Trigonometric Ratios of Compound Angles
Subject: Optional Mathematics
Overview
Let A and B two angles. Then their sum A + B or the difference A - B is called a compound angle. The sum or difference of any two or more than two angles is called a compound angle.
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) = \(\frac{cos(A - B)}{sin(A - B)}\)
=\(\frac{cosA\; cosB + sinA\; sinB}{sinA \;cosB - cosA \;sinB}\)
=\(\frac{\frac{cosA \;cosB}{sinA\; sinB} + \frac{sinA \;sinB}{sinA \;sinB}}{\frac{sinA \;cosB}{sinA \;sinB} - \frac{cosA\; sinB}{sinA \;sinB}}\)
| Trigonometric Ratios of Compound Angles | |
| sin(A + B) = sinA cosB + cosA sinB |
sin(A - B) = sinA cosB - cosA sinB |
| cos(A + B) = cosA cosB - sinA sinB | cos(A - B) = cosA cosB + sinA sinB |
| tan(A + B) = \(\frac{tan A + tan B}{1 - tanA\; tanB}\) | tan(A - B) =\(\frac{tanA - tanB}{1 + tanA \;tanB}\) |
| cot(A + B) =\(\frac{cotA\; cotB - 1}{cotB + cotA}\) | cot(A - B) =\(\frac{cotA \;cotB + 1}{cotB - cotA}\) |
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)
= \(\frac{tan(A + B) + tanC}{1 - tan (A + B) tanC}\)
= \(\frac{\frac{tanA + tanB}{1 - tanA tanB} + tanC}{1 -(\frac{tanA + tanB}{1 - tanA tanB}) tanC}\)
=\(\frac{tanA + tanB + tanC - tanA tanB tanC}{1 - tanB tanC - tanC tanA - tanA tanB}\)
Things to remember
| Trigonometric Ratios of Compound Angles | |
| sin(A + B) = sinA cosB + cosA sinB |
sin(A - B) = sinA cosB - cosA sinB |
| cos(A + B) = cosA cosB - sinA sinB | cos(A - B) = cosA cosB + sinA sinB |
| tan(A + B) = \(\frac{tan A + tan B}{1 - tanA tanB}\) | tan(A - B) =\(\frac{tanA - tanB}{1 + tanA tanB}\) |
| cot(A + B) =\(\frac{cotA cotB - 1}{cotB + cotA}\) | cot(A - B) =\(\frac{cotA cotB + 1}{cotB - cotA}\) |
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Videos for Trigonometric Ratios of Compound Angles
Evaluating Trig Ratios for Any Angle Between 0 and 360
How to solve problems based on trigonometric ratios of compound angles?
Questions and Answers
Without using calculator or table, find the value of cos75°.
Here,
cos75°
= cos (45° + 30°)
= cos45° ⋅ cos30° - sin45°⋅ sin30°
= \(\frac {1}{\sqrt 2}\) ⋅\(\frac {\sqrt 3}{2}\) - \(\frac 1{\sqrt 2}\) ⋅\(\frac 12\)
= \(\frac {\sqrt 3}{2\sqrt 2}\) - \(\frac 1{2\sqrt 2}\)
= \(\frac {\sqrt 3 - 1}{2\sqrt 2}\) Ans
Find the value of tan15° without using calculator or table.
Here,
tan15°
= tan (60° - 45°)
= \(\frac {tan60° - tan45°}{1 + tan60° ⋅tan45°}\)
= \(\frac {\sqrt 3 - 1}{1 + \sqrt 3 ⋅ 1}\)
= \(\frac {\sqrt 3 - 1}{\sqrt 3 + 1}\)× \(\frac {\sqrt 3 - 1}{\sqrt 3 - 1}\)
= \(\frac {(\sqrt 3 - 1)^2}{{(\sqrt 3)^2}-{1^2}}\)
= \(\frac {3 - 2 {\sqrt 3 + 1}}{3 - 1}\)
= \(\frac {4 -2{\sqrt 3}}{2}\)
= \(\frac {2(2 - \sqrt 3)}{2}\)
= 2 - \(\sqrt 3\)Ans
Find the value of sin 75° sin 15° without using calculator or table.
