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```Trigonometric Ratios of Compound Angles

An angle made up of the algebraic sum of two or more angles is called a compound angle. Some of the formulae and results regarding compound angles are:

sin(A + B) = sinA cosB + coA 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) = tanA + tan B/ 1 - tan A tanB

tan(A – B) = tanA - tan B/ 1 + tan A tan B

sin(A + B) sin(A – B) = sin2A – sin2B = cos2B – cos2A.

cos(A + B) = cos(A – B) = cos2A – sin2A – sin2B = cos2B – sin2A.

Illustration:

Prove that tan70o = 2 tan50o + tan20o.

Solution:

tan70o = tan(50o + 20o)

= tan 50o + tan 20o/1 - tan 50o. tan 20o

or, tan70o (1 – tan 50o tan20o) = tan50o + tan20o

or, tan70o = tan70o tan50o tan20o + tan50o + tan20o

= cot20o tan50o tan20o + tan50o + tan20o

= 2 tan50o + tan20o.

Illustration:

If A + B = 45o, show that (1 + tanA) (1 + tanB) = 2.

Solution:

tan(A + B) = tan A + tan B / 1 - tan A tan B = 1

tanA + tanB + tanA tanB + 1 = 1 + 1

tanA (1 + tanB) + (1 + tanB) = 2

(1 + tanA) (1 + tanB) = 2

Trigonometric Ratios of Multiples of an Angle

sin2A = 2sinA cosA = 2tan A/ 1 + tan2 A

cos2A = cos2A – sin2A = 1 – 2sin2A = 2 cos2A – 1 = 1 - tan2 A/1 + tan2 A

tan2A = 2tan A/ 1 - tan2 A

sin3A = 3sinA – 4sin3A = 4sin(60o – A) sinA sin(60o + A)

cos3A = 4cos3A – 3cosA = 4cos(60o A) cosA cos(60o + A)

tan3A = 3tan A - tan3 A/1 - 3 tan2 A = tan(60o – A) tan A tan (60o + A)

Illustration:

Find the values of (i) sin 18o (ii) tan 15o

Solution:

(i) sin 18o

Let θ = 18o then 2θ = 36o = 90o – 3θ.

Now sin2θ = 2sinθ cosθ and

sin(90o – 3θ) = cos3θ = 4cos3θ – 3cosθ

Hence we have 2sinθcosθ = cosθ (4cos2θ – 3) = cosθ (1 – 4sin2θ).

Hence, 2 sinθ = 1 – 4sin2θ (as cosθ # 0)

⇒ 4sin2θ + 2sinθ – 1 = 0 ⇒ sinθ = -2 + √4 + 16 /2.4 = -1 + √5/4

But as sinθ > 0 we have sinθ = √5 - 1/4 i.e. sin18o = √5 + 1/4

(ii)  tan 150o

tan 150o = tan (60o – 45o) = √3 - 1/ 1 + √3 = (√3 - 1)2/3 -1 = 4 - 2 √3/2 = 2 - √3 .

Sum of sines/Cosines in Terms of Products

sin A + sinB = 2sin A + B/2  cos A - B/2

sin A – sin B = 2sin A - B/2  cos A + B/2

cos A + cosB = 2cos A - B/2  cos A + B/2

cos A – cos B = –2sin A - B/2  sin A + B/2  (here notice (B – A))

tan A + tanB = sin (A + B )/ cos A. cos B

Conversely

2sinA cosB = sin(A + B) + sin (A – B)

2cosA sinB = sin(A + B) – sin (A – B)

2cosA cosB = cos(A + B) + cos (A – B)

2sinA sinB = cos(A – B) – cos (A + B)

Illustration:

If α, ß and g are in A.P., show that cotß = sin α - sinγ/ cosγ - cosα .

Solution:

Since α, ß and y are in A.P., 2ß = α + y ⇒ cotß = α+y/2

= (cosα+y/2) / sinα+y/2 = (2cosα+y/2 sinα+y/2)/ 2sinα+y/2sinα+y/2 = sin α - siny/ cos y - cosα  .

Illustration:

Show that sin 12o.sin48o.sin54o = 1/8.

Solution:

L.H.S. =  [cos 36o – cos 60o]sin 54o = 1/2  [cos 36o sin 54o  sin 54o]

= 1/4  [2 cos 36o sin 54o – cos 54o] = 1/4 [sin 90o + sin 18o – sin 54o]

= 1/4  [1 – (sin 54o– cos 18o)] = 1/4  [1 – 2sin 18o cos 36o]

= 1/4 [ 1 - 2sin 18° cos36°] = 1/4 [1 - sin36° cos 36°/ cos18° ]

= 1/4 [ 1 - 2sin 36° cos6°/2cos18°] = 1/4 [ 1 - sin 72°/2sin72°] = 1/4 [ 1 - 1/2 ] = 1/8 =  R.H.S.

Alternative Method

Let θ = 12o.

L.H.S. = 1/sin72° sin 12o sin 48o sin 72o sin 54o

= 1/ 4sin 72° 4 sin(60o – 12o) sin 12o sin (60o + 12o) sin 54o.

= 1/4 sin3 (12°) sin 54°/ sin72° = sin 36° sin 54°/ 8 sin 36° cos36° = 1/8  = R.H.S.

To read more, Buy study materials of Inverse Trigonometric Functions comprising study notes, revision notes, video lectures, previous year solved questions etc. Also browse for more study materials on Mathematics here.
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USE CODE: CHEM20