Values Of Trigonometric Functions At Multiples And Submultiple Of An Angles

Proof:

Take LHS:

Identities used:

cos 2x = 1 – 2 sin² x

= 2cos² x – 1

Therefore,

= tan x

= RHS

Hence Proved

Proof:

Take LHS:

Identities used:

cos 2x = 1 – 2 sin² x

sin 2x = 2 sin x cos x

Therefore,

= cot x

= RHS

Hence Proved

Proof:

Take LHS:

Identities used:

cos 2x = 2 cos² x – 1

sin 2x = 2 sin x cos x

Therefore,

= tan x

= RHS

Hence Proved

Proof:

Take LHS:

{∵ cos 2x = 2 cos² x – 1 ⇒ cos 4x = 2 cos² 2x -1}

{∵ cos 2x = 2 cos² x – 1}

= 2 cos x

= RHS

Hence Proved

Proof:

Take LHS

Identities used:

cos 2x = 2 cos² x – 1

= 1 – 2 sin² x

sin 2x = 2 sin x cos x

Therefore,

= tan x

= RHS

Hence Proved

Proof:

Take LHS:

Identities used:

cos 2x = cos² x – sin² x

sin 2x = 2 sin x cos x

Therefore,

= tan x

= RHS

Hence Proved

Proof:

Take LHS:

Identities used:

cos 2x = cos² x – sin² x

sin 2x = 2 sin x cos x

Therefore,

{∵ a² – b² = (a - b)(a + b) & sin² x + cos² x = 1}

{∵ a² + b² + 2ab = (a + b)²}

Multiplying numerator and denominator by

{∵ sin (A – B) = sin A cos B – sin B cos A

cos (A – B) = cos A cos B + sin A sin B}

= RHS

Hence Proved

Proof:

Take LHS:

Identities used:

cos 2x = cos² x – sin² x

sin 2x = 2 sin x cos x

Therefore,

{∵ a² – b² = (a - b)(a + b) & sin² x + cos² x = 1}

{∵ a² + b² + 2ab = (a + b)²}

Multiplying numerator and denominator by

{∵ sin (A – B) = sin A cos B – sin B cos A

cos (A – B) = cos A cos B + sin A sin B}

= RHS

Hence Proved

Proof:

Take LHS:

Identities used:

cos 2x = 2 cos² x – 1

⇒ 2 cos² x = 1 + cos 2x

Therefore,

{∵ cos (π – θ) = - cos θ, cos (π + θ) = - cos θ & cos(2π – θ) = cos θ}

= 2

= RHS

Hence Proved

Proof:

Take LHS:

Identities used:

cos 2x = 1 – 2 sin² x

⇒ 2 sin² x = 1 – cos 2x

Therefore,

{∵ cos (π – θ) = - cos θ,

cos (π + θ) = - cos θ &

cos(2π – θ) = cos θ}

= 2

= RHS

Hence Proved

Proof:

Take LHS:

{∵ cos (A – B) = cos A cos B + sin A sin B}

{∵ cos2x = 2cos² x – 1}

= RHS

Hence Proved

Proof:

Take LHS:

Identities used:

sin² A - sin² B = sin (A + B) sin(A – B)

Therefore,

= RHS

Hence Proved

Proof:

Take LHS:

{∵ cos2x = cos² x – sin² x & cos² x + sin² x = 1}

= RHS

Proof:

Take RHS:

{∵ a³ – b³ = (a – b) (a² + b² + ab)}

{∵ cos 2x = cos² x – sin² x}

{∵ a² + b² + 2ab = (a + b)²}

{∵ cos² x + sin² x = 1}

{∵ sin 2x= 2 sin x cos x}

{∵ sin² x = 1 – cos² x}

= LHS

Hence Proved

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

Proof:

Take LHS:

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

= (sin 3x)(sin x) + sin² x + (cos 3x)(cos x) – cos² x

= [(sin 3x)(sin x) + (cos 3x)(cos x)] + (sin² x – cos² x)

