Trigonometric Ratios of Compound Angles

Let, A and B be two angles. Then their sum A + B or the difference A - B is called a compound angle.

Trigonometric ratios of A + B (Addition formula)

Let a revolving line start from OX and trace out an angle XOY = A and revolve further through an angle YOZ = B

∴ ∠XOZ = A + B

Let P be any point in OZ. Draw PM perpendicular to OX and PN perpendicular to OY From N draws NQ perpendicular to OX ad NR perpendicular to MP.

Here, ∠RPN = 90 - ∠PNR

= ∠RNO

= ∠NOQ

= A

Again, RMQN is a rectangle, so, MR = QN and RN = MQ

Now, sin(A + B) =\(\frac{MP}{OP}\)

                         = \(\frac{MR + RP}{OP}\)

                         = \(\frac{QN + RP}{OP}\)

                         = \(\frac{QN}{OP}\) + \(\frac{RP}{OP}\)

                         = \(\frac{QN}{ON}\) \(\frac{ON}{OP}\) + \(\frac{RP}{NP}\) \(\frac{NP}{OP}\)

cos(A + B) = \(\frac{OM}{OP}\)

                = \(\frac{OQ - MQ}{OP}\)

                = \(\frac{OQ - RN}{OP}\)

                = \(\frac{OQ}{OP}\) - \(\frac{RN}{OP}\)

                = \(\frac{OQ}{ON}\) \(\frac{ON}{OP}\) - \(\frac{RN}{NP}\) \(\frac{NP}{OP}\)

                = cosA cosB - sinA sinB                                 

Hence, sin formula of compound angle (A + B) is sin (A + B) = sinA cosB + cosA sinB and consine formula of compound angle (A + B) and cos(A + B) = cosA cosB - sinA sinB

Trigonometric Ratios of A - B(Subtraction formula)

Let a revolving line start from OX and trace out an angle XOY = A and then revolve back through an angle YOZ = B

∴ ∠XOZ = A - B

Let P be any point in the Line OZ. Draw PM perpendicular to OX and PN perpendicular to OY.

From N Draw NQ perpendicular to OX and perpendicular to MP produced.

Here, ∠RPN = 90⁰ - ∠PNR

= ∠PNY

= ∠XOY

= A

Again QMRN is a rectangle. So, QN = MR and QM = NR

Now sin(A-B) =\(\frac{PM}{OP}\)

                    = \(\frac{MR - PR}{OP}\)

                    =\(\frac{QN - PR}{OP}\)

                    = \(\frac{QN}{OP}\) - \(\frac{PR}{OP}\)

                    =\(\frac{QN}{ON}\) \(\frac{ON}{OP}\) - \(\frac{PR}{NP}\) \(\frac{NP}{OP}\)

                    = sinA cosB - cosA sinB

cos(A - B) = \(\frac{OM}{OP}\)

                =\(\frac{OQ + QM}{OP}\)

                =\(\frac{OQ + NR}{OP}\)

                =\(\frac{OQ}{OP}\) + \(\frac{NR}{OP}\)

                =\(\frac{OQ}{ON}\) \(\frac{ON}{OP}\) - \(\frac{NR}{NP}\) \(\frac{NP}{OP}\)

                = cosA cosB + sinA sinB

Hence, sine formula of compound angle (A - B) is sin (A - B) = sinA cosB - cosA sinB and cosine formula of compound angle (A - B) is cos (A - B) = cosA cosB + sinA sinB

Alternative Method

Take a unit circle with centre at the origin. Let the circle intersect the X-axis at the point P. Then the coorinates of P are (1.0)

Let Q be another point on the circumference of the circle such that∠POQ = A. Then the coordinates of Q are (cosA, sinA).

Let R be another point on the cirumference of the circle such that ∠QOR = B

Then ∠POR = ∠POQ + ∠QOR = A + B

So coordinates of R are ( cos(A + B) , sin(A + B)).

Take a point S on the circumference such that ∠POS = -B.

