Trigonometric, Or Circular, Functions

Given:

Since, θ is in III^rd Quadrant. So, sin and cos will be negative but tan will be positive.

We know that,

cos² θ + sin² θ = 1

Putting the values, we get

[given]

Since, θ in III^rd quadrant and sinθ is negative in III^rd quadrant

Now,

Putting the values, we get

Now,

Putting the values, we get

= -2

Now,

Putting the values, we get

Now,

Putting the values, we get

=√3

Hence, the values of other trigonometric Functions are:

Given:

Since, θ is in IV^th Quadrant. So, sin and tan will be negative but cos will be positive.

We know that,

sin² θ + cos² θ = 1

Putting the values, we get

[given]

Since, θ in IV^th quadrant and cosθ is positive in IV^th quadrant

Now,

Putting the values, we get

Now,

Putting the values, we get

= -2

Now,

Putting the values, we get

Now,

Putting the values, we get

=-√3

Hence, the values of other trigonometric Functions are:

Given:

Since, θ is in II^nd Quadrant. So, cos and tan will be negative but sin will be positive.

Now, we know that

Putting the values, we get

…(i)

We know that,

sin² θ + cos² θ = 1

Putting the values, we get

[from (i)]

Since, θ in II^nd quadrant and cosθ is negative in II^nd quadrant

Now,

Putting the values, we get

Now,

Putting the values, we get

Now,

Putting the values, we get

Hence, the values of other trigonometric Functions are:

Given: sec θ = √2

Since, θ is in IV^th Quadrant. So, sin and tan will be negative but cos will be positive.

Now, we know that

Putting the values, we get

…(i)

We know that,

cos² θ + sin² θ = 1

Putting the values, we get

[given]

Since, θ in IV^th quadrant and sinθ is negative in IV^th quadrant

Now,

Putting the values, we get

=-1

Now,

Putting the values, we get

= -√2

Now,

Putting the values, we get

= -1

Hence, the values of other trigonometric Functions are:

Given:

To find: cos x and cot x

Since, x is in III^rd Quadrant. So, sin and cos will be negative but tan will be positive.

We know that,

sin² x + cos² x = 1

Putting the values, we get

[given]

Since, x in III^rd quadrant and cos x is negative in III^rd quadrant

Now,

Putting the values, we get

= 2√6

Now,

Putting the values, we get

Hence, the values of other trigonometric Functions are:

Given:

To find: value of sinx

Given that:

So, x lies in II^nd quadrant and sin will be positive.

We know that,

cos² θ + sin² θ = 1

Putting the values, we get

[given]

Since, x in II^nd quadrant and sinθ is positive in II^nd quadrant

Given: sec x = -2

Given that:

So, x lies in III^rd Quadrant. So, sin and cos will be negative but tan will be positive.

Now, we know that

Putting the values, we get

…(i)

We know that,

cos² x + sin² x = 1

Putting the values, we get

[given]

Since, x in III^rd quadrant and sinx is negative in III^rd quadrant

Now,

Putting the values, we get

=√3

Now,

Putting the values, we get

Now,

Putting the values, we get

Hence, the values of other trigonometric Functions are:

To find: Value of

Value of sin x repeats after an interval of 2π, hence ignoring 5 × (2π)

= sin 60°

To find: Value of

Value of cos x repeats after an interval of 2π, hence ignoring 4 × (2π)

= cos 90°

= 0 [∵ cos 90° = 1]

To find: Value of

We know that,

tan(-θ) = - tan θ

Value of tan x repeats after an interval of 2π, hence ignoring 4 × (2π)

= - tan 60°

= -√3

[∵ tan 60° = √3]

To find: Value of

We have,

Putting π = 180°

= cot (13 × 45°)

= cot (585°)

= cot [90° × 6 + 45°]

= cot 45°

[Clearly, 585° is in III^rd Quadrant and the multiple of 90° is even]

= 1 [∵ cot 45° = 1]

To find: Value of

We have,

[∵ sec(-θ) = sec θ]

Putting π = 180°

= sec[25 × 60°]

= sec[1500°]

= sec [90° × 16 + 60°]

Clearly, 1500° is in I^st Quadrant and the multiple of 90° is even

= sec 60°

= 2

To find: Value of

We have,

[∵ cosec(-θ) = -cosec θ]

Putting π = 180°

= -cosec[41 × 45°]

= -cosec[1845°]

= -cosec [90° × 20 + 45°]