sin 75° sin 15°
= sin (45° + 30°) sin (45° - 30°)
= (sin 45° cos 30° + cos45° sin30°) (sin 45° cos 30° - cos 45° sin30°)
= (\(\frac 1{\sqrt 2}\)⋅\(\frac {\sqrt 3}{2}\) + \(\frac 1{\sqrt 2}\)⋅\(\frac {1}{2}\))(\(\frac 1{\sqrt 2}\)⋅\(\frac {\sqrt 3}{2}\) -\(\frac 1{\sqrt 2}\)⋅\(\frac {1}{2}\))
= (\(\frac {\sqrt 3}{2\sqrt 2}\))2 -(\(\frac {1}{2\sqrt 2}\))2
= \(\frac 38\) - \(\frac 18\)
= \(\frac {3 - 1}{8}\)
= \(\frac 28\)
= \(\frac 14\) Ans
Find the value of cos 105° cos 15° without using calculator or table.
cos 105° cos 15°
= cos (60° + 45°) cos (60° - 45°)
= (cos 60° cos 45° - sin 60° sin 45°) (cos 60° cos 45° + sin 60° sin 45°)
= (cos 60° cos 45°)2 - (sin 60° sin 45°)2
= (\(\frac 12\) × \(\frac {1}{\sqrt 2}\))2 - (\(\frac {\sqrt 3}{2}\) × \(\frac {1}{\sqrt 2}\))2
= \(\frac 18\) - \(\frac 38\)
= \(\frac {1 - 3}{8}\)
= \(\frac {-2}{8}\)
= \(\frac {-1}{4}\) Ans
Without usinga calculator or a table, calculate the value of cos 105°.
cos 105°
= cos (60° + 45°)
= cos 60° cos 45° - sin 60° sin 45°
= \(\frac 12\)⋅ \(\frac 1{\sqrt 2}\) -\(\frac {\sqrt 3}2\)⋅ \(\frac 1{\sqrt 2}\)
= \(\frac {1 - \sqrt 3}{2\sqrt 2}\) Ans
Prove that:
1 - tan 35° tan 10° = tan 35° + tan 10°
Here,
10° + 35° = 45°
Putting tan on both sides,
tan (10° + 35°) = tan 45°
or, \(\frac {tan 10° + tan 35°}{1 - tan 10° tan 35°}\) = 1
or, tan 10° + tan 35° = 1 - tan 10° tan35°
or, 1 - tan 10° tan 35° = tan 10° + tan 35°
Hence, L.H.S. = R.H.S. Proved
If A + B = 45° , prove that:
(1 + tan A) (1 + tan B) = 2
Here,
A + B = 45°
Putting tan on both;
tan (A + B) = tan 45°
or, \(\frac {tan A + tan B}{1 - tan A tan B}\) = 1
or, tan A + tan B = 1 - tan A tan B
or, tan A + tan B + tan A tan B = 1
or, tan A + tan B + tan A tan B + 1 = 1 + 1
or, tan A + tan A tan B + 1 + tan B = 2
or, tan A (1 + tan B) + 1 (1 + tan B) = 2
or, (1 + tan B) (tan A + 1) = 2
∴ (1 + tan A) (1 + tan B) = 2
Hence, L.H.S. = R.H.S. Proved
Prove that:
cot (A - B) = \(\frac {cot A cot B + 1}{cot B - cot A}\)
L.H.S.
