**Example 1: Counter Example**

Provide counter–example to show that sin(A +B)=sin A + sin B is not an identity.

**Solution 1: Counter Example**

For identities, the equation is true for every real number in the domain of the functions. Counter example to show that the equation is not true for even one value is an effective way to answer at times.

To show that, sin(A+B)=sin A+sin B is not an Identity let’s take A =30° and B =60°

RHS=sin30°+sin60°= \(\frac{1}{2} +\frac{\sqrt{3}}{2}\) = \(\frac{1 +\sqrt{3}}{2}\)

LHS =sin(30°+ 60°)=sin90°=1

Since for A =30° and B =60° the equation sin(A + B)≠sin A + sin B the given equation is not an identity.

**Example 2: Identify Mistake**

Steps are shown below to prove that sin θ =\(\sqrt{1 -cos^2 θ }\) is an identity. Find the mistake and provide a counter-example to show that the trigonometric equation is not an identity.

sin^{2}θ+ cos^{2}θ=1 Pythagorean Identity

sin^{2} θ = 1 – cos^{2}θ Rearrange terms

sin θ = \(\sqrt{1 -cos^2θ }\) Take square–root on both sides

LHS= RHS so sin θ = \(\sqrt{1 -cos^2θ }\) is an identity

**Example 2: Identify Mistake**

Equation sinθ = \(\sqrt{1 -cos^2θ }\) is not an identity since the right side is always positive but the left side (value of ) can be negative also.

Mistake is in the third step. Whenever square-root is done we need to take both positive and negative signs. The third step should have been: sin =±\(\sqrt{1 -cos^2 θ }\)

**Counter-example**: θ =240°,LHS =sin 240°=−0.5, RHS=\(\sqrt{1 - cos ^2240° }\) = 0.5

LHS≠RHS, the equation is not an identity.

**Explore**

Explain the missing steps in the following proof of an Identity.

\(\frac{sin A + cos A}{sin A - cos A} + \frac{sin A - cos A}{sin A + cos A }\) = \(\frac{2}{sin^2 A - cos^2 A}\) = \(\frac{2}{2 sin ^2 A - 1}\) = \(\frac{2}{1 - 2 cos^2 A}\)