COMPOUND ANGLES

Below are the list of compound angles formula needed for this topic

1.       sin (A+B) = sin A cos B + sin B cos A

2.       sin (A – B) = sin A cos B – sin B cos A

3.       cos(A+B) = cos A cos B – sin A sin B

4.       cos(A – B) = cos A cos B + sin A sin B

5.       tan (A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}

6.       tan (A – B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}

Master them, as we will use them to solve the questions below.

 

                           Example

If cos P = 3/5 and sin Q = 5/13, evaluate

 

1 . sin (P+Q)                         2 . cos (P – Q)                      3 .   tan (P – Q)

                         Solution

We need to get sin P and cos Q first.

** we will get sin P from cos P = 3/5 using

sin2P + cos2P = 1

sin2P + (3/5)2 = 1

sin2P + 9/25 = 1

sin2P = 1 – 9/25

sin2P = 16/25

sin P = √(16/25)

sinP = 4/5

** we also get cos Q from sin Q

sin2Q + cos2Q = 1

(5/13)2 + cos2Q = 1

25/169 + cos2Q = 1

cos2Q = 1 – 25/169

cos2Q = 144/169

cosQ = √(144/169)

cos Q = 12/13

we can now proceed to evaluate

 

1 .  sin (P+Q) = sin P cos Q + sin Q cos P

= 4/5 x 12/13 + 5/13 x 3/5

= 48/65 + 15/65

= 63/65

 

2 .   cos (P – Q) = cos P cos Q + sin P sin Q

= 3/5 x 12/13 + 4/5 x 5/13

= 36/65 + 20/65

= 56/65

 

3 .  tan (P – Q) =

** Since the value of either tan P or tan Q is not given, we calculate it using

Tan P = sin P/ cos P                        or                    tan2P + 1 = sec2P

tan P = (4/5) ÷ (3/5)                                              tan2P + 1 = 1/cos2P

tan P = (4/5) x (5/3)                                               tan2P + 1 = \frac{1}{9/25}

5 cancels 5                                                               tan2P + 1 = 25/9

tan P = 4/3                                                               tan2P = \frac{25}{9} – 1

tan2P = \frac{25 - 9}{9}

tan2P = 16/9

tan P = √(16/9)

tan P = 4/3

** for tan Q, we use

Tan Q = sin Q / cos Q                                 or        1 + cot2Q = cosec2Q

tan Q = (5/13) ÷ (12/13)                                        1 + \frac{1}{\tan^2 Q} = \frac{1}{\sin^2 Q}

tan Q = (5/13) x (13/12)                                         1 + \frac{1}{\tan^2 Q} = \frac{1}{25/169}

tan Q = 5/12                                                           1 + \frac{1}{\tan^2 Q} = \frac{169}{25}

\frac{1}{\tan^2 Q} = \frac{169}{25} - 1

\frac{1}{\tan^2 Q} = \frac{169}{25} - \frac{1}{1}

\frac{1}{\tan^2 Q} = \frac{169 - 25}{25} \\ \\ \frac{1}{\tan^2 Q} = \frac{144}{25} \\ \\ \tan^2 Q = \frac{25}{144} \\ \\ \tan Q = \frac{5}{12}

now that we have our tan P and tan Q, we can now evaluate

tan (P - Q) = \frac{\tan P - \tan Q}{1 + \tan P \tan Q} \\ \\

= \frac{\frac{4}{3} - \frac{5}{12}}{1 + \frac{4}{3} \times \frac{5}{12}} \\ \\ \\ = \frac{\frac{16 - 5}{12}}{1 + \frac{1}{3} \times \frac{5}{3}} \\ \\ \\ = \frac{\frac{16 - 5}{12}}{1 + \frac{1}{3} \times \frac{5}{3}} \\ \\ \\ = \frac{\frac{11}{12}}{1 + \frac{5}{9}} \\ \\ \\