= cot (A - B)
= \(\frac {cos (A - B)}{sin (A - B)}\)
= \(\frac {cos A cos B + sin A sin B}{sin A cos B - cos A sin B}\)
=\(\frac {\frac {cos A cos B}{sin A sin B}+ \frac{sin A sin B}{sin A sin B}}{\frac {sin A cos B}{sin A sin B}+ \frac{cos A sin B}{sin A sin B}}\)
= \(\frac {cot A cot B + 1}{cot B - cot A}\)
= R.H.S Proved
If A + B = 45°, show that:
tan A + tan B + tan A⋅tan B = 1
Here,
A + B = 45°
Taking tan on both sides:
tan (A + B) = tan 45°
or, \(\frac {tan A + tan B}{1 - tan A tan B}\) = 1
or, tan A + tan B = 1 - tan A tan B
or, tan A + tan B + tan A tan B = 1
Hence, L.H.S. = R.H.S. Proved
Prove that:
1 - tan 20° tan 25° = tan 20° + tan 25°
Here,
20° + 25° = 45°
Taking tan on both sides,
tan (20° + 25°) = tan 45°
or, \(\frac {tan 20° + tan 25°}{1 - tan 20° tan 25°}\) = 1
or, tan 20° + tan 25° = 1 - tan 20° tan 25°
∴ 1 - tan 20° tan 25° = tan 20° + tan 25°
Hence, L.H.S. = R.H.S. Proved
If tan α = \(\frac 56\) and tan β = \(\frac 1{11}\), prove that:
α + β = \(\frac {π^2}4\)
tan (α + β) = \(\frac {tanα + tanβ}{1 - tan α tan β}\)
or, tan (α + β) =\(\frac {\frac 56 + \frac 1{11}}{1 - \frac56 × \frac {1}{11}}\)
or, tan (α + β) =\(\cfrac {\frac {55 + 5}{66}}{\frac {66 - 5}{66}}\)
or, tan (α + β) =\(\cfrac {\frac {61}{66}}{\frac {61}{66}}\)
or, tan (α + β) = \(\frac {61}{66}\)× \(\frac {66}{61}\)
or, tan (α + β) = 1
or, tan (α + β) = tan \(\frac {π^c}{4}\)
∴ (α + β) = \(\frac {π^c}{4}\) Proved
Without using table, prove that:
sin 105° + cos 105° = \(\frac {1}{\sqrt 2}\)
L.H.S.
= sin 105° + cos 105°
= sin (60° + 45°) + cos (60° + 45°)
= sin 60° cos 45° + cos 60° sin 45° + cos 60° cos 45° - sin 60° sin 45°
= \(\frac {\sqrt3}{2}\)⋅\(\frac 1{\sqrt 2}\) + \(\frac 12\)⋅\(\frac 1{\sqrt 2}\) + \(\frac 12\)⋅\(\frac 1{\sqrt 2}\) - \(\frac {\sqrt 3}2\)⋅\(\frac 1{\sqrt 2}\)
= \(\frac {\sqrt 3}{2\sqrt 2}\) + \(\frac 1{2\sqrt 2}\) + \(\frac 1{2\sqrt 2}\) - \(\frac {\sqrt 3}{2\sqrt 2}\)
= \(\frac {1 + 1}{2\sqrt 2}\)
= \(\frac 2{2\sqrt 2}\)
= \(\frac 1{\sqrt 2}\)
= R.H.S.
Hence, L.H.S. = R.H.S. Proved
If m sin (α + θ) = n sin (β + θ), prove that :
cot θ = \(\frac {m cos α - n cos β}{n sin β - m sin α}\)
Given,
m sin(α + θ) = n sin (β + θ)
or, m (sin α cos θ + cos α sin θ) = n (sin β cos θ + cos β sin θ)
or, msin α cos θ + mcos α sin θ = nsin β cos θ + ncos β sin θ
or, msin α cos θ -nsin β cos θ =ncos β sin θ -mcos α sin θ
or,cos θ (msin α - nsin β) = sin θ (ncos β - mcos α)
or, \(\frac {cos θ}{sin θ}\) = \(\frac {ncos β - mcos α}{msin α - nsin β}\)
∴ cot θ = \(\frac {ncos β - mcos α}{msin α - nsin β}\) Proved
Prove that:
tan 20° + tan 72° + tan 88° = tan 20° ⋅ tan 72° ⋅ tan 88°
Here,
20° + 72° + 88° = 180°
20° + 72° = 180° - 88°
Putting tan on both,
tan (20° + 72°) = tan (180° - 88°)
or, \(\frac {tan 20° + tan 72°}{1 - tan 20° tan 72°}\) = 0 - tan 88°
or, tan 20° + tan 72° = - tan 88° (1 - tan 20° tan 72°)
or, tan 20° + tan 72° = - tan 88° + tan 20° tan 72° tan 88°
or, tan 20° + tan 72° + tan 88° = tan 20° tan 72° tan 88°
Hence L.H.S. = R.H.S. Proved
Prove that:
sin (x + y) - sin (x - y) = 2cos x siny
Here,
L.H.S.
