Trigonometric Functions

(i) Given: 25°

We know that 1° = radians

∴ 25° = 25 × = radians

∴ 25° = radians

(ii) Given: – 47°30′

We know that 1° = 60’

⇒ 1’ = °

∴ -30’ = - ° = - 0.5°

∴ -47° 30’ = -47.5°

We know that 1° = radians

∴ -47.5° = -47.5 × = radians

∴ -47.5° = radians

(iii) Given: 240°

We know that 1° = radians

∴ 240° = 240 × = radians

∴ 240° = radians

(iv) Given: 520°

We know that 1° = radians

∴ 520° = 520 × = radians

∴ 520° = radians

(i) Given: radians

We know that 1 radian = °

∴ = × = × = 39° + 3/8°

Therefore,

radian = 39° 22' 30''

(ii) Given: - 4 radians

We know that 1 radian = °

∴ -4 = - 4 × = -4 × = -229.09°

Now 1º = 60'

0.09º = 0.09 × 60 = 5.4'

Therefore, - 4 radians = - 229º 5.4'

∴ -4 radian = 229.09°

(iii) Given: radian

We know that 1 radian = °

∴ = × = 300°

∴ radian = 300°

(iv) Given: radian

We know that 1 radian = °

∴ = × = 210°

∴ radian = 210°

Given: A wheel makes 360 revolutions in one minute.

Here, Number of revolutions made in 1 second = = 6 revolutions/sec (∵ 1 min = 60 sec)

We know that a wheel is in circular shape and any circular structure makes 360° in 1 complete revolution.

∴ 360° = 360 × = 2π radians. (for 1 revolution)

Therefore, for 6 revolutions

2π × 6 radians = 12π radians

∴ A wheel revolves 12π radians in 1 second.

Given: l = 22 cm (length of arc) and r = 100 cm (radius of circle)

We know that θ = (here, θ is angle subtended by arc)

∴ θ = = 0.22 radians

We know that,

∴ 0.22 radian = 0.22 × = 0.22 ×

0.22 radian = (12.6)°

Also we know that, 1° = 60'

So 0.6° = 0.6 × 60 = 36'

Hence, 12°36'

∴ Angle subtended by arc of length 22 cm at the centre of the circle of radius 100 cm is 12°36’.

Given: A circle of diameter 40 cm ⇒ radius = 20cm

The length of a chord is 20 cm.

Here, we can see that, the figure forms a equilateral triangle with side length 20 cm

(both radii = 20cm and length of chord = 20cm)

∴ ∠AOB = 60° (∵ Angles in an equilateral triangle are 60° each)

⇒ 60° = 60 × = radians

That is, The Chord AB makes radians at the centre of the circle.

Now, We know that θ = (here, θ is angle subtended by arc)

∴

⇒ l =

∴ Length of the arc AB is cm.

Given: Two circles with arcs of the same length and subtend angles 60° and 75° at the centre respectively.

Let L be the length of the arcs..

Let r₁ be the radius of the first circle and let r₂ be the radius of the second circle.

Now for Circle with centre O,


And Circle with Centre O',

r₁: r₂ = 5:4

(i) Given: l = 10cm (length of arc) and r = 75cm(radius of circle)

We know that θ = (here, θ is angle subtended by arc)

∴ θ = = radians

∴The angle by which the pendulum of length 75 cm swings if it describes an arc of 21 cm is radians.

(ii) Given: l = 15cm (length of arc) and r = 75cm(radius of circle)

We know that θ = (here, θ is angle subtended by arc)

∴ θ = = radians.

∴The angle by which the pendulum of length 75 cm swings if it describes an arc of 21 cm is radians.

(iii) Given: l = 21 cm (length of arc) and r = 75 cm (radius of circle)

We know that θ = (here, θ is angle subtended by arc)

∴ θ = = radians

∴ The angle by which the pendulum of length 75 cm swings if it describes an arc of 21 cm is radians.

Given:, x lies in 3^rd quadrant

Sign Convention: For knowing the sign of Identities we will take the Graph Shown below which shows in which Quadrant which Trigonometric Ratio is Positive.