Clearly, 1845° is in I^st Quadrant and the multiple of 90° is even

= -cosec 45°

= -√2

To find: Value of sin 405°

We have,

sin 405° = sin [90° × 4 + 45°]

= sin 45°

[Clearly, 405° is in I^st Quadrant and the multiple of 90° is even]

To find: Value of sec (-1470°)

We have,

sec (-1470°) = sec (1470°)

[∵ sec(-θ) = sec θ]

= sec [90° × 16 + 30°]

Clearly, 1470° is in I^st Quadrant and the multiple of 90° is even

= sec 30°

To find: Value of tan (-300°)

We have,

tan (-300°) = - tan (300°)

[∵ tan(-θ) = -tan θ]

= - tan [90° × 3 + 30°]

Clearly, 300° is in IV^th Quadrant and the multiple of 90° is odd

= - cot 30°

= -√3

To find: Value of

We have,

cot (585°) = cot [90° × 6 + 45°]

= cot 45°

[Clearly, 585° is in III^rd Quadrant and the multiple of 90° is even]

= 1 [∵ cot 45° = 1]

To find: Value of cosec (-750°)

We have,

cosec (-750°) = - cosec(750°)

[∵ cosec(-θ) = -cosec θ]

= - cosec [90° × 8 + 30°]

Clearly, 405° is in I^st Quadrant and the multiple of 90° is even

= - cosec 30°

= -2 [∵ cosec 30° = 2]

To find: Value of cos 2220°

We have,

cos (-2220^°) = cos 2220°

[∵ cos(-θ) = cos θ]

= cos [2160 + 60°]

= cos [360° × 6 + 60°]

= cos 60°

[Clearly, 2220° is in I^st Quadrant and the multiple of 360° is even]

To prove:

Taking LHS,

Putting π = 180°

= tan² 60° + 2 cos² 45° + 3 sec² 30° + 4 cos² 90°

Now, we know that,

Putting the values, we get

= 3 + 1 + 4

= 8

= RHS

∴ LHS = RHS

Hence Proved

To prove:

Taking LHS,

Putting π = 180°

= sin 30° cos 0° + sin 45° cos 45° + sin 60° cos 30°

Now, we know that,

Putting the values, we get

= RHS

∴ LHS = RHS

Hence Proved

To prove:

Taking LHS,

Putting π = 180°

= 4 sin 30° sin² 60° + 3 cos 60° tan 45° + cosec² 90°

Now, we know that,

tan 45° = 1

cosec 90° = 1

Putting the values, we get

= 4

= RHS

∴ LHS = RHS

Hence Proved

(i)

Cos840° = Cos(2.360° + 120°) …………(using Cos(2ϖ + x) = Cosx)

= Cos(120°)

= Cos(180° - 60°)

= - Cos60° ……………(using Cos(ϖ - x) = - Cosx)

(ii) sin870° = sin(2.360° + 150°) ………….(using sin(2ϖ + x) = sinx)

= sin150°

= sin(180° - 30°) ………(using sin(ϖ - x) = sinx)

= sin30°

(iii)tan( - 120°) = - tan12 …….(tan( - x) = tanx)

= - tan(180° - 60°) ……. (in II quadrant tanx is negative)

= - ( - tan60°)

= tan60°

(iv)

= ………..(using cos( - x) = - cosx)

= ………...(using cos(2ϖ + x) = cosx)

=

(v)

……..(IV quadrant sinx is negative)

(vi)tan225° = tan(180° + 45°) …………(in III quadrant tanx is positive)

(vii)

.….(tan( - x) = - tanx)

…..(in IV quadrant tanx is negative)

(viii)sin( - 1230°) = sin1230° ………….(using sin( - x) = sinx)

= sin(3.360° + 150°)

= sin150°

= sin(180° - 30°) ………….(using sin(180° - x) = sinx)

= sin30°

(ix)cos495° = cos(360° + 135°) …………(using cos(360° + x) = cosx)

= cos135°

= cos(180° - 45°) ………….(using cos(180° - x) = - cosx)

= - cos45°

Sin135° = sin(180° - 45°) …….....(using sin(180° - x) = sinx)

Cos135° = cos(180° - 45°) .….…..(using cos(180° - x) = - cosx)

Tan135° =

Cosec135° =

Sec135° =

Cot135° =

(i)sin80°cos20° - cos80°sin20° = sin(80° - 20°)

(using sin(A - B) = sinAcosB - cosAsinB)

= sin60°

(ii)cos45°cos15° - sin45°sin15° = cos(45° + 15°)

(using cos(A + B) = cosAcosB - sinAsinB)