= sin (x + y) - sin ( x - y)
= sin x cos y + cos x sin y - (sin x cos y - cos x sin y)
=sin x cos y + cos x sin y - sin x cos y + cos x sin y
= 2 cos x sin y
= R.H.S.
Hence L.H.S. = R.H.S. Proved
If A + B = \(\frac {π^c}{4}\), prove that:
(cot A - 1) (cot B - 1) = 2
Here,
A + B = \(\frac {π^c}{4}\)
Taking cot on both,
cot (A + B) = cot \(\frac {π^c}{4}\)
or, \(\frac {cot A cot B - 1}{cot B + cot A}\) = 1
or, cot A cot B - 1 = cot B + cot A
or, cot A cot B - cot A - cot B = 1
or, cot A cot B - cot A - cot B + 1 = 1 + 1
or, cot A (cot B - 1) -1 (cot B - 1) = 2
or, (cot A - 1) (cot B - 1) = 2
∴ L.H.S. = R.H.S. Proved
Without using calculator or a table, prove that:
tan 55° - tan 35° = 2 tan 20°
Here,
55° - 35° = 20°
Taking tan on both sides,
tan (55° - 35°) = tan 20°
or, \(\frac {tan 55° - tan 35°}{1 + tan 55° tan35°}\) = tan 20°
or, tan 55° - tan 35° = tan 20° (1 + tan 55° tan35°)
or,tan 55° - tan 35° = tan 20° + tan 20° tan35° tan 55°
or, tan 55° - tan 35° = tan 20° + tan 20° tan (90° - 55°) tan 55°
or, tan 55° - tan 35° = tan 20° + tan 20° cot 55° tan 55°
or, tan 55° - tan 35° = tan 20° + tan 20°
∴ tan 55° - tan 35° = 2 tan 20°
Hence, L.H.S. = R.H.S. Proved
Prove that:
cos (A+ B + C) = cos A cos B cos C (1 - tan B tan C - tan C tan B - tan A tan B)
L.H.S.
= cos (A + B + C)
= cos (A + B) cos C - sin (A + B) sin C
= (cos A cos B - sin A sin B) cos C - (sin A cos B + cos A sin B) sin C
multiply by \(\frac {cos A cos B cos C}{cos A cos B cos C}
= cos A cos B cos C (\(\frac {cos A cos B cos C}{cos A cos B cos C}\) - \(\frac {sin A sin B sin C}{cos A cos B cos C}\) - \(\frac {sin A cos Bsin C}{cos A cosB cos C}\) - \(\frac {cos A sin B sin C}{cos A cos B cos C}\))
= cos A cos B cos C (1 - tan A tan B - tan A tan C - tan B tan C)
= cos A cos B cos C (1 - tan B tan C - tan C tan A - tan A tan B)
Hence, L.H.S. = R.H.S. Proved
Prove that:
\(\frac {cos 17° + sin 17°}{cos 17° - sin 17°}\) = tan 62°
R.H.S.
= tan 62°
= \(\frac {sin 62°}{cos 62°}\)
= \(\frac {sin (45° + 17°)}{cos (45° + 17°)}\)
= \(\frac {sin 45° cos 17° + cos 45° sin 17°}{cos 45° cos 17° - sin 45° sin 17°}\)
= \(\frac {\frac {1}{\sqrt 2} cos 17° + \frac {1}{\sqrt 2} sin 17°}{\frac {1}{\sqrt 2} cos 17°- \frac {1}{\sqrt 2} sin 17°}\)
= \(\frac {\frac {1}{\sqrt 2} (cos 17° + sin 17°)}{\frac {1}{\sqrt 2} (cos 17°- sin 17°)}\)
= \(\frac {cos 17° + sin 17°}{cos 17° - sin 17°}\)
= L.H.S. Proved
Prove that:
\(\frac {cos 8° + sin 8°}{cos 8° - sin 8°}\) = tan 53°
R.H.S.