We know the identity:

sin ² x + cos ² x = 1

(∵ x lies in 3^rd quadrant)

Now,

Given, , x lies in 2^nd quadrant

sin²x + cos²x = 1

(∵ x lies in 2^nd quadrant)

Now,

Given, , x lies in 3^rd quadrant

To find: sin x, cos x, tan x, sec x, cosec x




Applying the formula : cosec² x - cot² x = 1

1 + cot²x = cosec²x

(∵ x lies in 3^rd quadrant)

Now,

∴ sin²x + cos²x = 1

( As it is the third quadrant)

So all the trigonometric Ratios are

Given, , x lies in 4^th quadrant

1 + tan²x = sec²x

tan²x = sec²x – 1

(∵ x lies in 4^th quadrant)

Now,

∴ sin²x + cos²x = 1

Given, , x lies in 2^nd quadrant

sec²x – tan²x = 1

(∵ x lies in 2^nd quadrant)

∴ sin²x + cos²x = 1

We know that,

Value of sinx repeat after an interval of 2π or 360⁰.

Sin765⁰ = sin(2 × 360⁰ + 45⁰)

= sin45⁰ [∵ sin(2nπ + θ = sinθ)]

We know that,

Value of cosecx repeat after an interval of 2π or 360⁰.

cosec(-1410⁰) = cosec(90⁰ × 15 + 60⁰)

= -(-)sec60⁰ [∵ cosec(90⁰ + θ) = secθ]

= 2

We know that,

Value of tanx repeat after an interval of 2π or 360⁰.

= tan60⁰

We know that,

Value of sin x repeat after an interval of 2π or 360⁰.



This means that: f( x + 2 n π) = f(x)

We know that,

Value of cot x repeat after an interval of 2π or 360⁰.

= 1

To Prove

∴ L.H.S = R.H.S

Hence, Proved.

To Prove:

∴ LHS = RHS

Hence, Proved

To Prove

RHS = 6

LHS = 3 + 1 + 2 = 6

∴ LHS = RHS

Hence, proved.

To Prove

RHS = 10

LHS = 1 + 1 + 8 = 10

∴ LHS = RHS

Hence, Proved.

sin 75° = sin (30° + 45°)

we know, sin (x + y) = sin x × cos y + cos x × sin y

so, sin 75° = sin (30° + 45°)

sin 75° = sin 30° × cos 45° + cos 30° × sin 45°

∴

tan 15° = tan (45° - 30°)

tan 15° = tan (45° -30°)

tan 15° = 2 - √3

To Prove:

Formula used : cos A cos B - sin A sin B = cos(A + B)


Proof :

Since, [cos A cos B – sin A sin B = cos (A + B)]

Here

∴

[ By complementary angle formula cos(90° - A) = sin A]

∴ LHS = RHS

Hence, proved.

To Prove:

Proof:

Using formula:



And we know that:



So L.H.S becomes,

∴ LHS = RHS

Hence, proved.

To Prove

RHS = cot² x

∴ LHS = RHS

Hence, Proved.

To Prove:

we know that,



Therefore, L.H.S becomes,

L.H.S = sin x. sin x + cos x. cos x

L.H.S = sin² x + cos² x

Since, sin² x + cos² x = 1

So, LHS = 1

∴ LHS = RHS

Hence, proved.

To Prove: sin (n + 1)x sin (n + 2)x + cos (n + 1)x cos (n + 2)x = cos x

Proof:

R.H.S = cos x

L.H.S = sin (n + 1)x sin (n + 2)x + cos (n + 1)x cos (n + 2)x

[Since, cos (A - B)=cos A cos B + sin A sin B ]

sin (n + 1)x sin (n + 2)x + cos (n + 1)x cos (n + 2)x = cos[(n + 1) x - (n + 2)x] ......[ A = (n +1)x and B = (n + 2)x ]

sin (n + 1)x sin (n + 2)x + cos (n + 1)x cos (n + 2)x = cos(-x) = cos x

[As cos (- x) = cos x]

L.H.S= R.H.S

Hence, proved

To Prove:



Formula to use :

Proof:

We know,

LHS = -√2 sin x

∴ LHS = RHS

Hence, proved

To Prove: sin² 6x – sin² 4x = sin 2x sin 10x
Proof:

RHS = sin 2x sin 10x

LHS = sin² 6x – sin²4x

we know that,

[a² – b² = (a – b)(a + b)]

Rearranging we get

LHS = (2 × sin x × cos x)(2 × sin 5x × cos 5x)

LHS = sin 2x sin 10x [∵2sin A cos A = sin 2A]

∴ LHS = RHS

Hence, proved

To prove cos² 2x – cos² 6x = sin 4x sin 8x

RHS = sin 4x sin 8x

LHS = cos² 2x – cos² 6x

= (1 – sin² 2x) – (1 – sin² 6x)

= 1 – sin² 2x – 1 + sin² 6x

= sin² 6x – sin² 2x

= (sin 6x – sin 2x) (sin 6x + sin 2x)

[a² – b² = (a – b)(a + b)]

Rearranging we get

LHS = (2 × sin 2x × cos 2x)(2 × sin 4x × cos 4x)

LHS = sin 4x sin 8x


[∵2sin A cos A = sin 2A]

∴ LHS = RHS

Hence, proved.