= cos60°

(iii)cos75°cos15° + sin75°sin15° = cos(75° - 15°)

(using cos(A - B) = cosAcosB + sinAsinB)

= cos60°

(iv)sin40°cos20° + cos40°sin20° = sin(40° + 20°)

(using sin(A + B) = sinAcosB + cosAsinB)

= sin60°

(v)cos130°cos40° + sin130°sin40° = cos(130° - 40°)

(using cos(A - B) = cosAcosB + sinAsinB)

= cos90°

= 0

(i)sin(50° + θ)cos(20° + θ) - cos(50° + θ)sin(20° + θ)

= sin(50° + θ - (20° + θ))(using sin(A - B) = sinAcosB - cosAsinB)

= sin(50° + θ - 20° - θ)

= sin30°

(ii)cos(70° + θ)cos(10° + θ) + sin(70° + θ)sin(10° + θ)

= cos(70° + θ - (10° + θ))(using cos(A - B) = cosAcosB + sinAsinB)

= cos(70° + θ - 10° - θ)

= cos60°

(i)cos(n + 2)x.cos(n + 1)x + sin(n + 2)x.sin(n + 1)x

= sin((n + 2)x + (n + 1)x)(using cos(A - B) = cosAcosB + sinAsinB)

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

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

= cosx

(ii)

(using cos(A + B) = cosAcosB - sinAsinB)

)

= sin(x + y)

Hence, Proved.

(i)sin75° = sin(90° - 15°) .…….(using sin(A - B) = sinAcosB - cosAsinB)

= sin90°cos15° - cos90°sin15°

= 1.cos15° - 0.sin15°

= cos15°

Cos15° = cos(45° - 30°) …………(using cos(A - B) = cosAcosB + sinAsinB)

= cos45°.cos30° + sin45°.sin30°

(ii)(using sin(180° - x) = sinx)

(using cos(180° - x) = - cosx)

=

(iii)tan15° + cot15° =

First, we will calculate tan15°,

………………….(1)

Putting in eq(1),

(i) cos15⁰

Sin15⁰

Cos15⁰ - sin15⁰

(ii)cot105° - tan105° = cot(180° - 75°) - tan(180° - 75°)

(II quadrant tanx is negative and cotx as well)

= - cot75° - ( - tan75°)

= tan75° - cot75°

Tan75° =

(using sin(90° - x) = - cosx and cos(90° - x) = sinx)

Cot75° =

Cot105° - tan105° =

(iii)

(II quadrant tanx negative)

- tan45° = - 1

First we will take out cos9°common from both numerator and denominator,

°

First we will take out cos8° common from both numerator and denominator,

Using sin(90° + θ) = cosθ and sin( - θ) = sinθ,tan(90° + θ) = - cotθ

Sin(180° + θ) = - sinθ(III quadrant sinx is negative)

Using cos(90° + θ) = - sinθ(I quadrant cosx is positive

cosec( - θ) = - cosecθ

tan(270° - θ) = tan(180° + 90° - θ) = tan(90° - θ) = cotθ

(III quadrant tanx is positive)

Similarily sin(270° + θ) = - cosθ (IV quadrant sinx is negative

cot(360° - θ) = cotθ(IV quadrant cotx is negative)

(i)sin(θ - Φ) = sinθcosΦ + cosθsinΦ

=

(ii) cos(θ - Φ) = cosθ.cosΦ + sinθ.sinΦ

(iii)We will first find out the Values of tanθ and tanΦ,

tan(θ - Φ) =

,

Now we will calculate value of cos x and cosy

Sin(x + y) = sinx.cosy + cosx.siny

Now we will calculate value of sinx and siny

Hence,

Cos(x - y) = cosx.cosy + sinx.siny

,

Here we will find values of cosx and cosy

(i) sin(x + y) = sinx.cosy + cosx.siny

(ii)cos(x + y) = cosx.cosy + sinx.siny

(iii)Here first we will calculate value of tanx and tany,

We will first find out value of sinx and siny,

(i)sin(x + y) = sinx.cosy + cosx.siny

(ii)cos(x - y) = cosx.cosy + sinx.siny

=

(iii)Here first we will calculate value of tanx and tany,

(i)

(ii)

(iii)

(iv)

(i)

………[Using –2sinx.siny = cos(x + y)–cos (x–y)]

(ii)

………..[using 2cosx.cosy = cos(x + y) + cos(x–y)]

(iii)

...[Using2sinx.cosy = sin(x + y) + sin(x–y)]