= [(sin 3x)(sin x) + (cos 3x)(cos x)] – (cos² x – sin² x)

= cos(3x – x) – cos 2x

{∵ cos 2x = cos² x – sin² x &

cos A cos B + sin A sin B = cos(A – B)}

= cos 2x – cos 2x

= 0

= RHS

Hence Proved

Proof:

Take LHS:

Identities used:

cos² A – sin² A = cos 2A

Therefore,

= RHS

Hence Proved

To prove: cos 4x = 1 – 8 cos² x + 8 cos⁴ x

Proof:

Take LHS:

cos 4x

Identities used:

cos 2x = = 2 cos² x – 1

Therefore,

= 2 cos² 2x – 1

= 2(2 cos² 2x – 1)² – 1

= 2{(2 cos² 2x}² + 1² – 2×2 cos² x} – 1

= 2(4 cos⁴ 2x + 1 – 4 cos² x) – 1

= 8 cos⁴ 2x + 2 – 8 cos² x – 1

= 8 cos⁴ 2x + 1 – 8 cos² x

= RHS

Hence Proved

To prove: sin 4x = 4 sin x cos³ x – 4 cos x sin³ x

Proof:

Take LHS:

sin 4x

Identities used:

sin 2x = 2 sin x cos x

cos 2x = cos² x – sin² x

Therefore,

= 2 sin 2x cos 2x

= 2 (2 sin x cos x) (cos² x – sin² x)

= 4 sin x cos x (cos² x – sin² x)

= 4 sin x cos³ x – 4 sin³ x cos x

= RHS

Hence Proved

To prove: 3(sin x – cos x)⁴ + 6 (sin x + cos x)² + 4 (sin⁶ x + cos⁶ x) = 13

Proof:

Take LHS:

3(sin x – cos x)⁴ + 6 (sin x + cos x)² + 4 (sin⁶ x + cos⁶ x)

Identities used:

(a + b)² = a² + b² + 2ab

(a – b)² = a² + b² – 2ab

a³ + b³ = (a + b) (a² + b² – ab)

Therefore,

= 3{(sin x – cos x)²}² + 6 {(sin x)² + (cos x)² + 2 sin x cos x) + 4 {(sin² x)³ + (cos² x)³}

= 3{(sin x)² + (cos x)² – 2 sin x cos x)}² + 6 (sin² x + cos² x + 2 sin x cos x) + 4{(sin² x + cos² x) (sin⁴ x + cos⁴ x – sin² x cos² x)}

= 3(1 – 2 sin x cos x) ² + 6 (1 + 2 sin x cos x) + 4{(1) (sin⁴ x + cos⁴ x – sin² x cos² x)}

{∵ sin² x + cos² x = 1}

= 3{1² + (2 sin x cos x)² – 4 sin x cos x}+ 6 (1 + 2 sin x cos x) + 4{(sin² x)² + (cos² x)² + 2 sin² x cos² x – 3 sin² x cos² x)}

= 3{1 + 4 sin² x cos² x – 4 sin x cos x}+ 6 (1 + 2 sin x cos x) + 4{(sin² x + cos² x)² – 3 sin² x cos² x)}

= 3 + 12 sin² x cos² x – 12 sin x cos x + 6 + 12 sin x cos x + 4{(1)² – 3 sin² x cos² x)}

= 9 + 12 sin² x cos² x + 4(1 – 3 sin² x cos² x)

= 9 + 12 sin² x cos² x + 4 – 12 sin² x cos² x

= 13

= RHS

Hence Proved

To prove: 2(sin⁶ x + cos⁶ x) – 3(sin⁴ x + cos⁴ x)+ 1 = 0

Proof:

Take LHS:

2(sin⁶ x + cos⁶ x) – 3(sin⁴ x + cos⁴ x)+ 1

Identities used:

(a + b)² = a² + b² + 2ab

a³ + b³ = (a + b) (a² + b² – ab)