= \frac{\frac{11}{12}}{\frac{14}{9}} \\ \\ = \frac{11}{12} \div \frac{14}{9} \\ \\ = \frac{11}{12} \times \frac{9}{14} \\ \\ = \frac{11}{12} \times \frac{9}{14} \\ \\ = \frac{11}{4} \times \frac{3}{14} \\ \\ = \frac{33}{56}

 

Exercise

1.     If sinα = 7/25 and cos β = 35/37, evaluate the following

a.     sin (α – β)        b.     cos (α + β)           c.  tan (α + β)

2.    Given that cos θ = 8/17 and sin α = 9/41, evaluate

a.    sin (α + θ)         b.     cos (α – θ)             c.    tan (α – θ)

3.    if sin x = 0.8 and sin y = 0.3, find the value of

a.  cos (x – y)           b.   sin (x – y)            c.   tan (x + y)

 

Example 2

Without using tables evaluate the following

1.    cos 75°               2.   Sin 105°              3.   Tan 15°

 

Solution

Without using tables means we should use our special angle values.

I hope you have mastered how to draw the triangles (30° by 60°) and 45°?

 

1.      cos 75°

75° is not a special angle, so we split it to form two special angles of 45° and 30°.

= cos (45° + 30°)

= cos 45° cos 30° – sin 45° sin 30°

= \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} \times \frac{1}{2} \\ \\ = \frac{\sqrt{3}}{2 \sqrt{2}} - \frac{1}{2 \sqrt{2}} \\ \\ = \frac{\sqrt{3} - 1}{2 \sqrt{2}} \\ \\ rationalize \\ \\ = \frac{\sqrt{3} - 1}{2 \sqrt{2}} \times \frac{2 \sqrt{2}}{2 \sqrt{2}} \\ \\ = \frac{2 \sqrt{6} - 2 \sqrt{2}}{4 \times 2}

= \frac {2 (\sqrt{6} - \sqrt{2})}{8} \\ \\ = \frac {\sqrt{6} - \sqrt{2}}{4}

 

2.   Sin 105°

= sin (60 + 45)

= sin 60 cos 45 + sin 45 cos 60

= √3/2 x 1/√2 + 1/√2 x ½

= √3/2√2 + 1/2√2

= (√3 + 1)/2√2

rationalize

= \frac{\sqrt{3} + 1}{2\sqrt{2}} \times \frac{2\sqrt{2}}{2\sqrt{2}} \\ \\ = \frac{2 \sqrt{6} + 2\sqrt{2}}{8} \\ \\ \\ = \frac{\sqrt{6} + \sqrt{2}}{4}

 

3 . tan 15°

= tan (45 – 30)

\small = \frac{\tan 45 - \tan 30}{1 + \tan 45 \tan 30} \\ = \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \times \frac{1}{\sqrt{3}}} \\ = \frac{1 - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}} \\ = \frac{\frac{\sqrt{3} - 1}{\sqrt{3}}}{\frac{\sqrt{3} + 1}{\sqrt{3}}} \\ = \frac{\sqrt{3} - 1}{\sqrt{3}} \div \frac{\sqrt{3} + 1}{\sqrt{3}} \\ = \frac{\sqrt{3} - 1}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3} + 1} \\

\small = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \\ = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \times \frac{\sqrt{3} - 1}{\sqrt{3} - 1} \\ = \frac{3 - \sqrt{3} - \sqrt{3} + 1}{3 - 1} \\ = \frac{4 - 2\sqrt{3}}{2} \\ = \frac{2(2 - \sqrt{3})}{2} \\ = 2 -\sqrt{3}

 

                                    Exercise

1 .     Evaluate the following without using tables

a .  sin 15°   b . cos 105°    c . tan 75°   d .  cos 195°

2 .    If tan A = 2/3 and tan B = ¼, find the values of

a . tan (A – B)      b .   sin (A + B)              c . cos (A – B)

 

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