MULTIPLE ANGLES

If sin (A + B) = sinA cosB + sinB cosA

Then,

sin (A + A) = sinA cosA + sinA cosA

Notice that we have replaced the B with A  

sin (A + A) = 2 sinA cosA

          A + A = 2A

sin 2A = 2 sinA cosA

 

also, from

cos (A + B) = cosA cosB – sinA sinB

cos (A + A) = cosA cosA – sinA sinA

cos (A + A) = cos2A – sin2A

cos 2A = cos2A – sin2A ---------------------------------------- (equation J)

          but

sin2A + cos2a = 1

sin2A = 1 – cos2a

            substitute 1 – cos2A for sin2A in (J

cos 2A = cos2A – sin2A

cos 2A = cos2A – (1 – cos2A)

cos 2A = cos2A – 1 + cos2A

cos 2A = cos2A + cos2A – 1

cos2A = 2cos2A – 1

 

we can also make cos2a the subject of sin2A + cos2a = 1 to give cos2a = 1 – sin2A.

 

substitute 1 – sin2A for cos2a in (J)

 

cos 2A = cos2A – sin2A

cos 2A = 1 – sin2A – sin2A

cos 2A = 1 – 2sin2A

 

In summary,

          sin 2A = 2 sinA cosA

          cos 2A = cos2A – sin2A

          cos 2A = 2cos2A – 1

          cos 2A = 1 – 2sin2A

 

Same goes for tan.

      if \ tan (A+B) = \frac{tan A + tan B}{1 - tanA \ tanB} \ then \\ \\ tan (A+A) = \frac{tan A + tan A}{1 - tanA \ tanA} \\ \\ tan (2A) = \frac{2tan A}{1 - tan^2A} \\ \\

the above proofs also applies to half angles 

sin (A/2 + A/2) = 2 sin (A/2) cos (A/2) 

sin A = 2 sin (A/2) cos (A/2) 

             Note that: A/2 + A/2 = A 

cos A = cos2(A/2) – sin2(A/2)          or       cosA = cos2(A/2) – 1          or        cosA = 1 – 2sin2(A/2) 

 

                           Note:   sinA x sinA = sin2A

                                      sin2A x sinA = sin3A

                                      sin2A x sin3A = sin5A

                                      sin4A x sin2A = sin6A

                      Add the powers.

 

    Examples

1.   Find the values of the following  (a) sin 3A    (b) cos 4A  

 

Solution

a.         Sin 3A

= sin (2A + A)

= sin 2A cos A + sin A cos 2A

            Sin 2A = 2 sin A cos A           and      cos 2A = cos2A – sin2A

= (2sin A cos A) cos A + sin A (cos2A – sin2A)

= 2 sinA cos2A + sin A (cos2A – sin2A)

            From sin2A + cos2A = 1,

             cos2A = 1 - sin2A

= 2 sin A (1 - sin2A) + sin A (1 - sin2A – sin2A)

=2 sin A (1 - sin2A) + sin A (1 – 2sin2A)

= 2 sin A – 2 sin3A + sin A – 2 sin3A

= 2 sin A + sin A – 2 sin3A – 2 sin3A

= 3 sin A – 4 sin3A

 

b.         cos 4A

= cos (2A + 2A)

= cos2 2A – sin2 2A

= (cos 2A)2 – (sin 2A)2

= (2 cos2A – 1)2 – (2 sin A cos A) 2

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

= 4cos4A – 4cos2A + 1 – 4 sin2A cos2A

= 4cos4A – 4cos2A + 1 – 4 (1 – cos2A) cos2A

= 4cos4A – 4cos2A + 1 – 4 cos2A (1 – cos2A)

= 4cos4A – 4cos2A + 1 – 4 cos2A + 4 cos4A

= 4cos4A + 4 cos4A – 4cos2A – 4cos2A + 1

= 8 cos4A – 8cos2A + 1

 

2.         Find without using tables the value of sin 2P when sin P = 3/5.

 

            Solution

Sin 2P = 2 sin P cos P

First, We need to get cos p from sin P = 3/5

sin2 P + cos2 P = 1

(3/5) 2 + cos2 P = 1

9/25 + cos2 P = 1

cos2 P = 1 – 9/25

cos2 P = 16/25

cos P = 4/5

sin 2P = 2 sin P cos P

            = 2 x (3/5) x (4/5)

            = 24/25.  