Then coordinates of the points S are (cos(-B), sin(-B)) = (cosB, sin(-B))

Here, ∠SOQ = ∠SOP + ∠POQ = A + B and ∠POR = ∠POQ + ∠QOR = A + B

∴ ∠SOQ = ∠POR

So, arc QS = arc PR

∴ Chord QS = Chord PR.

Now by distance formula

PR² = [cos(A + B)-1]² + [sin(A + B) - 0]²

       = cos² (A + B) - 2cos(A + B) + 1 + sin² (A + B)

       = 2 - 2cos (A + B)

QS² =(cosA - cosB)² + [sinA - sin(-B)]² = (cosA - cosB)² + (sinA + sinB)²

       = cos²A - 2cosA.cosB + cos²B + sin²A + 2sinA.sinB + sin²B

       = 2 - 2cosA.cosB + 2sinA.sinB

Now, PR² = QS²

or, 2 - 2cos(A + B) = 2 - 2cosA.cosB + 2sinA.sinB

or, cos(A + B) = cosA.cosB - sinA.sinB ........(i)

If the angle B is replaced by (-B), Then

Cos(A-B) = cosA.cos(-B) - sinA.sin(-B) = cosA.cosB + sinA.sinB .........(ii)

Again, cos[\(\frac{\pi}{2}\) - (A+B)] = cos[(\(\frac{\pi}{2}\) - A) - B)]

or, sin(A + B) = cos (\(\frac {\pi}{2}\) - A) cos B + sin (\(\frac{\pi}{2}\) - A) sin B = sinA cosB + cosA sinB ....... (iii)

Similarly, cos[\(\frac{\pi}{2}\) - (A+B)] = cos[(\(\frac{\pi}{2}\) - A) + B)]

or, sin(A - B) = cos ( \(\frac{\pi}{2}\) - A) cosB - sin( \(\frac{\pi}{2}\) - A) sinB = sinA cosB - cosA sinB .......... (iv)

Tangent formula of compound angle (A + B)

tan (A + B) = \(\frac{sin(A + B)}{cos(A + B)}\)

                  = \(\frac{sinA\; cosB + cosA\; sinB}{cosA \;cosB - sinA \;sinB}\)

                  = \(\frac {\frac {sinA\; cosB}{cosA \;cosB} + \frac {cosA \;sinB}{cosA \;cosB}}{\frac {cosA \;cosB}{cosA\; cosB} - \frac {sinA\; sinB}{cosA \;cosB}}\)

                  = \(\frac{tan A + tan B}{1 - tanA\; tanB}\)

Tangent formula of compound angle (A - B)

tan (A - B) = \(\frac{sin(A - B)}{cos(A - B)}\)

                 =\(\frac{sinA\; cosB - cosA \;sinB}{cosA \;cosB + sinA \;sinB}\)

                 =\(\frac{\frac{sinA\; cosB}{cosA\; cosB} - \frac{cosA\; sinB}{cosA \;cosB}}{\frac{cosA \;cosB}{cosA \;cosB} + \frac{sinA\; sinB}{cosA\; cosB}}\)

                 = \(\frac{tan A - tan B}{1 + tanA \;tanB}\)

Cotangent formula of compound angle (A + B)

cot (A + B) = \(\frac{cos(A + B)}{sin(A + B)}\)

                 =\(\frac{cosA\; cosB - sinA \;sinB}{sinA\; cosB + cosA\; sinB}\)

                 =\(\frac{\frac{cosA \;cosB}{sinA\; sinB} - \frac{sinA\; sinB}{sinA\; sinB}}{\frac{sinA \;cosB}{sinA\; sinB} + \frac{cosA\; sinB}{sinA\; sinB}}\)

                 =\(\frac{cotA \;cotB - 1}{cotB + cotA}\)

Cotangent formula of compound angle (A - B)

cot(A - B) = \(\frac{cos(A - B)}{sin(A - B)}\)

                =\(\frac{cosA\; cosB + sinA\; sinB}{sinA \;cosB - cosA \;sinB}\)