In this question the following formula will be used:

Sin( A +B)= sinA cos B + cosA sinB

Sin( A - B)= sinA cos B - cosA sinB

= sin150 cosx + cos 150 sinx + sin150 cosx – cos150 sinx

=2sin150cosx

=2sin(90 + 60)cosx

=2cos60cosx

=2cosx

=cosx

In this question the following formulas will be used:

cos (A + B) = cosAcosB – sinAsinB

cos (A - B) = cosAcosB+ sinAsinB

= cos x + cos 120⁰ cosx – sin120sinx + cos 120⁰cosx+sin120sinx

= cosx + 2cos120 cosx

=cosx + 2cos (90 + 30) cosx

= cosx + 2 (-sin30) cosx

=cosx - 2 cosx

=

=0.

In this question the following formulas will be used:

sin (A - B) = sinA cos B - cosA sinB

cos (A - B) = cosAcosB+ sinAsinB

=

=

=

=

=.

In this question the following formulas will be used:

=

=

In this question the following formulas will be used:

=

=

1. = 2

=2

=2

Using,

sin( A +B)= sinA cos B + cosA sinB

= 2

= 2

= 2cos5x sin2x

Using,

sin( A - B)= sinA cos B - cosA sinB

= 2

= 2

= 2

Using,

cos (A + B) = cosAcosB – sinAsinB

4.= -2

= -2

= 2sin3x sinx

Using,

cos (A - B) = cosAcosB+ sinAsinB

i) 2sin 6x cos 4x = sin (6x+4x) + sin (6x-4x)

= sin 10x + sin 2x

Using,

2sinAcosB = sin (A+ B) + sin (A - B)

ii) 2cos 5x sin3x = sin (5x + 3x) – sin (5x – 3x)

= sin8x – sin2x

Using,

2cosAsinB = sin(A + B) – sin (A - B)

iii) 2cos7xcos3x = cos (7x+3x) + cos (7x – 3x)

=cos10x + cos 4x

Using,

2cosAcosB = cos (A+ B) + cos (A - B)

iv)2sin8xsin2x = cos (8x - 2x) – cos (8x + 2x)

= cos6x – cos10x

Using,

2sinAsinB = cos (A - B) – cos (A+ B)

=

=

=

= cotx

Using the formula,

sinA + sinB = 2sin cos

cosA - cosB = -2sinsin

=

=

=

= tanx

Using the formula,

sinA - sinB = 2cos sin

cosA + cosB = 2coscos

=

=

= tan4x

Using the formula,

sinA + sinB = 2sin cos

cosA + cosB = 2coscos

=

=

=

=

Using the formula,

cosA - cosB = -2sinsin

sinA - sinB = 2cos sin

=

=

=

=

=

= tan3x.

Using the formula,

sinA + sinB = 2sin cos

cosA + cosB = 2coscos

=

=

=

=

=

=

Using the formula,

sinA + sinB = 2sin cos

cosA + cosB = 2coscos

L.H.S

cot 4x (sin 5x + sin3x)

= cot 4x (2)

= cot 4x (2 sin4x cosx)

= (2 sin4x cosx)

= 2cos4xcosx

R.H.S

cot x (sin 5x - sin3x)

= cot x (2)

= cot x (2 cos4x sinx)

= (2 cos4x sinx)

= 2cos4xcosx

L.H.S=R.H.S

Hence, proved.

Using the formula,

sinA + sinB = 2sin cos

sinA - sinB = 2cos sin

= (sin 3x + sin x) sin x + (cos 3x – cos x) cos x

= (2sincos) sin x + (-2sin sin) cosx

= (2sin2x cosx) sinx-(2sin2x sinx) cosx

= 0.

Using the formula,

sinA + sinB = 2sin cos

cosA - cosB = -2sinsin

= (cos x – cos y)² + (sin x – sin y)²

=(-2sinsin) ² +(2cossin) ²

=4 sin²(sin²+cos²)

=4 sin²

Using the formula,

cosA - cosB = -2sinsin

sinA - sinB = 2cos sin

=

=

=

=

=)

Using the formula,

cosA - cosB = -2sinsin

sinA - sinB = 2cos sin

=

=

=

=

=

Using the formula,

cosA - cosB = -2sinsin

cosA + cosB = 2coscos

=

=

=

Using the formula,

sinA + sinB = 2sin cos

sinA - sinB = 2cos sin

=Sin3x+sin2x-sinx

= (sin3x- sinx)+sin2x

= ( 2cossin) + sin2x

= 2cos2xsinx +sin2x

= 2cos2xsinx + 2sinxcosx

= 2sinx (cos2x + cosx )