 

3.         Find without using tables the value of cos 2θ when tan θ = 2/3.

           

Solution

cos 2θ = 2cos2θ – 1

          from tan2θ = 2/3, get cos2θ.

          sin2θ + cos2θ = 1

          dividing through by cos2θ

          tan2θ + 1 = 1/cos2θ

          (2/3)2 + 1 = 1/cos2θ

          4/9 + 1 = 1/cos2θ

          13/9 = 1/cos2θ

          9/13 = cos2θ

cos 2θ = 2cos2θ – 1

          = 2 (9/13) – 1

          = (18/13) – 1

          = 5/13.

 

4.       If cos 2α = 9/49, find without using tables the value of

(a) sin α       (b) cos α      (c) tan α

 

Solution

a          cos 2α = 1 – 2sin2α

            9/49 = 1 – 2sin2α

            2sin2α = 1 – 9/49

            2sin2α = 40/49

            Divide through by 2

            sin2α = 20/49

            \sin \alpha = \sqrt{\frac{20}{49}} \ \ \ \ = \ \frac{\sqrt{20}}{7} \ or \ \frac{2\sqrt{5}}{7}

            sinα = √20/7        or        2√5/7

 

b          cos α

            cos 2α = 9/49

            2cos2 α – 1 = 9/49

          2cos2 α = 9/49 + 1

          2cos2 α = 58/49

          cos2 α = 29/49 

         \cos \alpha = \sqrt{\frac{29}{49}} \ \ = \ \frac{\sqrt{29}}{7}

         cos α = 5.385/7

         cos α = 0.769   

 

c          tan α

            the easiest way to get tan α is to divide sin α above by cos α

            i.e.

            tan α = sinα/cosα

            tan α = sinα ÷ cosα 

            \tan \alpha = \frac{\sqrt{20}}{7} \div \frac{\sqrt{29}}{7} \\ \\ \tan \alpha = \frac{\sqrt{20}}{7} \times \frac{7}{\sqrt{29}} \\ \\ \tan \alpha = \frac{\sqrt{20}}{\sqrt{29}} \ rationalize \\ \\ \tan \alpha = \frac{\sqrt{20}}{\sqrt{29}} \times \frac{\sqrt{29}}{\sqrt{29}} \\ \\ \tan \alpha = \frac{\sqrt{580}}{29}

           

When the values of sin α and cos α are not initially known, we have to find the tan direct instead of getting it from sin and cos.

 

cos 2α = 9/49

            from sin2α + cos2α = 1

            divide through by cos2α

            tan2 α + 1 = 1/cos2α

                we will now get cos2α from the original question and substitute it in the equation above.

cos 2α = 9/49

2cos2 α – 1 = 9/49

2cos2 α = 9/49 + 1

2cos2 α = 58/49

cos2 α = 29/49

1/cos2 α = 49/29

          The 49/29 above will be substituted for 1/cos2 α in tan2 α + 1 = 1/cos2α.

tan2 α + 1 = 1/cos2α

tan2 α + 1 = 49/29

tan2 α = (49/29) – 1

tan2 α = (49 – 29)/29

tan2 α = 20/29

tanα = √(20/29)

 

          our answer is the same as the first we got from sin and cos. 

 

                                  Exercise

 

1.       Find the value of the following

(a)      cos 3A        (b)     sin 4A

 

2.       Find without using tables the value of

(i) sin 2θ when 

     (a) sin θ = 3/5  (b) sin θ = √3/2  (c) cos θ = 5/13 (d) tan θ = 2/3

(ii) cos 2θ when

     (a) sin θ = 3/5  (b) sin θ = √3/2  (c) cos θ = 5/13 (d) tan θ = 2/3

(iii) tan 2θ when

      (a) sin θ = 3/5  (b) sin θ = √3/2  (c) cos θ = 5/13 (d) tan θ = 2/3

(iv) sin 3θ when 

     (a) sin θ = 3/5  (b) sin θ = √3/2  (c) cos θ = 5/13 (d) tan θ = 2/3

 

3.       If cos 2θ = 18/25, find without using tables the values of

          (a) tan θ       (b) sin θ       (c) cos θ      (d) tan 2θ     (e) sin 2θ 

 

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