Trigonometric Ratios of Compound Angles

Trigonometric Ratios of Compound Angles.

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

(a) 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

trigonometric ratios of A+B (Addition formula)
trigonometric ratios of A+B (Addition formula)

= ∠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

(b) 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 \begin{bmatrix} \frac {\pi}2 - (A + B)\\ \end{bmatrix} = cos \begin{bmatrix}( \frac{\pi}2 - A) - B\\ \end{bmatrix}

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 \begin{bmatrix} \frac {\pi}2 - (A + B)\\ \end{bmatrix} = cos \begin{bmatrix}( \frac{\pi}2 - A) + B\\ \end{bmatrix}

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

(c) Tangent formula of compound angle (A + B)

tan (A + B) = \(\frac{sin(A + B)}{cos(A + B)}\) = \(\frac{sinA cosB + cosA sinB}{cosA cosB - sinA sinB}\) = \(\frac {\frac {sinA cosB}{cosA cosB} + \frac {cosA sinB}{cosA cosB}}{\frac {cosA cosB}{cosA cosB} - \frac {sinA sinB}{cosA cosB}}\)

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

(d) Tangent formula of compound angle (A - B)

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

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

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

(e) Cotangent formula of compound angle (A + B)

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

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

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

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

(f) 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:

LHS = 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:

L.H.S = sin(A + B). sin(A - B) = cos2B - cos2A = 1 - sin2B - (1 - sin2A) = sin2A - sin2B

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

Proof :

LHS = 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 :

LHS = cos(A + B) . cos(A - B) = cos2A - sin2B = 1 - sin2A - (1 - cos2B) = cos2B - sin2A

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

LHS = 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 :

LHS = 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 :

LHS = 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:

LHS = 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:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

0%
  • Find the value without using a calculator .

    sin 15(^o)

    90(^o)


    (frac{sqrt{3}-1}{2 sqrt{2} })


    (frac{sqrt{2}-1}{2 sqrt{3} })


    (frac{sqrt{3}-3}{3 sqrt{3} })


  • Find the value without using a calculator .

    Sin 75(^o)

    (frac{sqrt{3}+1}{3sqrt{1}})


    (frac{sqrt{3}+3}{2sqrt{3}})


    (frac{sqrt{3}+1}{2sqrt{2}})


    (frac{sqrt{5}+2}{3sqrt{1}})


  • Find the value without using a calculator .

    Sin 105(^o)

    (frac{sqrt{2+1}}{3sqrt{1}})


    (frac{sqrt{3+1}}{2sqrt{2}})


    (frac{sqrt{3+1}}{4sqrt{2}})


    (frac{sqrt{3+1}}{3sqrt{4}})


  • Find the value without using a calculator .

    Sin 135(^o)

    (frac{2}{sqrt{4}})


    (frac{1}{sqrt{2}})


    (frac{2}{sqrt{1}})


    (frac{2}{sqrt{-2}})


  • Find the value without using calculator .

    Cos 105(^0)

    (frac{1-sqrt{1}}{1sqrt{1}})


    (frac{1-sqrt{3}}{2sqrt{2}})


    (frac{1-sqrt{2}}{2sqrt{2}})


    (frac{1-sqrt{2}}{2sqrt{3}})


  • Find the value without using calculator . 

    tan 15(^o)

    -2 + (sqrt{-3})


    3 - (sqrt{-3})


    2 - (sqrt{3})


    2 - (sqrt{-3})


  • Evaluate without using a calculator.

    tan 165 (^o)

    (sqrt{2}) - 3


    1.005423


    (sqrt{3}) - 2


    (sqrt{2}) - 1


  • tan 70(^o) = tan 20(^o) + 2 tan 50 (^o)

     1 (frac{24}{24})


     1.009 , 12


    1 (frac{25}{24})


    1 (frac{24}{25})


  • tan 50 (^o) = tan 40 (^o) + 2 tan 10 (^o)

     - (frac{119}{169}) , (frac{120}{169})


     - (frac{119}{119}) , (frac{120}{119})


     - (frac{112}{162}) , (frac{122}{129})


    tan 60


  • tan 65 (^0) - tan 25 (^0) = 2 tan 40 (^o)

     - (frac{119}{169}) , (frac{120}{169})


    11


    1


    0.554


  • If cot α = (frac{1}{2}) and sec β = (frac{5}{3}) , find tan ( α + β).

    -1.009


    (frac{1}{2})


    -1.224


    3


  • If sin A = (frac{3}{5}) & sin B = (frac{12}{13}) , find the value of cos (A - B).

    (frac{54}{65})


    (frac{56}{65})


    (frac{66}{65})


    (frac{55}{55})


  • If cosα = (frac{4}{5}) and cosβ = (frac{12}{13}) then find the value of cos (α+β).

     

    (frac{36}{35})


    (frac{33}{35})


    (frac{33}{65})


    1.090


  • If sin A = (frac{3}{5}) and cos B = (frac{15}{17}) then find the value of sin (A + B).

    (frac{77}{85})


    (frac{77}{77})


    (frac{80}{84})


    (frac{85}{85})


  • If tan A = (frac{3}{4}) and tan B = (frac{5}{12}) then find the value of sin (A - B).

    (frac{15}{65})


    (frac{65}{15})


    (frac{10}{60})


    (frac{12}{60})


  • You scored /15


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Discussions about this note

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Arun

what is the value of tan 3A

Samiksha yogi

Find the value of:Sin70-cos80 cos140

shruti

what is value of cot3A=?

sasmeeta bhandari

2tan50 tan20 is equals to cot20