

In a right triangle

**Sin =P/H, Cos =B/H, Tan= P/B, Cosec= H/P, Sec =H/B, Cot= B/P,**

Mathematics tutors use these three concepts to solve trigonometric problems

Concept-1

**In a trigonometric problem, If we are given a trigonometric ratio or asked about it, we will first find all three sides of the triangle**

#### Using the above concept and above formulas, we can solve the first type of trigonometric problems

#### Sample Problems

**Example-1 If sin θ = 8/17, find other trigonometric ratios of θ.**

Ans-

Let us draw a ∆ OMP in which ∠M = 90°

Then sin θ = MP/OP = 8/17.

Let MP = 8k and OP = 17k, where k is positive.

By Pythagoras’ theorem, we get

OP^{2} = OM^{2} + MP^{2}

⇒ OM^{2} = OP^{2} – MP^{2}

⇒ OM^{2} = [(17k)^{2} – (8k)^{2}]

⇒ OM^{2} = [289k^{2} – 64k^{2}]

⇒ OM^{2} = 225k^{2}

⇒ OM = √(225k^{2})

⇒ OM = 15k

Therefore, sin θ = MP/OP = 8k/17k = 8/17

**cos θ = OM/OP = 15k/17k = 15/17**

**tan θ = Sin θ/Cos θ = (8/17 × 17/15) = 8/15**

**csc θ = 1/sin θ = 17/8**

**sec θ = 1/cos θ = 17/15 and**

**cot θ = 1/tan θ = 15/8.**

**Example-2 If cos θ = 3/5, find the value of (5cosec θ – 4 tan θ)/(sec θ + cot θ)**

**Answer-**

Let us draw an ∆ ABC in which ∠B = 90°.

Let ∠A = θ°

Then cos θ = AB/AC = 3/5.

Let AB = 3k and AC = 5k, where k is positive.

By Pythagoras’ theorem, we get

AC^{2} = AB^{2} + BC^{2}

⇒ BC^{2} = AC^{2} – AB^{2}

⇒ BC^{2} = [(5k)^{2} – (3k)^{2}]

⇒ BC^{2} = [25k^{2} – 9k^{2}]

⇒ BC^{2} = 16k^{2}

⇒ BC = √(16k^{2})

⇒ BC = 4k

Therefore, sec θ = 1/cos θ = 5/3

tan θ = BC/AB =4k/3k = 4/3

cot θ = 1/tan θ = 3/4 and

csc θ = AC/BC = 5k/4k = 5/4

Now **(5cosec θ -4 tan θ)/(sec θ + cot θ)**

= (5 × 5/4 – 4 × 4/3)/(5/3 + 3/4)

= (25/4 -16/3)/(5/3 +3/4)

= 11/12 × 12/29

= 11/29

In next post of Mathematics tutors series, we will discuss the remaining two concepts of non-right triangle trigonometry

To get a better understanding of the concept you should check all my posts of Mathematics tutors series. you can check them by clicking on the links given below