Conditional Identities

A trigonometric relation says that two expressions made of angles are equal. You can test one by putting real angles into it and working out each side on its own. When the two sides come out equal every time, the relation is probably true for all angles. If they come out different even once, it is not.

Here are five relations to test. Put \( \theta \), A and B equal to 0°, 30°, 45°, 60° and 90° in turn:

• (a) \( \sin 2\theta = 2\sin\theta\cos\theta \)
• (b) \( \cos 3\theta = 4\cos^3\theta - 3\cos\theta \)
• (c) \( \sin(A + B) = \sin A\cos B + \cos A\sin B \)
• (d) \( \sin A + \cos B = 1 \)
• (e) \( \tan A = \cot B \)

Take relation (a) with \( \theta = 30^\circ \). The left side is \( \sin 60^\circ = \frac{\sqrt{3}}{2} \). The right side is \( 2 \times \frac{1}{2} \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2} \). They match. Try any other angle from the list and they match again. Relations (b) and (c) behave the same way: whatever angles you choose, both sides agree.

Relations (d) and (e) are different. With A = 0° and B = 0°, relation (d) gives \( 0 + 1 = 1 \), which is true. With A = 30° and B = 30°, it gives \( \frac{1}{2} + \frac{\sqrt{3}}{2} \), which is about 1.37 and not 1. Relation (e) is true for A = 30° and B = 60°, but false for A = 45° and B = 60°. The table shows these trials side by side.

For relation (b) at \( \theta = 60^\circ \), the left side is \( \cos 180^\circ = -1 \) and the right side is \( 4 \times \frac{1}{8} - 3 \times \frac{1}{2} = \frac{1}{2} - \frac{3}{2} = -1 \).

The trials split the relations into two kinds, and each kind has a name.

Look back at relation (e). It worked for A = 30° and B = 60°, and \( 30^\circ + 60^\circ = 90^\circ \). It failed for A = 45° and B = 60°, and \( 45^\circ + 60^\circ = 105^\circ \). The condition \( A + B = 90^\circ \) is exactly what decides whether it works. Relation (d), on the other hand, is simply not an identity.

The conditions you meet most often are about sums of angles. The two main ones are \( A + B = 90^\circ \), which is true for the two acute angles of a right-angled triangle, and \( A + B + C = 180^\circ \), which is true for the three angles of any triangle.

A condition works because it lets you write one angle in terms of the others. Then the rules for complementary and allied angles do the rest.

If \( A + B = 90^\circ \), then \( B = 90^\circ - A \). A and B are complementary angles, so:

\[ \cos B = \cos(90^\circ - A) = \sin A \]

\[ \cot B = \cot(90^\circ - A) = \tan A \]

If \( A + B + C = 180^\circ \), then \( A + B = 180^\circ - C \). The sine of \( 180^\circ - C \) equals \( \sin C \), while its cosine and tangent change sign:

\[ \sin(A + B) = \sin(180^\circ - C) = \sin C \]

\[ \cos(A + B) = \cos(180^\circ - C) = -\cos C \]

Halving the same condition gives \( \frac{A + B}{2} = 90^\circ - \frac{C}{2} \), so the half angles are complementary. The table collects what each condition gives.

Testing, or verifying, a relation for given angles always follows the same order. Work each side separately and never move terms across the equals sign.

• Step 1: Write down the given angles.
• Step 2: If the relation is a conditional identity, check that the angles meet the condition, for example by adding the three angles of a triangle.
• Step 3: Work out the left-hand side (L.H.S.), replacing each trigonometric ratio by its value.
• Step 4: Work out the right-hand side (R.H.S.) in the same way, on its own.
• Step 5: Compare the two results. If they are equal, write L.H.S. = R.H.S. and the relation is verified for those angles.

You need the values of the ratios at the standard angles, including the obtuse ones that appear in triangles. An angle between 90° and 180° has a positive sine but a negative cosine and a negative tangent.

Test \( \sin(A + B) = \sin A\cos B + \cos A\sin B \) when A = 30° and B = 60°.

Step 1: Adding the angles on the left-hand side and finding the sine:

\[ \text{L.H.S.} = \sin(30^\circ + 60^\circ) = \sin 90^\circ = 1 \]

Step 2: Putting the angles into the right-hand side:

\[ \text{R.H.S.} = \sin 30^\circ\cos 60^\circ + \cos 30^\circ\sin 60^\circ \]

Step 3: Replacing each ratio by its value from the table:

\[ = \frac{1}{2} \times \frac{1}{2} + \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} \]

Step 4: Multiplying each pair and adding the results:

\[ = \frac{1}{4} + \frac{3}{4} = 1 \]

Hence L.H.S. = R.H.S. This relation is an identity, so no condition had to be checked first. The test only confirms it for these two angles.

Test \( \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A\tan B} \) when A = 30° and B = 30°.