Therefore,

= 2{(sin² x)³ + (cos² x)³} – 3{(sin² x)² + (cos² x)²} + 1

= 2{(sin² x + cos² x)(sin⁴ x + cos⁴ x – sin² x cos² x} – 3{(sin² x)² + (cos² x)² + 2sin² x cos² x – 2sin² x cos² x }+ 1

= 2{(1)(sin⁴ x + cos⁴ x + 2 sin² x cos² x – 3 sin² x cos² x}–3{(sin² x + cos² x)² – 2sin² x cos² x} + 1

{∵ sin² x + cos² x = 1}

= 2{(sin² x + cos² x)² – 3 sin² x cos² x} – 3{(1)² – 2sin² x cos² x } + 1

= 2{(1)² – 3 sin² x cos² x} – 3(1 – 2sin² x cos² x) + 1

= 2(1 – 3 sin² x cos² x) – 3 + 6 sin² x cos² x + 1

= 2 – 6 sin² x cos² x – 2 + 6 sin² x cos² x

= 0

= RHS

Hence Proved

Proof:

Take LHS:

Identities used:

(a + b)² = a² + b² + 2ab

a³ – b³ = (a – b) (a² + b² + ab)

Therefore,

{∵ cos 2x = cos² x – sin² x}

{∵ sin² x + cos² x = 1}

{∵ sin 2x = 2 sin x cos x}

= RHS

Hence Proved

Proof:

Take LHS:

Identities used:

Therefore,

{∵ (a – b)(a + b) = a² – b²;

(a + b)² = a² + b² + 2ab &

(a – b)² = a² + b² – 2ab}

{∵ cos² x + sin² x = 1 & cos 2x = cos² x – sin² x}

= 2 sec 2x

= RHS

Hence Proved

To prove: cot² x – tan² x = 4 cot 2x cosec 2x

Proof:

Take LHS:

cot² x – tan² x

Identities used:

a² – b² = (a – b)(a + b)

Therefore,

= (cot x – tan x)(cot x + tan x)

{∵ cot² x + 1 = cosec² x}

{∵ sin 2x = 2 sin x cos x}

= 4 cot 2x cosec 2x

= RHS

Hence Proved

To prove: cos 4x – cos 4α = 8(cos x – cos α) (cos x + cos α) (cos x – sin α) (cos x + sin α)

Proof:

Take LHS:

Cos 4x – cos 4α

{∵ cos 2θ = 2 cos² θ – 1}

= 2 cos² 2x – 1 – (2 cos² 2α – 1)

= 2 cos² 2x – 1 – 2 cos² 2α + 1

= 2 cos² 2x – 2 cos² 2α

= 2(cos² 2x – cos² 2α)

{∵ (a – b)(a + b) = a² – b²}

= 2(cos 2x – cos 2α) (cos 2x + cos 2α)

{∵ cos 2θ = 2 cos² θ – 1 = 1 – 2 sin² θ}

= 2{2 cos² x – 1 – (2 cos² α – 1)}(2 cos² x – 1 +1 – 2 sin² α)

= 2{2 cos² x – 1 – 2 cos² α + 1}(2 cos² x – 2 sin² α)

= 2 × 2{2 cos² x – 2 cos² α}(cos² x – sin² α)

= 4 × 2{cos² x – cos² α}(cos² x – sin² α)

= 8(cos x – cos α)(cos x + cos α)(cos x – sin α)(cos x + sin α)

= RHS

Hence Proved

Proof:

Take LHS:

sin 3x + sin 2x – sin x

Identities used:

sin 2x = 2 sin x cos x

Therefore,

= RHS

Hence Proved

Proof:

Identities used:

Therefore,

tan 15° = tan (45° - 30°)

On rationalising:

{∵ (a – b)(a + b) = a² – b²}

On rationalising

{∵ (a – b)(a + b) = a² – b²}

Let 2θ = 15°

We know,

Formula used:

{∵ (a + b)² = a² + b² + 2ab}

cot θ < 0 as θ is in 1^st quadrant.

