Hello mathematicians, our topic for today on mathematics is multiple angles double angle and half angle formulae, mind you stick around because an ebook will be reveal on how to pass mathematics examination.

I hope you understand everything we expressed in our last class on trigonometric ratio formulae well if you don’t, you should always know that, you are free to ask any question

# Session Objectives

At the end of this session, candidates should be able to;

1) solve math problems using double, triple and half angle formulae.

# Double Angle

**First Expression**

Recall that;

sin (A + B) = sinAcosB + cosAsinB

Replacing B = A

then;

sin (A + A) = sin (2A) = sinAcosA + cosAsinA

sin 2A = 2sinAcosA

# Second Expression

cos (A + B) = cosAcosB – sinAsinB

Replacing B = A

then;

cos (A + A) = cos (2A) = cosAcosA – sinAsinA

Hence,

cos (2A) = cos²A – sin²A

= cos²A – (1 – cos²A)

= cos²A – 1 + cos²A

cos (2A) = 2cos²A – 1

**Or**

cos (2A) = cos²A – sin²A

= (1 – sin²A) – sin²A

= 1 – sin²A – sin²A

= 1 – 2sin²A

Hence,

cos (2A) = cos²A – sin²A = 2cos²A – 1 = 1 – 2sin²A

# Third Expression

Recall that,

then,

# Triple Angle Formulae

**Fourth Expression**

sin3A = sin(2A + A)

= sin2AcosA + cos2AsinA

= 2sinAcosAcosA + (1 –2sin²A)sinA

=2sinAcos²A + (1 –2sin²A)sinA

= 2sinA –2sin³A + sinA –2sin³A

= 3sinA –4sin³A

Hence,

sin 3A = 3sinA – 4sin³A

# Fifth Expression

cos 3A = cos (2 A + A)

= cos2AcosA –sin2AsinA

= (2cos²A – 1)cosA – 2sinAcosAsinA

= (2cos²A – 1)cosA – 2cosAsin²A

= (2cos²A – 1)cosA – 2cosA(1 – cos²A)

= 2cos³A – cosA – 2cosA + 2cos³A

= 4cos³A – 3cosA

Hence,

cos3A = 4cos³A – 3cosA

# Sixth Expression

tan 3A = tan (2A + A)

# Half Angle Formulae

Sin A = sin(½A + ½A)

= sin(½Acos½A + cos½Asin½A

Sin A = 2sin½Acos½A

Cos A = cos(½A + ½A)

= cos² ½A – sin² ½A

= 2cos² ½A – 1

= 1 – 2sin² ½A

Tan A = tan(½A + ½A)

## Second Expression

# Example 1

**Brief**Acute Angles are angles less than 90° ( θ < 90)

## Solution

Hyp² = Opp² + Adj²

= 12² + 5²

= 144 + 25

Hyp² = 169

Hyp = √169

= 13

Hence,

Hypotenus = 13

**Note :** whenever you see 12 and 5 in a triangle, the third side is always 13 irrespective of their placement and whenever you see 4 and 3 in a triangle, the third side is always 5 irrespective of their placement.

Hence,

**That is a brief explanation on SOH, CAH, TOA**

Back to the question,

# Example 1

**Brief**Acute Angles are angles less than 90° ( θ < 90)

## Solution

Hyp² = Opp² + Adj²

= 12² + 5²

= 144 + 25

Hyp² = 169

Hyp = √169

= 13

Hence,

Hypotenus = 13

**Note :** whenever you see 12 and 5 in a triangle, the third side is always 13 irrespective of their placement and whenever you see 4 and 3 in a triangle, the third side is always 5 irrespective of their placement.

Hence,

**That is a brief explanation on SOH, CAH, TOA**

Back to the question,

# Example 2

Prove that tan 22.5° = √2 − 1

## Solution

First use a calculator to solve for the above expression, you will found out that,

tan 22.5° = √2 − 1 = 0.414

Let x = 22.5°

Hence,

1 − t² = 2t

Equate the expression to zero

2t + t² − 1 = 0

Rewrite to form a quadratic equation

t² + 2t − 1 = 0

Using

**formula method**

If ax² + bx + c = 0

then x = − b ± √b² − 4ac

**Brief**

**Sum of roots is calculated by**: = − b

a = 1, b = 2 and c = − 1

Hence,

t = − 2 ± √2² − 4(1 × − 1)

Recall that,

22.5° is less (<) than 90° hence 22.5° is in first quadrant, hence it is positive.

# Quadrant

There are four quadrant in mathematics, the 1st, 2nd, 3rd and the 4th quadrant.

## 1st Quadrant

The 1st quadrant is represented by **A**, meaning all angles that are less than or equal to 90 is or are positive,

That is, < 90 or = 90 is positive.

### Example

Find sin 32°, cos 88°, and tan 66 using calculator or mathematical table into 3 decimal places.

#### Solution

sin 32° = 0.530

cos 88° = 0.035

tan 66° = 2.246

you can see that they are all positive.

Also,

cos 90° = 0

sin 90° = 1

tan 90° = math error.

# Conclusion

: All angles less than or equal to 90° are in 1st quadrant and are all positive.

## 2nd Quadrant

2nd quadrant is represented by **S** meaning only Sin angles that are not more than 180° is or are **positive.**

That is, Only sin angles that are (less than) < or = 180° are positive, other angles found here are negative

### Example

Find Sin 140, Sin 113, Cos 140, Cos 113, Tan 140 and Tan 113 using calculator or mathematical table into 3 decimal places.

#### Solution

sin 140° = 0.643

sin 113° = 0.921

cos 140° = − 0.766

cos 113° = − 0.391

tan 140° = − 0.839

tan 113° = − 2.356

# Conclusion

Only Sin angles that are not more than 180° are positive in 2nd quadrant.

## 3rd Quadrant

3rd quadrant is represented by **T** meaning **only tan** angles that are not more than 270° is or are positive.

That is, only tan angles that are (less than) < or = 270° are positive, other angles found here will be negative.

### Example

Find tan 266°, tan 181°, sin 266°, sin 181°, cos 266° and cos 181° using calculator or mathematical table into 3 decimal places.

#### Solution

tan 266° = 14.301

tan 181° = 0.017

sin 266° = − 0.998

sin 181° = − 0.017

cos 266° = − 0.070

cos 181° = − 0.100

# Conclusion

**Only tan angles** that is less than or equal to 270° is or are **positive** in the 3rd quadrant

## 4th Quadrant

The 4th quadrant is represented by **C** meaning **only cos angles** is or are **positive**.

That is, only cos angle that are less than or equal to 360° is or are positeve other angles will be negative.

### Example

Find cos 288°, cos 354°, sin 288°, sin 354°, tan 288°,and tan 354° with calculator or mathematical table into 3 decimal places.

#### Solution

cos 288° = 0.309

cos 354° = 0.995

sin 288° = − 0.951

sin 354° = − 0.105

tan 288° = − 3.078

tan 354° = − 0.105

By now you should be able to know whether an angle will be positive or negative without using mathematical table or calculator.

I hope you still remember that we are evaluating an expression before we branched into quadrant.

So since, 22.5° is in the first quadrant, then tan 22.5° is positive,

tan 22.5° = −1 + √2

= √2 − 1

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