1 Trigonometric identities

An identity is a relation which is always true. To emphasise this the symbol ‘ ≡ ’ is often used rather than ‘ = ’. For example, ( x + 1 ) 2 ≡ x 2 + 2 x + 1 (always true) but ( x + 1 ) 2 = 0 (only true for x = − 1 ).

Task!
  1. Using the exact values, evaluate sin 2 θ + cos 2 θ for (i) θ = 3 0 ∘ (ii) θ = 4 5 ∘

    [Note that sin 2 θ means ( sin θ ) 2 , cos 2 θ means ( cos θ ) 2 ]

  2. Choose a non-integer value for θ and use a calculator to evaluate sin 2 θ + cos 2 θ .
  1. (i)   sin 2 3 0 ∘ + cos 2 3 0 ∘ = 1 2 2 + 3 2 2 = 1 4 + 3 4 = 1

      (ii)   sin 2 4 5 ∘ + cos 2 4 5 ∘ = 1 2 2 + 1 2 2 = 1 2 + 1 2 = 1

  2. The answer should be 1 whatever value you choose.
Key Point 12

For any value of θ

sin 2 θ + cos 2 θ ≡ 1 ( 5 )

One way of proving the result in Key Point 12 is to use the definitions of sin θ and cos θ obtained from the circle of unit radius. Refer back to Figure 22 on page 23.

Recall that   cos θ = O Q , sin θ = O R = Q P . By Pythagoras’ theorem

( O Q ) 2 + ( Q P ) 2 = ( O P ) 2 = 1

hence cos 2 θ + sin 2 θ = 1.

We have demonstrated the result (5) using an angle θ in the first quadrant but the result is true for any θ i.e. it is indeed an identity.

Task!

By dividing the identity sin 2 θ + cos 2 θ ≡ 1 by

  1. sin 2 θ
  2. cos 2 θ obtain two further identities.

    [Hint: Recall the definitions of c o s e c θ , s e c θ , cot θ .]

  1. sin 2 θ sin 2 θ + cos 2 θ sin 2 θ = 1 sin 2 θ
  2. sin 2 θ cos 2 θ + cos 2 θ cos 2 θ = 1 cos 2 θ

    1 + c o t 2 θ ≡ c o s e c 2 θ tan 2 θ + 1 ≡ s e c 2 θ

Key Point 13 introduces two further important identities.

Key Point 13
sin ( A + B ) ≡ sin A cos B + cos A sin B ( 6 )
cos ( A + B ) ≡ cos A cos B − sin A sin B ( 7 )

Note carefully the addition sign in (6) but the subtraction sign in (7).

Further identities can readily be obtained from (6) and (7).

Dividing (6) by (7) we obtain

tan ( A + B ) ≡ sin ( A + B ) cos ( A + B ) ≡ sin A cos B + cos A sin B cos A cos B − sin A sin B

Dividing every term by cos A cos B we obtain

tan ( A + B ) ≡ tan A + tan B 1 − tan A tan B

Replacing B by − B in (6) and (7) and remembering that cos ( − B ) ≡ cos B , sin ( − B ) ≡ − sin B we find

sin ( A − B ) ≡ sin A cos B − cos A sin B

cos ( A − B ) ≡ cos A cos B + sin A sin B

Task!

Using the identities sin ( A − B ) ≡ sin A cos B − cos A sin B and

cos ( A − B ) ≡ cos A cos B + sin A sin B obtain an expansion for tan ( A − B ) :

tan ( A − B ) ≡ sin A cos B − cos A sin B cos A cos B + sin A sin B .

Dividing every term by cos A cos B gives

tan ( A − B ) ≡ tan A − tan B 1 + tan A tan B

The following identities are derived from those in Key Point 13.

Key Point 14

tan ( A + B ) ≡ tan A + tan B 1 − tan A tan B ( 8 )

sin ( A − B ) ≡ sin A cos B − cos A sin B ( 9 )

cos ( A − B ) ≡ cos A cos B + sin A sin B ( 10 )

tan ( A − B ) ≡ tan A − tan B 1 + tan A tan B ( 11 )