So,

{∵ (a + b)² = a² + b² + 2ab}

{∵ cot θ = tan(90° - θ)}

Hence Proved

Proof:

Take LHS:

Let 2θ = 45°

We know,

{∵ cot 45° = 1}

Formula used:

cot θ < 0 as θ is in 1^st quadrant.

So,

Hence Proved

Given:

We know,

cos 2x = 2 cos² x – 1

Since,

So,

We know,

cos 2x = 1 – 2 sin² x

Since,

So,

We know,

sin² x + cos² x = 1

⇒ sin² x = 1 – cos² x

Since,

So,

Now,

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

Given:

We know,

cos 2x = 1 – 2 sin² x

Since,

So,

We know,

sin² x + cos² x = 1

⇒ sin² x = 1 – cos² x

Since,

⇒sin x will be positive in second quadrant

So,

Now,

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

Given:

We know,

sin² x + cos² x = 1

⇒ cos² x = 1 – sin² x

Since,

⇒cosx will be negative in second quadrant

So,

We know,

cos 2x = 2 cos² x – 1

Since,

So,

We know,

cos 2x = 1 – 2 sin² x

Since,

So,

We know,

Given:

We know,

sin² x + cos² x = 1

⇒ cos² x = 1 – sin² x

Since,

⇒cosx will be negative in second quadrant

So,

We know,

cos 2x = 2 cos² x – 1

Since,

So,

We know,

cos 2x = 1 – 2 sin² x

Since,

So,

We know,

On rationalising:

{∵ (a + b)(a – b) = a² – b²}

Given:

We know,

sin² x + cos² x = 1

⇒ sin² x = 1 – cos² x

Since,

⇒sinx will be negative in first quadrant

So,

Now,

We know,

Hence, value of

Given:

To find: Values of sin4x

We know,

sin² x + cos² x = 1

⇒ cos² x = 1 – sin² x

Since,

⇒cosx will be negative in first quadrant

So,

We know,

sin 2x = 2 sin x cos x

cos 2x = 2 cos² x – 1

Therefore,

sin 4x = 2 sin 2x cos 2x

⇒ sin 4x = 2 (2 sin x cos x) (2 cos² x – 1)

Hence, value of

On taking LCM:

Dividing numerator and denominator by a:

{∵ (a + b)(a – b) = a² – b²}

To prove: cos 2A = sin 4B

We know,

Take LHS:

cos 2A

Now,

Take RHS:

sin 4B

Clearly, LHS = RHS

Hence Proved

Proof:

Take LHS:

Multiplying and Dividing 2⁴ sin 7°

{∵ sin 2x = 2 sin x cos x}

We know,

sin (180° - θ) = sin θ

sin (90° - θ) = cos θ

Now,

= RHS

Hence Proved

Proof:

Take LHS:

{∵ sin 2x = 2 sin x cos x}

{∵ sin (2π + θ) = sin θ}

= RHS

Hence Proved

Proof:

Take LHS:

{∵ sin 2x = 2 sin x cos x}

{∵ sin (3π + θ) = - sin θ}

= RHS

Hence Proved

Proof:

Take LHS:

{∵ sin 2x = 2 sin x cos x}

{∵ sin (π - θ) = sin θ}

= RHS

Hence Proved

Given: 2 tan α = 3 tan β

Proof:

Take LHS:

tan α – tan β

{∵ sin 2x = 2(sin x)(cos x)}

{∵ 2 cos² x = 1 + cos 2x & 2 sin² x = 1 – cos 2x}

= RHS

Hence Proved

Given: sin α + sin β = a & cos α + cos β = b

Proof:

sin α + sin β = a ………(3)

cos α + cos β = b ……(4)

Dividing equation 3 and 4:

We know,

Therefore,

Hence Proved

Given: sin α + sin β = a & cos α + cos β = b

Proof:

sin α + sin β = a

Squaring both sides, we get

(sin α + sin β)² = a²

⇒ sin² α + sin² β + 2 sin α sin β = a² ……(1)

cos α + cos β = b

Squaring both sides, we get

(cos α + cos β)² = a²

⇒ cos² α + cos² β + 2 cos α cos β = b² ………(2)

Adding equation 1 and 2, we get

sin² α + sin² β + 2 sin α sin β + cos² α + cos² β + 2 cos α cos β = a² + b²

⇒ sin² α + cos² α + sin² β + cos² β + 2 sin α sin β + 2 cos α cos β = a² + b²

⇒ 1 + 1 + 2 sin α sin β + 2 cos α cos β = a² + b²

{∵ sin² x + cos² x = 1}

⇒ 2 + 2 sin α sin β + 2 cos α cos β = a² + b²

⇒ 2(sin α sin β + cos α cos β) = a² + b² – 2

We know,

sin A sin B + cos A cos B = cos (A – B)

Therefore,

Hence Proved

Proof:

Take LHS:

cos α

Now, Take RHS:

= cos α

Hence Proved

We know,

Applying componendo and dividendo, we get

Taking reciprocal both sides:

Hence Proved

Given: sec (x + α) + sec(x - α) = 2 sec x

sec (x + α) + sec(x - α) = 2 sec x

{∵ 2 cos A cos B = cos (A + B) + cos (A – B)}

{∵ cos 2x = 2 cos² x – 1}

{∵ cos 2x = 2 cos² x – 1}

Hence Proved

Squaring both sides, we get

Squaring both sides, we get

Adding equation (1) and (2), we get

We know,

sin A sin B + cos A cos B = cos (A – B)

Therefore,

Hence Proved

Proof:

We know,

sin² α + cos² α = 1

⇒ cos² α = 1 – sin² α

Similarly,

sin² β + cos² β = 1

⇒ sin² β = 1 – cos² β

Identity used:

cos (α – β) = cos α cos β + sin α sin β

Hence Proved

Given: a cos 2x + b sin 2x = c

We know,

Therefore,

a cos 2x + b sin 2x = c

We know,

If m and n are roots of the equation ax² + bx + c = 0

then,

Sum of the roots(m+n)

Therefore,

If tan α and tan β are the roots of the equation

then,

Hence Proved

Given: a cos 2x + b sin 2x = c

We know,

Therefore,

a cos 2x + b sin 2x = c

We know,

If m and n are roots of the equation ax² + bx + c = 0

then,

Product of the roots(mn)

Therefore,

If tan α and tan β are the roots of the equation

then,

Hence Proved

We know,

Therefore,

From previous question:

Hence Proved

Proof:

cos α + cos β = 0

Squaring both sides:

⇒ (cos α + cos β)² = (0)²

⇒ cos² α + cos² β + 2 cos α cos β = 0 ……(1)

sin α + sin β = 0

Squaring both sides:

⇒ (sin α + sin β)² = (0)²

⇒ sin² α + sin² β + 2 sin α sin β = 0 ………(2)

Subtracting equation (1) from (2), we get

cos² α + cos² β + 2 cos α cos β – (sin² α + sin² β + 2 sin α sin β) = 0

⇒ cos² α + cos² β + 2 cos α cos β – sin² α – sin² β – 2 sin α sin β = 0

⇒ cos² α – sin² α + cos² β – sin² β + 2(cos α cos β – sin α sin β) = 0

{∵ cos² x – sin² x = 2x &

cos A cos B – sin A sin B = cos(A + B)}

⇒ cos 2α + cos 2β + 2 cos (α + β) = 0

⇒ cos 2α + cos 2β = - 2 cos (α + β)

Hence Proved

LHS is

sin 5x = sin(3x+2x)

But we know,

sin(x+y) = sin x cos y+cos x sin y…..(i)