Step 1: Working out the left-hand side:

\[ \text{L.H.S.} = \tan(30^\circ + 30^\circ) = \tan 60^\circ = \sqrt{3} \]

Step 2: Putting \( \tan 30^\circ = \frac{1}{\sqrt{3}} \) into the right-hand side:

\[ \text{R.H.S.} = \frac{\frac{1}{\sqrt{3}} + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}} \times \frac{1}{\sqrt{3}}} \]

Step 3: Adding in the numerator and multiplying in the denominator:

\[ = \frac{\frac{2}{\sqrt{3}}}{1 - \frac{1}{3}} = \frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}} \]

Step 4: Dividing by a fraction, which means multiplying by its reciprocal:

\[ = \frac{2}{\sqrt{3}} \times \frac{3}{2} = \frac{3}{\sqrt{3}} = \sqrt{3} \]

Hence L.H.S. = R.H.S. Notice that \( \frac{3}{\sqrt{3}} = \sqrt{3} \) because \( 3 = \sqrt{3} \times \sqrt{3} \).

The three angles of a triangle are \( \alpha = 90^\circ \), \( \beta = 30^\circ \) and \( \gamma = 60^\circ \). Test two relations: first \( \sin(\alpha + \beta) = \sin\gamma \), then \( \cos(\alpha + \beta) = -\cos\gamma \).

Step 1: Checking the condition by adding the three angles:

\[ 90^\circ + 30^\circ + 60^\circ = 180^\circ \]

Step 2: Working out the left-hand side of the sine relation. Since \( 120^\circ = 180^\circ - 60^\circ \), its sine equals \( \sin 60^\circ \):

\[ \sin(90^\circ + 30^\circ) = \sin 120^\circ = \frac{\sqrt{3}}{2} \]

Step 3: Working out the right-hand side of the sine relation:

\[ \sin\gamma = \sin 60^\circ = \frac{\sqrt{3}}{2} \]

So the sine relation is verified. Step 4: Working out the left-hand side of the cosine relation. The cosine of an obtuse angle is negative:

\[ \cos(90^\circ + 30^\circ) = \cos 120^\circ = -\frac{1}{2} \]

Step 5: Working out the right-hand side of the cosine relation:

\[ -\cos\gamma = -\cos 60^\circ = -\frac{1}{2} \]

Hence L.H.S. = R.H.S. for both relations. They hold here because \( \alpha + \beta = 180^\circ - \gamma \).

Test \( \sin 2A + \sin 2B + \sin 2C = 4\sin A\sin B\sin C \) when A = 45°, B = 45° and C = 90°.

Step 1: Checking the condition:

\[ 45^\circ + 45^\circ + 90^\circ = 180^\circ \]

Step 2: Doubling each angle:

\[ 2A = 90^\circ,\ 2B = 90^\circ,\ 2C = 180^\circ \]

Step 3: Working out the left-hand side:

\[ \sin 90^\circ + \sin 90^\circ + \sin 180^\circ = 1 + 1 + 0 = 2 \]

Step 4: Putting the values into the right-hand side:

\[ 4 \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} \times 1 = 4 \times \frac{1}{2} = 2 \]

Hence L.H.S. = R.H.S. Now see what happens when the condition is broken. Take A = B = C = 30°, whose sum is only 90°.

Step 5: Working out the left-hand side, where each double angle is 60°:

\[ 3\sin 60^\circ = \frac{3\sqrt{3}}{2} \]

Step 6: Working out the right-hand side, where each sine is \( \frac{1}{2} \):

\[ 4 \times \left(\frac{1}{2}\right)^3 = \frac{4}{8} = \frac{1}{2} \]

The two sides are not equal. So this relation is a conditional identity: it is true only when \( A + B + C = 180^\circ \).

The angles of triangle ABC are A = 45°, B = 60° and C = 75°. Test \( \tan A + \tan B + \tan C = \tan A\tan B\tan C \). You found \( \tan 75^\circ = 2 + \sqrt{3} \) earlier from \( \tan(45^\circ + 30^\circ) \).

Step 1: Checking the condition:

\[ 45^\circ + 60^\circ + 75^\circ = 180^\circ \]

Step 2: Adding the three tangents on the left-hand side:

\[ 1 + \sqrt{3} + (2 + \sqrt{3}) = 3 + 2\sqrt{3} \]

Step 3: Multiplying the three tangents on the right-hand side:

\[ 1 \times \sqrt{3} \times (2 + \sqrt{3}) \]

Step 4: Opening the bracket, using \( \sqrt{3} \times \sqrt{3} = 3 \):

\[ = 2\sqrt{3} + 3 \]

Hence L.H.S. = R.H.S. The sum of the tangents of a triangle's angles equals their product.

The angles of triangle ABC are A = 120°, B = 30° and C = 30°. Test \( \cos^2 A + \cos^2 B + \cos^2 C = 1 - 2\cos A\cos B\cos C \).

Step 1: Checking the condition:

\[ 120^\circ + 30^\circ + 30^\circ = 180^\circ \]

Step 2: Writing the left-hand side with the angles in place:

\[ \cos^2 120^\circ + \cos^2 30^\circ + \cos^2 30^\circ \]

Step 3: Squaring each value. Since \( \cos 120^\circ = -\frac{1}{2} \), its square is \( \frac{1}{4} \), and \( \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4} \):

\[ = \frac{1}{4} + \frac{3}{4} + \frac{3}{4} = \frac{7}{4} \]

Step 4: Putting the values into the right-hand side:

\[ 1 - 2\left(-\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) \]

Step 5: Multiplying. The product \( 2 \times \left(-\frac{1}{2}\right) \times \frac{3}{4} \) is \( -\frac{3}{4} \), and subtracting a negative number adds it:

\[ = 1 + \frac{3}{4} = \frac{7}{4} \]

Hence L.H.S. = R.H.S.