# Multiple Angles, Double Angle And Half Angle Formulae

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,

tan (A + B) = tanA + tanB
1 – tanAtanB
Replacing B = A

then,

tan (A + A) = tanA + tanA
1 – tanAtanA
= 2tanA
1 – tan²A

# 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)

= tan 2A + tan A
1 – tan 2AtanA

2tanA + tan A
1 – tan²A
1 – 2 tan A . tanA
1 – tan²A
2 tan A + tan A(1 – tan²A)
1 – tan²A
1 – tan²A – 2tan²A
1 – tan²A
= 2tan A + tan A – tan³A
1 – 3 tan²A
3 tanA – tan³A
1 – 3tan²A
Hence,

tan 3A = 3 tanA – tan³A
1 – 3tan²A

# 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)

= 2 tan½A
1 – tan² ½A
Let t = tan½A

Sin A = 2sin½Acos½A
1
= 2sin½Acos½A
sin² ½A + cos² ½A
Divide the numerator and denominator by cos² ½A

sin A = 2sin½Acos½A
cos²½A
sin² ½A + cos² ½A
cos² ½A
= 2sin½A
cos½A
1 + sin² ½A
cos² ½A
= 2tan ½A
1 + tan² ½A
= 2t
1 + t²

## Second Expression

cos A = cos(½A + ½A)
cos² ½A – sin² ½A
1
cos² ½A – sin² ½A
cos² ½A + sin² ½A
Divide the numerator and denominator by cos² ½A

cos A = cos² ½A – sin² ½A
cos² ½A
cos² ½A + sin² ½A
cos² ½A
1 – tan² ½A
1 + tan² ½A
= 1 – t²
1 + t²
cos A = 1 – t²
1 + t²

# Example 1

if cot θ = 12
5
, where θ is an acute angle
Brief Acute Angles are angles less than 90° ( θ < 90)

Evaluate cos 2θ
sin θ + cos θ

## Solution

In mathematics cot θ = 1
tan θ

And tan θ = Opposite
Hence,

Opposite
= 5
12
If Adjacent (adj)= 5 and Opposite (opp) = 12, then we can use Pythagoras theorem to find hypotenus (hyp);

= 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,

And sin θ = Opposite
Hypotenus
= 5
13
Hypotenus
= 12
13
That is a brief explanation on SOH, CAH, TOA

Back to the question,

Evaluate cos 2θ
sin θ + cos θ
Recall that in double Angle Formulae

cos 2θ = cos²θ − sin²θ
= 144 − 25
169
199
169
cos 2θ = =
sin θ + cos θ
119   169
5 + 12
13
199
169
17
13
=199 × 13
169 17
= 7
13

# Example 1

if cot θ = 12
5
, where θ is an acute angle
Brief Acute Angles are angles less than 90° ( θ < 90)

Evaluate cos 2θ
sin θ + cos θ

## Solution

In mathematics cot θ = 1
tan θ

And tan θ = Opposite
Hence,

Opposite
= 5
12
If Adjacent (adj)= 5 and Opposite (opp) = 12, then we can use Pythagoras theorem to find hypotenus (hyp);

= 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,

And sin θ = Opposite
Hypotenus
= 5
13
Hypotenus
= 12
13
That is a brief explanation on SOH, CAH, TOA

Back to the question,

Evaluate cos 2θ
sin θ + cos θ
Recall that in double Angle Formulae

cos 2θ = cos²θ − sin²θ
= 144 − 25
169
199
169
cos 2θ
sin θ + cos θ
119169
5 + 12
13
199
169
17
13
=199 × 13
169 17
= 7
13

# 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°

tan 45 = 2 tan 22.5°
1 − tan² 22.5°
= 2t
1 − t²
But tan 45° = 1
Hence,

1 = 2t
1 − t²

Cross multiply
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
2a
Brief

Sum of roots is calculated by : = − b
a

And Product of roots is calculated by = c
a
Back to the question,

t² + 2t − 1 = 0
a = 1, b = 2 and c = − 1
Hence,
t = − 2 ± √2² − 4(1 × − 1)
2(1)
t = −2 ± √4 + 4
2
= −2 ± √8
2
= −2 ± √4 × 2
2
= −2 ± 2√2
2
= 2( −1 ± √2)
2
= −1 ± √2
Hence,

tan 22.5° = −1 ± √2

Recall that,

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

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

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

Next Topic : Product Formulae

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