⇒ sin 5x = sin 3x cos 2x+cos 3x sin 2x

⇒ sin 5x = sin (2x+x) cos 2x+cos (2x+x) sin 2x……..(ii)

And

cos (x+y) = cos(x)cos(y) – sin(x)sin(y)……(iii)

Now substituting equation (i) and (iii) in equation (ii), we get

⇒ sin 5x = (sin 2x cos x+cos 2x sin x )cos 2x+( cos 2x cos x – sin 2x sin x) sin 2x

⇒ sin 5x = sin 2x cos 2x cos x+cos² 2x sin x+(sin 2x cos 2x cos x – sin² 2x sin x)

⇒ sin 5x = 2sin 2x cos 2x cos x+cos² 2x sin x– sin² 2x sin x …….(iv)

Now sin 2x = 2sin x cos x………(v)

And cos 2x = cos²x –sin²x………(vi)

Substituting equation (v) and (vi) in equation (iv), we get

⇒ sin 5x = 2(2sin x cos x)(cos²x –sin²x)cos x+(cos²x –sin²x)²sin x –(2sin x cos x)²sin x

⇒ sin 5x = 4(sin x cos² x)([1–sin²x] –sin²x)+([1–sin²x]–sin²x)²sin x –(4sin² x cos² x)sin x (as cos²x+sin²x=1 ⇒ cos²x=1–sin²x)

⇒ sin 5x = 4(sin x [1–sin²x])(1–2sin²x)+(1–2sin²x)²sin x–4sin³ x [1–sin²x]

⇒ sin 5x = 4sin x(1–sin²x)(1–2sin²x)+(1–4sin²x+4sin⁴x)sin x–4sin³ x +4sin⁵x

⇒ sin 5x = (4sin x–4sin³x)( 1–2sin²x) +sin x–4sin³x+4sin⁵x–4sin³ x +4sin⁵x

⇒ sin 5x = 4sin x–8sin³x–4sin³x+8sin⁵x+sin x–8sin³x+8sin⁵x

⇒ sin 5x = 5sin x–20sin³x+16sin⁵x

Hence LHS = RHS

[Hence proved]

We know that

⇒ sin (3× 20°)=cos (3× 10°)

⇒ 3sin 20°–4sin³20°=4cos³10°–3cos 10°

(as sin 3θ=3sin θ–4sin³ θ and cos 3θ =4cos³θ–3cosθ)

⇒ 4(cos³10°+sin³20°)=3(sin 20°+cos 10°)

LHS=RHS

Hence proved

We know that,

cos 3θ =4cos³θ–3cosθ

⇒4 cos³θ=cos3θ+3cosθ

And similarly

sin 3θ=3sin θ–4sin³ θ

⇒4 sin³θ=3sinθ–sin 3θ

Now,

Substituting the values from equation (i) and (ii), we get

(as sin(x+y) = sin x cos y+cos x sin y)

RHS

Hence Proved

≠ –3

Hence LHS≠ RHS

Hence proved

Hence proved

We know,

(as –cot θ=cot (180°–θ))

Hence the above LHS becomes

Hence proved

LHS is

sin 5x = sin(3x+2x)

But we know,

sin(x+y) = sin x cos y+cos x sin y…..(i)

⇒ sin 5x = sin 3x cos 2x+cos 3x sin 2x

⇒ sin 5x = sin (2x+x) cos 2x+cos (2x+x) sin 2x……..(ii)

And

cos (x+y) = cos(x)cos(y) – sin(x)sin(y)……(iii)

Now substituting equation (i) and (iii) in equation (ii), we get

⇒ sin 5x = (sin 2x cos x+cos 2x sin x )(cos 2x)+( cos 2x cos x – sin 2x sin x) (sin 2x)……..(iv)

Now sin 2x = 2sin x cos x………(v)

And cos 2x = cos²x –sin²x………(vi)

Substituting equation (v) and (vi) in equation (iv), we get

⇒ sin 5x =[(2 sin x cos x)cos x+(cos²x–sin²x)sin x]( cos²x–sin²x)+[( cos²x–sin²x)cos x – (2 sin x cos x) sin x)]( 2 sin x cos x)