To prove: sin 2x + 2 sin 4x + sin 6x = 4 cos²x sin 4x
Formula to use:

Proof:

L.H.S = sin 2x + 2 sin 4x + sin 6x

L.H.S = 2 sin 4x + (sin 6x + sin 2x)

We know,

L.H.S = 2 sin 4x + 2 sin 4x cos 2x

= 2 sin 4x × (1 + cos 2x)

= 2 sin 4x × 2 cos² x [∵cos2x = 2cos² x – 1]

L.H.S = 4 cos² x sin 4x
R.H.S = 4 cos² x sin 4x

∴ L.H.S = R.H.S

Hence, proved.

To prove cot4x (sin5x + sin3x) = cot x (sin5x – sin3x)

LHS = cot 4x (sin 5x + sin 3x)

We know,

LHS = cot 4x × 2 sin 4x × cos x

LHS = 2 cos 4x cos x

RHS = cot x (sin 5x – sin 3x)

We know,

RHS = cot x × 2 cos 4x × sin x

RHS = 2 cos 4x cos x

∴ LHS = RHS

Hence, proved

To prove

∴ LHS = RHS

Hence, proved.

To Prove

RHS = tan 4x

LHS = tan 4x

∴ LHS = RHS

Hence, proved.

To prove

∴ LHS = RHS

Hence, proved.

To prove

RHS = tan 2x

.

LHS = tan 2x

∴ LHS = RHS

Hence, proved.

To prove:

Proof:

∴ LHS = RHS

Hence, proved.

To prove:

RHS = cot 3x

LHS = cot 3x

∴ LHS = RHS

Hence, proved.

To prove: cot x cot 2x – cot 2x cot 3x – cot 3x cot x = 1

Formula to use:

LHS = cot x cot 2x – cot 2x cot 3x – cot 3x cot x

= cot x cot 2x – cot 3x (cot 2x + cot x)

= cot x cot 2x – cot (2x + x) (cot 2x + cot x)

LHS = cot x cot 2x – (cot x cot 2x – 1)

= cot x cot 2x – cot x cot 2x + 1

= 1

∴ LHS = RHS

Hence, proved.

To prove

LHS = tan 4x = tan 2(2x)

∴ LHS = RHS

Hence, proved.

To prove cos 4x = 1 – 8sin² x cos² x

RHS = 1 – 8sin² x cos² x

LHS = cos 4x

= cos 2(2x)

= 1 – 2 sin² 2x [cos 2A = 1 – 2 sin2 A]

= 1 – 2(2 sin x cos x)²[sin2A = 2sin A cosA]

= 1 – 8 sin² x cos² x

∴ LHS = RHS

Hence, proved.

To prove cos 6x = 32 cos⁶ x – 48cos⁴ x + 18 cos² x – 1

RHS = 32 cos⁶ x –48cos⁴ x +18 cos² x – 1

LHS = cos 6x

= cos 3(2x)

= 4 cos³ 2x – 3 cos 2x

[As cos 3A = 4 cos3 A – 3 cos A]

= 4 [(2 cos² x – 1)³ – 3 (2 cos² x – 1) [cos 2x = 2 cos2 x – 1]

= 4 [(2 cos² x)³ – (1)³ – 3 (2 cos² x)² + 3 (2 cos² x)] – 6cos² x + 3

= 4 [8cos⁶ x – 1 – 12 cos⁴ x + 6 cos² x] – 6 cos² x + 3

= 32 cos⁶ x – 4 – 48 cos⁴ x + 24 cos² x – 6 cos² x + 3

= 32 cos⁶ x – 48 cos⁴ x + 18 cos² x – 1

∴ LHS = RHS

Hence, Proved.