= 2sinx (2cos cos )

= 4sinxcos cos

Using the formula,

sinA - sinB = 2cos sin

cosA + cosB = 2coscos

=

=

=

=

=

=

=

=

Using the formulas,

2cosAsinB = sin (A + B) – sin (A - B)

2cosAcosB = cos (A + B) + cos (A - B)

2sinAsinB = cos (A - B) – cos (A + B)

=

=

=

=

=

=

=

=

=

Using the formulas,

2cosAsinB = sin (A + B) – sin (A - B)

2sinAsinB = cos (A - B) – cos (A + B)

L.H.S

=

=

={}

={}

={}}

={}

={}

={}

={}

={}

={}

=

=R.H.S

L.H.S

=

=

={}

={}

={}}

={}

={}

={}

={}

={}

={}

=

=R.H.S

L.H.S

=

=

={}

={}

={}}

={}

={

={}

={}

={}

={}

=

cosx+cosy = --------- i

sinx+siny = ----------ii

dividing ii by I we get,

=

=

=

Using the formula,

sinA + sinB = 2sin cos

cosA + cosB = 2coscos

L.H.S

= 2

=2

=2)

=

=

=

=

L.H.S

= 2

=2

=)

=-

=

=

L.H.S

) - )

Given:

To find: sin2x

We know that,

sin2x = 2 sinx cosx …(i)

Here, we don’t have the value of cos x. So, firstly we have to find the value of cosx

We know that,

sin²x + cos²x = 1

Putting the values, we get

Putting the value of sinx and cosx in eq. (i), we get

sin2x = 2sinx cosx

Given:

To find: cos2x

We know that,

cos 2x = 1 – 2sin²x

Putting the value, we get

To find: tan2x

From part (i) and (ii), we have

and

We know that,

Replacing x by 2x, we get

Putting the values of sin 2x and cos 2x, we get

∴ tan 2x = -4√5

Given:

To find: sin2x

We know that,

sin2x = 2 sinx cosx …(i)

Here, we don’t have the value of sin x. So, firstly we have to find the value of sinx

We know that,

cos²x + sin²x = 1

Putting the values, we get

Putting the value of sinx and cosx in eq. (i), we get

sin2x = 2sinx cosx

Given:

To find: cos2x

We know that,

cos 2x = 2cos²x – 1

Putting the value, we get

To find: tan2x

From part (i) and (ii), we have

and

We know that,

Replacing x by 2x, we get

Putting the values of sin 2x and cos 2x, we get

Given:

To find: sin 2x

We know that,

Putting the values, we get

Given:

To find: cos 2x

We know that,

Putting the values, we get

Given:

To find: tan 2x

We know that,

Putting the values, we get

Given:

To find: sin 3x

We know that,

sin 3x = 3 sinx – sin³x

Putting the values, we get

Given:

To find: cos 3x

We know that,

cos 3x = 4cos³x – 3 cosx

Putting the values, we get

cos 3x = 1

To Prove:

Taking LHS,

[∵ cos 2x = cos²x – sin²x]

Using, (a² – b²) = (a – b)(a + b)

= cos x + sin x

= RHS

∴ LHS = RHS

Hence Proved

To Prove:

Taking LHS,

[∵ sin 2x = 2 sinx cosx]

[∵ 1 + cos 2x = 2 cos²x]

= tan x

= RHS

∴ LHS = RHS

Hence Proved

To Prove:

Taking LHS,

[∵ sin 2x = 2 sinx cosx]

[∵ 1 – cos 2x = 2 sin²x]

= cot x

= RHS

∴ LHS = RHS

Hence Proved

To Prove:

Taking LHS,

[∵ sin 2x = 2 sinx cosx]

[∵ 1 + cos 2x = 2 cos²x]

= tan x

= RHS

∴ LHS = RHS

Hence Proved

To Prove: sin 2x(tan x + cot x) = 2

Taking LHS,

sin 2x(tan x + cot x)

We know that,

We know that,

sin 2x = 2 sinx cosx

= 2(sin²x + cos²x)

= 2 × 1 [∵ cos² θ + sin² θ = 1]

= 2

= RHS

∴ LHS = RHS

Hence Proved

To Prove: cosec 2x + cot 2x = cot x

Taking LHS,

= cosec 2x + cot 2x …(i)

We know that,

Replacing x by 2x, we get

So, eq. (i) becomes

[∵ 1 + cos 2x = 2 cos²x]