⇒ sin 5x =[2 sin x cos² x+sin xcos²x–sin³x]( cos²x–sin²x)+[cos³x–sin²xcos x – 2 sin² x cos x]( 2 sin x cos x)

⇒ sin 5x =cos²x[3 sin x cos² x –sin³x]–sin²x[3 sin x cos² x–sin³x]+2 sin x cos⁴x–2 sin³ x cos² x – 4 sin³ x cos² x

⇒ sin 5x = 3 sin x cos⁴ x –sin³xcos²x– 3 sin³ x cos² x–sin⁵x +2 sin x cos⁴x–2 sin³ x cos² x – 4 sin³ x cos² x

⇒ sin 5x = 5 sin x cos⁴ x –10sin³xcos²x +sin⁵x

Hence LHS = RHS

[Hence proved]

sin 3θ=3sin θ–4sin³ θ

⇒4 sin³θ=3sinθ–sin 3θ

Now,

Substituting equation (i) in above LHS, we get

We know,

(as sin θ =sin (180°–θ))

Similarly,

(as –sin θ =sin (180°+θ))

Substituting the equation (iii) and (iv) in equation (ii), we get

We know,

Substituting this in the above equation, we get

Hence proved

We know

sin (A+B)sin (A–B)=sin²A–sin²B

So the above LHS becomes,

But 3sin x–4 sin³x=sin 3x

But |sin θ|≤ 1 for all values of x

Hence

Therefore For all values of x

We know

cos (A+B)cos (A–B)=cos²A–sin²B

So the above LHS becomes,

But 4cos³x–3cos x=cos 3x

But |cos θ|≤ 1 for all values of x

Hence

Therefore For all values of x

But sin (90°–θ)=cos θ

Then the above equation becomes,

And

Hence the above equation becomes,

Hence proved

But sin (A+B)sin(A–B)=sin²A–sin²B

Then the above equation becomes,

And

Hence the above equation becomes,

Hence proved

But cos (A+B)cos(A–B)=cos²A–sin²B

Then the above equation becomes,

And

Hence the above equation becomes,

Hence proved

Multiply and divide by 2, we get

But 2cos A cos B = cos(A+B)+cos(A–B)

Then the above equation becomes,

But cos(180°–θ)=–cos θ

So the above equation becomes,

And

Hence the above equation becomes,

Hence proved

Multiply and divide by , we get

But 2sin A cos A = sin 2A

Then the above equation becomes,

Multiply and divide by 2, we get

But 2sin A cos A = sin 2A

Then the above equation becomes,

Multiply and divide by 2, we get

But 2sin A cos A = sin 2A

Then the above equation becomes,

Multiply and divide by 2, we get

But 2sin A cos B = sin (A+B) +sin(A–B), so the above equation becomes,

Hence proved

By regrouping the LHS and multiplying and dividing by 4 we get,

But 2cos A cos B = cos (A+B) +cos (A–B)

Then the above equation becomes,

But cos(90°–θ)=sin θ and cos(180°–θ)=–cos(θ).

Then the above equation becomes,

Now,

Substituting the corresponding values, we get

Hence proved

By regrouping the LHS and multiplying and dividing by 4 we get,

But 2sin A sin B = cos (A–B) –cos (A+B)

Then the above equation becomes,

But cos(90°–θ)=sin θ and cos(180°–θ)=–cos(θ).

Then the above equation becomes,

Now,

Substituting the corresponding values, we get

Hence proved

By regrouping the LHS and multiplying and dividing by 2 we get,

But 2cos A cos B = cos (A+B) +cos (A–B)

Then the above equation becomes,

But cos(90°–θ)=sin θ and cos(180°–θ)=–cos(θ).