Given

It`s known that and

∴ the principal solutions of the equation are

The general solution is

where n ϵ Z and Z denotes sets of integer

∴ the general solution of the equation is where n ϵ Z and Z denotes sets of integer

Given equation is sec x= 2

Now

∴ the principal solutions of the equation are

The general solution is given by

⇒

⇒ where n ϵ Z and Z is set of integers

∴ General solution of the equation is

where n ϵ Z and Z is set of integers

Given equation is

It`s known that

Also

⇒

∴ the principal solutions of the equation are

Now when,

⇒

⇒

Hence the general solution of the equation is where n ϵ Z and Z is set of integers

Given equation is cosec x = -2

We know that

∴

⇒

∴ the principal solutions are

Now considering

⇒

⇒

Given cos 4 x = cos 2 x

⇒ cos 4x – cos 2x = 0

⇒

⇒ sin 3x sin x = 0

⇒ sin 3x = 0 or sin x = 0

∴

⇒

Given equations is cos 3x + cos x – cox 2x = 0

⇒

⇒ 2 cos2x cos x – cos 2x = 0

⇒ cos 2x( 2cos x – 1) = 0

⇒ cos 2x = 0 or 2 cos x – 1 = 0

⇒ cos 2x = 0 or cos x = 1/2

We know that the general solution for cosx = 0 is x = (2n + 1) π/2, where n ϵ Z and Z is set of integers

Therefore,

∴where n ϵ Z and Z is set of integers

⇒ where n ∈Z and z is set of integers

Given equation: sin 2 x + cos x= 0

⇒ 2 sin x cos x + cos x = 0 (∵ sin 2A = 2 sin A cosA)

⇒ cos x (2 sin x + 1) = 0

⇒ cos x = 0 or 2 sin x +1 = 0

Now cos x = 0

⇒

And 2 sin x + 1= 0

⇒

⇒

∴ the general solution of the equation is

Given equation is

⇒

⇒

⇒ tan 2x (tan 2x +1) = 0

⇒ tan 2x = 0 or (tan 2x +1) = 0

Now tan 2x = 0

⇒ tan 2x = tan 0

⇒

⇒

Now (tan 2x +1) = 0

⇒

⇒ where n ∈Z

⇒ where n ∈Z and Z is set of integers

The general solution of the equation is

where n ∈Z and Z is set of integers

Given equation is

sin x + sin 3x + sin 5x = 0

⇒ (sin x+ sin 5x) + sin 3x = 0

⇒ + sin 3x = 0

⇒ 2 sin 3x cos (-2x) + sin 3x= 0

⇒ sin 3 x (2 cos 2 x + 1) = 0

⇒ sin 3x = 0 or 2 cos 2x +1 = 0

Now,

⇒

Now, 2 cos 2x +1 = 0

⇒

⇒

⇒ where n ∈ Z and Z is set of integers

⇒ where n ∈ Z

∴ the general solution of the equation is

where n ∈ Z and Z is set of integers

We know that-

2 cos x cos y = cos(x + y) + cos(x - y)

Putting x = (9π/13) and y = (π/13) in (1), we get-

…(1)

∴ L.H.S

[From (1)]

since cos (π/2) = 0

∴ L.H.S = 0 + 0 = 0 = R.H.S

Hence, L.H.S = R.H.S…Proved

To Prove: (sin 3x + sin x) sin x + (cos 3x – cos x) cos x = 0

Formulas to use:

Proof:

Replacing x with 3x and y with x in the formula, we get-

∴ sin 3x + sin x = 2 sin 2x cos x …(1)

Similarly,

We know that

Replacing x with 3x and y with x, we get-

∴ cos 3x - cos x = -2 sin 2x sin x …(2)

Now,

L.H.S = (sin 3x + sin x) sin x + (cos 3x – cos x) cos x

Putting the values obtained from equations 1 and 2 we get,

= (2 sin 2x cos x) sin x + (-2 sin 2x sin x) cos x

= 2 sin 2x cos x sin x - 2 sin 2x sin x cos x

= 0

= R.H.S

Hence, L.H.S = R.H.S…Proved

To Prove: (cos x + cos y)² + (sin x - sin y)² = 4 cos²[(x + y)/2]

Formulas Used:

L.H.S = (cos x + cos y)² + (sin x - sin y)²


Applying the above formulas we get,

= 4 cos²[(x + y)/2]×1 [∵ sin²x + cos²x = 1]

= 4 cos²[(x + y)/2]

= R.H.S

Hence, L.H.S = R.H.S…Proved

Proof:

We know that:

Applying the formulas and putting in L.H.S we get,

L.H.S = (cos x - cos y)² + (sin x - sin y)²

[∵ sin²x + cos²x = 1]

= 4 sin²[(x-y)/2]×1

= 4 sin²[(x-y)/2]

= R.H.S

Hence, L.H.S = R.H.S…Proved

We know that-

Now,

L.H.S = sin x + sin 3x + sin 5x + sin 7x

= (sin x + sin 7x) + (sin 3x + sin 5x)

= 2 sin 4x cos 3x + 2 sin 4x cos x

= 2 sin4x (cos 3x + cos x)

= 2 sin 4x (2 cos 2x cos x)

= 4 cos x cos 2x sin 4x

= R.H.S

Hence, L.H.S = R.H.S…Proved

We know that-

Now,

L.H.S

= tan 6x

= R.H.S

Hence, L.H.S = R.H.S…Proved

To Prove:

Formulas to use for proof:

Proof:

L.H.S = sin 3 x + sin 2 x - sin x

= (sin 3 x - sin x) + sin 2 x

= 2 cos 2x sin x + 2 sin x cos x [∵ sin 2x = 2 sin x cos x]

= 2 sin x (cos 2x + cos x)

= R.H.S

Hence, L.H.S = R.H.S…Proved

Given that x is in quadrant II

So,

90° < x < 180°

Dividing with 2 all sides

(90°/2) < x/2 < (180°/2)

45° < x/2 < 90°

∴ lies in 1^st quadrant

In 1^st quadrant,

sin, cos & tan are positive

⇒ are positive

Given

tan x = -(4/3)

We know that

Replacing x with x/2

Replacing by a

⇒ 4a²-6a-4 = 0

⇒ 4a²-8a+2a-4 = 0

⇒ 4a(a-2)+2(a-2) = 0

⇒ (4a+2)(a-2) = 0

∴ a = -1/2 or a = 2

Hence,

or

∵ lies in 1^st quadrant

is positive

∴

Now,

We know that

1 + tan²x = sec²x

Replacing x with x/2

∵ lies in 1^st quadrant.

is positive in 1^st quadrant.

We know that-

Replacing x with x/2

Hence,

Given that x is in quadrant II

So,

180° < x < 270°

Dividing with 2 all sides

(180°/2) < x/2 < (270°/2)

90° < x/2 < 135°

∴ lies in 2^nd quadrant

In 2^nd quadrant,

sin is positive and cos, tan are negative

⇒ is positive and are negative

Given

cos x = -(1/3)

We know that

cos 2x = 2 cos²x - 1

Replacing x with x/2

cos 2(x/2) = 2 cos²(x/2) - 1

⇒ cos x = 2 cos²(x/2) - 1

⇒ -1/3 = 2 cos²(x/2) - 1

⇒ 2/3 = 2 cos²(x/2)

⇒ cos²(x/2) = 1/3

∴ cos(x/2) = (1/√3)

∵ lies in 2^nd quadrant

is negative

∴

Now,

We know that

1 + tan²x = sec²x

Replacing x with x/2

∵ lies in 2^nd quadrant

is positive in 2^nd quadrant

We know that-

Replacing x with x/2

Hence,

Given that x is in quadrant II

So,

90° < x < 180°

Dividing with 2 all sides

(90°/2) < x/2 < (180°/2)

45° < x/2 < 90°

∴ lies in 1^st quadrant

In 1^st quadrant,

sin, cos & tan all are positive

⇒ all are positive

Given

sin x = (1/4)

We know that

cos²x = 1 - sin²x

= 1 - (1/4)²

= 1 - (1/16)

∴ cos²x = 15/16

⇒ cos x = (√15)/4

Since x is in II^nd quadrant

∴ cos x is negative

⇒ cos x = - (√15)/4

Also,

cos 2x = 2 cos²x - 1

Replacing x with x/2

cos 2(x/2) = 2 cos²(x/2) - 1

⇒ cos x = 2 cos²(x/2) - 1

⇒ - (√15)/4 = 2 cos²(x/2) - 1

⇒ 1- (√15)/4 = 2 cos²(x/2)

⇒ cos²(x/2) = [4 - (√15)]/8

∵ lies in 1^st quadrant

is positive

Now,

We know that

1 + tan²x = sec²x

Replacing x with x/2

∵ lies in 1^st quadrant

is positive in 1^st quadrant

We know that-

Replacing x with x/2

Hence,