[∵ sin 2x = 2 sinx cosx]

= cot x

= RHS

Hence Proved

To Prove: cos 2x + 2sin²x = 1

Taking LHS,

= cos 2x + 2sin²x

= (2cos²x – 1) + 2sin²x [∵ 1 + cos 2x = 2 cos²x]

= 2(cos²x + sin²x) – 1

= 2(1) – 1 [∵ cos² θ + sin² θ = 1]

= 2 – 1

= 1

= RHS

∴ LHS = RHS

Hence Proved

To Prove: (sin x – cos x)² = 1 – sin 2x

Taking LHS,

= (sin x – cos x)²

Using,

(a – b)² = (a² + b² – 2ab)

= sin²x + cos²x – 2sinx cosx

= (sin²x + cos²x) – 2sinx cosx

= 1 – 2sinx cosx [∵ cos² θ + sin² θ = 1]

= 1 – sin2x [∵ sin 2x = 2 sinx cosx]

= RHS

∴ LHS = RHS

Hence Proved

To Prove: cot x – 2cot 2x = tan x

Taking LHS,

= cot x – 2cot 2x …(i)

We know that,

Replacing x by 2x, we get

So, eq. (i) becomes

[∵ sin 2x = 2 sinx cosx]

[∵ 1 + cos 2x = 2 cos²x]

[∵ cos² θ + sin² θ = 1]

= tan x

= RHS

∴ LHS = RHS

Hence Proved

To Prove:

Taking LHS,

= cos⁴x + sin⁴x

Adding and subtracting 2sin²x cos²x, we get

= cos⁴x + sin⁴x + 2sin²x cos²x – 2sin²x cos²x

We know that,

a² + b² + 2ab = (a + b)²

= (cos²x + sin²x) – 2sin²x cos²x

= (1) – 2sin²x cos²x [∵ cos² θ + sin² θ = 1]

= 1 – 2sin²x cos²x

Multiply and divide by 2, we get

[∵ sin 2x = 2 sinx cosx]

= RHS

∴ LHS = RHS

Hence Proved

To Prove:

Taking LHS,

…(i)

We know that,

a³ – b³ = (a – b)(a² + ab + b²)

So, cos³x – sin³x = (cosx – sinx)(cos²x + cosx sinx + sin²x)

So, eq. (i) becomes

= cos²x + cosx sinx + sin²x

= (cos²x + sin²x) + cosx sinx

= (1) + cosx sinx [∵ cos² θ + sin² θ = 1]

= 1 + cosx sinx

Multiply and Divide by 2, we get

[∵ sin 2x = 2 sinx cosx]

= RHS

∴ LHS = RHS

Hence Proved

To prove:

Taking LHS,

We know that,

1 – cos 2x = 2 sin²x & sin 2x = 2 sinx cosx

Taking sinx common from the numerator and cosx from the denominator

= tan x

= RHS

∴ LHS = RHS

Hence Proved

To Prove:

Taking LHS,

= cosx cos2x cos4x cos8x

Multiply and divide by 2sinx, we get

[∵ sin 2x = 2 sinx cosx]

Multiply and divide by 2, we get

We know that,

sin 2x = 2 sinx cosx

Replacing x by 2x, we get

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

or sin 4x = 2 sin 2x cos 2x

Multiply and divide by 2, we get

We know that,

sin 2x = 2 sinx cosx

Replacing x by 4x, we get

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

or sin 8x = 2 sin 4x cos 4x

Multiply and divide by 2, we get

We know that,

sin 2x = 2 sinx cosx

Replacing x by 8x, we get

sin 2(8x) = 2 sin(8x) cos(8x)

or sin 16x = 2 sin 8x cos 8x

= RHS

∴ LHS = RHS

Hence Proved

To Prove:

Taking LHS,

…(i)

We know that,

2sinx cosx = sin 2x

Here,

So, eq. (i) become

= sin 45^°

= RHS

∴ LHS = RHS

Hence Proved

To Prove:

Taking LHS,

= 2 cos² 15° - 1 …(i)

We know that,

1 + cos 2x = 2 cos²x

Here, x = 15°

So, eq. (i) become

= [1 + cos 2(15°)] – 1

= 1 + cos 30° - 1

= cos 30°

= RHS

∴ LHS = RHS

Hence Proved

To Prove: 8 cos³ 20° - 6 cos 20° = 1

Taking LHS,

= 8 cos³ 20° - 6 cos 20°

Taking 2 common, we get

= 2(4 cos³ 20° - 3 cos 20°) …(i)