= tan 53°
= \(\frac {sin 53°}{cos 53°}\)
= \(\frac {sin (45° + 8°)}{cos (45° + 8°)}\)
= \(\frac {sin 45° cos 8° + cos 45° sin 8°}{cos 45° cos 8° - sin 45° sin 8°}\)
= \(\frac {\frac {1}{\sqrt 2} cos 8° + \frac {1}{\sqrt 2} sin 8°}{\frac {1}{\sqrt 2} cos 8°- \frac {1}{\sqrt 2} sin 8°}\)
= \(\frac {\frac {1}{\sqrt 2} (cos 8° + sin 8°)}{\frac {1}{\sqrt 2} (cos 8°- sin 8°)}\)
= \(\frac {cos 8° + sin 8°}{cos 8° - sin 8°}\)
= L.H.S. Proved
If sin A = \(\frac 35\) and cos B = \(\frac 5{13}\), find the value of cos (A +B).
Here,
sin A = \(\frac 35\) and cos B = \(\frac 5{13}\)
cos A = \(\sqrt {1 - sin^2 A}\)
= \(\sqrt {1 - (\frac 35)^2}\)
= \(\sqrt {1 - \frac 9{25}}\)
= \(\sqrt {\frac {25 - 9}{25}}\)
= \(\sqrt {\frac {16}{25}}\)
= \(\frac 45\)
sin B = \(\sqrt {1 - cos^2 B}\)
= \(\sqrt {1 - (\frac 5{13})^2}\)
= \(\sqrt {1 - \frac {25}{169}}\)
= \(\sqrt {\frac {169 - 25}{169}}\)
= \(\sqrt {\frac {144}{169}}\)
= \(\frac {12}{13}\)
Now,
cos (A + B) = cos A cos B - sin A sin B
= \(\frac 45\)⋅ \(\frac 5{13}\) - \(\frac 35\)⋅ \(\frac {12}{13}\)
= \(\frac {20}{65}\) - \(\frac {36}{65}\)
= \(\frac {20 - 36}{65}\)
= -\(\frac {16}{65}\) Ans
If sin A = \(\frac 1{\sqrt {10}}\), sin B = \(\frac 1{\sqrt 5}\), prove that:
A + B = \(\frac {π^2}{4}\)
Here,
sin A = \(\frac 1{\sqrt {10}}\), sin B = \(\frac 1{\sqrt 5}\)
cos A = \(\sqrt {1 - sin^2 A}\)
= \(\sqrt {1 - (\frac 1{\sqrt {10}})^2}\)
= \(\sqrt {1 - \frac 1{10}}\)
= \(\sqrt {\frac {10 - 1}{10}}\)
= \(\sqrt {\frac {9}{10}}\)
= \(\frac 3{\sqrt {10}}\)
cos B = \(\sqrt {1 - sin^2 B}\)
= \(\sqrt {1 - (\frac 1{\sqrt 5})^2}\)
= \(\sqrt {1 - \frac 15}\)
= \(\sqrt {\frac {5 - 1}{5}}\)
= \(\sqrt {\frac {4}{5}}\)
= \(\frac 2{\sqrt 5}\)
sin (A + B) = sin A cos B + cos A sin B
or, sin (A + B) = \(\frac 1{\sqrt {10}}\)× \(\frac 2{\sqrt 5}\) + \(\frac 3{\sqrt {10}}\)× \(\frac 1{\sqrt 5}\)
or, sin (A + B) = \(\frac 2{\sqrt {50}}\) + \(\frac 3{\sqrt {50}}\)
or, sin (A + B) = \(\frac {2 + 3}{\sqrt {50}}\)
or, sin (A + B) = \(\frac 5{\sqrt {50}}\)
or, sin (A + B) = \(\frac 5{5\sqrt 2}\)
or, sin (A + B) = \(\frac 1{\sqrt 2}\)
or, sin (A + B) = sin \(\frac {π^2}4\)
∴ A + B =\(\frac {π^2}4\) Proved
Prove that:
\(\frac {cos 35° + sin 35°}{cos 35° - sin 35°}\) = cot 10°
L.H.S.