                =\(\frac{\frac{cosA \;cosB}{sinA\; sinB} + \frac{sinA \;sinB}{sinA \;sinB}}{\frac{sinA \;cosB}{sinA \;sinB} - \frac{cosA\; sinB}{sinA \;sinB}}\)

Some more results :

1. sin(A + B). sin(A - B) = cos²B - cos²A

Proof:

sin(A + B) .sin(A - B)

= (sinA cosB + cosA sinB) . (sinA cosB - cosA sinB)

= sin²A cos²B - cos²A sin²B

= (1 - cos²A) cos²B - cos²A(1 - cos²B)

= cos²B - cos²A cos²B - cos²A + cos²A cos²B

= cos²B - cos²A

2. sin(A + B). sin(A - B) = sin²A - sin²B

proof:

sin(A + B). sin(A - B)

= cos²B - cos²A

= 1 - sin²B - (1 - sin²A)

= sin²A - sin²B

3. cos(A + B). cos(A - B) = cos²A - sin²B

Proof :

cos (A + B). cos(A - B)

= (cosA cosB - sinA sinB) (cosA cosB + sinA sinB)

= cos²A cos²B - sin²A sin²B

= cos²A(1 - sin²B) - (1 - cos²A) sin²B

= cos²A - cos²A sin²B - sin²B + cos²A sin²B

= cos²A - cos²A sin²B - sin²B + cos²A sin²B

= cos²A - sin²A

4. cos(A + B) . cos(A - B) = cos²B - sin²A

Proof :

cos(A + B) . cos(A - B)

= cos²A - sin²B

= 1 - sin²A - (1 - cos²B)

= cos²B - sin²A

5. cot(A + B) .cot(A - B) =\(\frac{cot^{2} A. cot^{2} B - 1}{cot^{2} B - cot^{2} A}\)

Proof:

cot(A + B). cot(A - B)

= ( \(\frac{cotA. cotB - 1}{cotB + cotA}\)) ( \(\frac{cotA .cotB + 1}{cotB - cotA}\))

=\(\frac{cot^{2}A . cot^{2}B}{cot^{2}B - cot^{2}A}\)

6. tan(A + B). tan(A - B) =\(\frac{tan^{2}A - tan^{2}B}{1 - tan^{2}A . tan^{2}B}\)

Proof :

tan(A + B). tan(A - B)

= (\(\frac{tanA + tanB}{1 - tanA.tanB}\)) (\(\frac{tanA - tanB}{1 + tanA tanB}\))

= \(\frac{tan^{2}A - tan^{2}B}{1 - tan^{2}A tan^{2}B}\)

7. sin(A + B + C) = sinA cosB cosC + cosA sinB cosC + cosA cosB sinC - sinA sinB sinC

Proof :

sin(A + B + C)

= sin(A + B) cosC + cos(A + B) sinC

= (sinA cosB + cosA sinB) cosC + (cosA cosB - sinA sinB) sinC

= sinA cosB cosC + cosA sinB cosC + cosA cosB sinC - sinA sinB sinC

8. cos(A + B + C) = cosA.cosB.cosC - cosA.sinB.sinC - sinC.cosB.sinA - sinA.sinB.cosC

Proof:

cos(A + B + C)

= cos(A + B) cosC - sin(A + B) sinC

= (cosA cosB - sinA sinB) cosC - (sinA cosB + cosA sinB) sinC

= cosA.cosB.cosC - sinA sinB cosC - sinC.cosB.sinA - cosA.sinB.sinC

9. tan (A + B + C) =\(\frac{tanA + tanB + tanC - tanA tanB tanC}{1 - tanB tanC - tanC tanA - tanA tanB}\)

Proof:

tan(A + B + C)

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

= \(\frac{\frac{tanA + tanB}{1 - tanA tanB} + tanC}{1 -(\frac{tanA + tanB}{1 - tanA tanB}) tanC}\)

=\(\frac{tanA + tanB + tanC - tanA tanB tanC}{1 - tanB tanC - tanC tanA - tanA tanB}\)