We know that,

cos 3x = 4cos³x – 3 cosx

Here, x = 20°

So, eq. (i) becomes

= 2[cos 3(20°)]

= 2[cos 60°]

= 1

= RHS

∴ LHS = RHS

Hence Proved

To prove:

Taking LHS,

= 3 sin 40° - sin³ 40° …(i)

We know that,

sin 3x = 3 sinx – sin³x

Here, x = 40°

So, eq. (i) becomes

= sin 3(40°)

= sin 120°

= sin (180° - 60°)

= sin 60° [∵sin (180° - θ) = sin θ]

= RHS

∴ LHS = RHS

Hence Proved

To Prove:

Taking LHS,

= sin²24° - sin²6°

We know that,

sin²A – sin²B = sin(A + B) sin(A – B)

= sin(24°+ 6°) sin(24° - 6°)

= sin 30° sin 18° …(i)

Now, we will find the value of sin 18°

Let x = 18°

so, 5x = 90°

Now, we can write

2x + 3x = 90°

so 2x = 90° - 3x

Now taking sin both the sides, we get

sin2x = sin(90° - 3x)

sin2x = cos3x [as we know, sin(90°- 3x) = Cos3x ]

We know that,

sin2x = 2sinxcosx

Cos3x = 4cos³x - 3cosx

2sinxcosx = 4cos³x - 3cosx

⇒ 2sinxcosx - 4cos³x + 3cosx = 0

⇒ cosx (2sinx - 4cos²x + 3) = 0

Now dividing both side by cosx we get,

2sinx - 4cos²x + 3 = 0

We know that,

cos²x + sin²x = 1

or cos²x = 1 – sin²x

⇒ 2sinx – 4(1 – sin²x) + 3 = 0

⇒ 2sinx – 4 + 4sin²x + 3 = 0

⇒ 2sinx + 4sin²x – 1 = 0

We can write it as,

4sin²x + 2sinx - 1 = 0

Now applying formula

Here, ax² + bx + c = 0

So,

now applying it in the equation

Now sin 18° is positive, as 18° lies in first quadrant.

Putting the value in eq. (i), we get

= sin 30° sin 18°

= RHS

∴ LHS = RHS

Hence Proved

To Prove:

Taking LHS,

= sin²72° - cos²30°

= sin²(90° - 18°) - cos²30°

= cos² 18° - cos²30° …(i)

Here, we don’t know the value of cos 18°. So, we have to find the value of cos 18°

Let x = 18°

so, 5x = 90°

Now, we can write

2x + 3x = 90°

so 2x = 90° - 3x

Now taking sin both the sides, we get

sin2x = sin(90° - 3x)

sin2x = cos3x [as we know, sin(90°- 3x) = Cos3x ]

We know that,

sin2x = 2sinxcosx

Cos3x = 4cos³x - 3cosx

2sinxcosx = 4cos³x - 3cosx

⇒ 2sinxcosx - 4cos³x + 3cosx = 0

⇒ cosx (2sinx - 4cos²x + 3) = 0

Now dividing both side by cosx we get,

2sinx - 4cos²x + 3 = 0

We know that,

cos²x + sin²x = 1

or cos²x = 1 – sin²x

⇒ 2sinx – 4(1 – sin²x) + 3 = 0

⇒ 2sinx – 4 + 4sin²x + 3 = 0

⇒ 2sinx + 4sin²x – 1 = 0

We can write it as,

4sin²x + 2sinx - 1 = 0

Now applying formula

Here, ax² + bx + c = 0

So,

now applying it in the equation

Now sin 18° is positive, as 18° lies in first quadrant.

Now, we know that

cos²x + sin²x = 1

or cosx = √1 – sin²x

∴cos 18° = √1 –sin² 18°

⇒

⇒

Putting the value in eq. (i), we get

= cos² 18° - cos²30°

= RHS

∴ LHS = RHS

Hence Proved

To Prove: tan 6^° tan 42^° tan 66^° tan 78^° = 1

Taking LHS,

= tan 6^° tan 42^° tan 66^° tan 78^°

Multiply and divide by tan 54^° tan 18^°

…(i)

We know that,

tan x tan(60° – x) tan (60° + x) = tan 3x

So, eq. (i) becomes

= 1

= RHS

∴ LHS = RHS

Hence Proved

Given:

To Prove: a sin 2θ + b cos 2θ = b

Given:

We know that,

By Pythagoras Theorem,

(Perpendicular)² + (Base)² = (Hypotenuse)²

⇒ (a)² + (b)² = (H)²

⇒ a² + b² = (H)²

So,

Taking LHS,

= a sin 2θ + b cos 2θ

We know that,

sin 2θ = 2 sin θ cos θ

and cos 2θ = 1 – 2 sin²θ

= a(2 sin θ cos θ) + b(1 – 2 sin²θ)

Putting the values of sinθ and cosθ, we get

= b

= RHS

∴ LHS = RHS

Hence Proved

Given: sin x = and <x< i.e, x lies in the Quadrant II .