= \(\frac {cos 35° + sin 35°}{cos 35° - sin 35°}\)
=\(\frac {cos (45° - 10°) + sin (45° - 10°)}{cos (45° - 10°) - sin (45° - 10°)}\)
= \(\frac {cos 45° cos 10° + sin 45° sin 10° + sin 45° cos 10° - cos 45° sin 10°}{cos 45° cos 10° + sin 45° sin 10° - sin 45° cos 10° + cos 45° sin 10°}\)
= \(\frac {\frac 1{\sqrt 2} cos 10° + \frac 1{\sqrt 2} sin 10° + \frac 1{\sqrt 2} cos 10° - \frac 1{\sqrt 2} sin 10°}{\frac 1{\sqrt 2} cos 10° + \frac 1{\sqrt 2} sin 10° - \frac 1{\sqrt 2} cos 10° + \frac 1{\sqrt 2} sin 10°}\)
=\(\frac {\frac 2{\sqrt 2} cos 10°}{\frac 2{\sqrt 2} sin 10°}\)
= \(\frac {cos 10°}{sin 10°}\)
= cot 10°
= R.H.S.Proved
Prove that:
cos (A + B)⋅cos (A - B) = cos2A - sin2B = cos2B - sin2A
L.H.S.
=cos (A + B)⋅cos (A - B)
= (cos A cos B - sin a sin B)(cos A cos B + sin A sin B)
= (cos A cos B)2 - (sin A sin B)2
= cos2A cos2B - sin2A sin2B
= cos2A (1 - sin2B) - (1 - cos2A) sin2B
= cos2A - cos2A sin2B - sin2B + cos2A sin2B
= cos2A - sin2B (M.H.S.)
Again,
cos2A - sin2B
= (1 - sin2A) - (1 - cos2B)
= 1 - sin2A - 1 + cos2B
= cos2B - sin2A
∴ L.H.S. = M.H.S. = R.H.S. Proved
An angle θ is divided into two parts α and β such that tanα:tanβ = x:y show that:
sin(α - β) = \(\frac {x - y}{x + y}\) sinθ
Given,
θ =α +β
sinθ = sin(α + β)............................(1)
And
\(\frac {tanα}{tanβ}\) = \(\frac xy\)
\(\frac {tanα - tanβ}{tanα + tanβ}\) = \(\frac {x - y}{x + y}\).............................(2)
Now,
R.H.S.
= \(\frac {x - y}{x + y}\) sinθ
=\(\frac {tanα - tanβ}{tanα + tanβ}\) sinθ
=\(\frac {\frac {sinα}{cosα} -\frac {sinβ}{cosβ}}{\frac {sinα}{cosα} + \frac {sinβ}{cosβ}}\) sinθ
= \(\frac{\frac{sin\alpha.cos\beta - cos\alpha.sin\beta}{cos\alpha.cos\beta}}{\frac{sin\alpha.cos\beta + cos\alpha.sin\beta}{cos\alpha.cos\beta}}\). sin\(\theta\)
= \(\frac {sin\alpha cos\beta - cos\alpha sin\beta}{sin\alpha cos\beta + cos\alpha sin\beta}\) . sin\(\theta\)
= \(\frac {sin (\alpha - \beta)}{sin(\alpha + \beta)}\) . sin\(\theta\)
= \(\frac {sin(\alpha - \beta)}{sin \theta}\). sin\(\theta\)
= sin (α - β)
Hence, L.H.S. = R.H.S. Proved
Prove that:
sin(A + B + C) = cosA . cosB . cosC(tanA + tanB + tanC - tanA . tanB . tanC)
L.H.S.
= sin (A + B + C)
= sin {A + (B + C)}
= sinA . cos(B + C) + cos A . cos (B + C)
= sinA {cosB cosC - sinB sinC} + cosA {sinB cosC + cosB sinC}
= sinA cosB cosC - sinA sinB sinC + cosA sinB cosC + cosA cosB sinC
= cosA cosB cosC[\(\frac {sinA cosB cosC - sinA sinB sinC + cosA sinB cosC + cosA cosB sinC}{cosA cosB cosC}\)]
= cosA cosB cosC [tanA - tanA tanB tanC + tanB + tanC]
= cosA cosB cosC [tanA + tanB + tanC- tanA tanB tanC ]
∴ L.H.S. = R.H.S. Proved
Quiz
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