To Find: i)sin ii)cos iii)tan

Now, since sin x =

We know that cos x =

cos x =

cos x =

cos x =

since cos x is negative in II quadrant, hence cos x = -

i) sin

Formula used:

sin =

Now, sin = = =

Since sinx is positive in II quadrant, hence sin

ii)cos

Formula used:

cos =

now, cos = = = = =

since cosx is negative in II quadrant, hence cos =

iii)tan

Formula used:

tan x =

hence, tan = = = = -

Here, tanx is negative in II quadrant.

Given: cos x = = and <x<.i.e, x lies in II quadrant

To Find: i)sin ii)cos iii)tan

i) sin

Formula used:

sin =

Now, sin = = =

Since sinx is positive in II quadrant, hence sin

ii)cos

Formula used:

cos =

now, cos = = = = =

since cosx is negative in II quadrant, hence cos =

iii)tan

Formula used:

tan x =

hence, tan = = = = -

Here, tanx is negative in II quadrant.

Given: sin x = and x lies in Quadrant IV.

To Find: i)sin ii)cos iii)tan

Now, since sin x =

We know that cos x =

cos x =

cos x =

cos x =

since cos x is positive in IV quadrant, hence cos x =

i) sin

Formula used:

sin =

Now, sin = = = = =

Since sinx is negative in IV quadrant, hence sin

ii)cos

Formula used:

cos =

now, cos = = = = =

since cosx is positive in IV quadrant, hence cos =

iii)tan

Formula used:

tan x =

hence, tan = = = = -1

Here, tanx is negative in IV quadrant.

Given: cos = and x lies in Quadrant I i.e, All the trigonometric ratios are positive in I quadrant

To Find: i)sin x ii)cos x iii)cot x

i)sin x

Formula used:

We have, Sin x =

We know that, cos = (cos x is positive in I quadrant)

2 – 1 = cos x

2 – 1 = cos x

2 – 1 = cos x

cos x =

Since, Sin x =

Sin x =

Sin x =

Hence, we have Sin x = .

ii)cos x

Formula used:

We know that, cos = (cos x is positive in I quadrant)

2 – 1 = cos x

2 – 1 = cos x

2 – 1 = cos x

cos x =

iii) cot x

Formula used:

cot x =

cot x = = =

Hence, we have cot x =

Given: sin x = and 0< x< i.e, x lies in Quadrant I and all the trignometric ratios are positive in quadrant I.

To Find: tan

Formula used:

tan =

Now, cos x = (cos x is positive in I quadrant)

cos x = =

Since, tan =

Hence, tan

To Prove: cot - tan = 2cot x

Proof: Consider L.H.S,

cot - tan -

=

= ()

Here multiply and divide L.H.S by 2

=

= (2sinxcosx = sin2x)

()

cot - tan = 2cotx = R.H.S

L.H.S = R.H.S, Hence proved

To Prove: tan() = tan x + sec x

Proof: Consider L.H.S,

tan() = ()

= =

=

Multiply and divide L.H.S by

=

=

= ()

=

= ()

=

tan( ) = sec x + tan x = R.H.S

L.H.S = R.H.S, Hence proved

To Prove: = tan( )

Proof: Consider, L.H.S =

Multiply and divide L.H.S by

= =

= ()

= ()

=

= ()

=

Multiply and divide the above with cos

=

Here, since tan = 1

= tan() = R.H.S

Since, L.H.S = R.H.S, Hence proved.

To prove: tan( )+ tan( ) = 2secx

Proof: Consider, L.H.S = tan( )+ tan( )

tan( )+ tan( ) = +

()

= +

= +

= +

=

By Expanding the numerator we get,

= ()

tan( )+ tan( ) = 2secx = R.H.S

since L.H.S = R.H.S, Hence proved.

To Prove: = tan

Proof: consider, L.H.S =

= ()

= ()

=

= R.H.S

Since L.H.S = R.H.S, Hence proved.