Then the above equation becomes,

Now,

Substituting the corresponding values, we get

Hence proved

This can be rewritten as,

But so the above equation becomes,

This can be rewritten as,

But sin (90°–θ)=cos θ

Then the above equation becomes,

Now,

Hence the above equation becomes,

Hence proved

Multiply and divide by , we get

But 2sin A cos A = sin 2A

Then the above equation becomes,

Multiply and divide by 2, we get

But 2sin A cos A = sin 2A

Then the above equation becomes,

Multiply and divide by 2, we get

But 2sin A cos A = sin 2A

Then the above equation becomes,

Multiply and divide by 2, we get

But 2sin A cos B = sin (A+B) +sin(A–B), so the above equation becomes,

Multiply and divide by , we get

But 2sin A cos A = sin 2A

Then the above equation becomes,

Multiply and divide by 2, we get

But 2sin A cos A = sin 2A

Then the above equation becomes,

Hence proved

Given equation is

cos 4x = 1 + k sin²x cos²x

Now consider the LHS of the equation,

cos 4x = 2cos² 2x – 1

[Formula for Cos 2x = 2cos² x – 1]

= 2[2cos² x - 1]² – 1

= 2[(2cos² x)² – 2 × (2 cos² x) × (1) + (1)²] – 1

[Applying (a-b)² = a² – 2ab + b² formula]

= 2[4cos⁴ x – 4cos² x +1] -1

= 8 cos⁴ x – 8cos² x +2 – 1

= 8cos² x (cos² x – 1) + 1

= 8cos² x (-sin² x) +1

= - 8cos² x sin² x + 1

Now as per the LHS cos 4x = - 8cos² x sin² x + 1 -------- (1)

Comparing LHS with the RHS,

cos 4x = 1 - 8cos² x sin² x = 1 + k sin²x cos²x

by comparing we get k = -8

Given,

We need to find the value of m sin x + n cos x

Now consider

m sinx + n cos x

[ using the formulas sin 2x & cos 2x in terms of tan x

and ]

[Substituting ]

= n

Hence the value of m sin x + n cos x = n.

Given then the value of

Hence

But as given,

This states that, 90° < x < 270°, which means x lies between 2^nd and 3^rd quadrants.

In the 2^nd and 3^rd quadrants, the cosine function is negative, so the value of

Given,

To find the value of

[using the formula cos 2x = 2cos² x – 1]

[using the formula cos 2x = 2cos² x –1, here 2x = θ so x = ]

As given, now by dividing the whole inequation with 2 we get, .

This clearly state that lies in the 1^st quadrant and between 45° and 90°.

So

Given, for the value of

Consider,

[by using the formula cos 2x = cos² x – sin² x]

[by using the formula cos² x + sin² x = 1]

As already mentioned in the question, , x is in the 2nd quadrant, where tangent function is negative.

Therefore,

Given, for the value of

Consider,

[by using the formula cos 2x = cos² x – sin² x]

[by using the formula cos² x + sin² x = 1]

As already mentioned in the question, , x is in the 3^rd quadrant, where tangent function is positive.

Therefore,

Given, triangle ABC is right angle.

So, let ∠ B = 90°

Then as per the property of angles in a triangle

∠ A + ∠ B + ∠ C = 180°

As ∠ B = 90°

∠ A + 90° + ∠ C = 180°

Then ∠ A + ∠ C = 180° - 90° = 90°

Now, consider sin ²A + sin ²B + sin ²C

As ∠ B = 90°

sin²A + sin²B + sin²C = sin²A + sin²(90°) + sin²C

= sin²A + 1 + sin²C

From before, we know that ∠ A + ∠ C = 90° ; ∠ C = 90° - ∠ A

sin²A + sin²B + sin²C = sin²A + 1 + sin²( 90° - A)

= sin²A + cos²(A) + 1

[by using the identity cos x = sin ( 90° - x)]

sin²A + sin²B + sin²C = (sin²A + cos²A) + 1

= 1 + 1

= 2

[by using the identity sin²θ + cos²θ = 1]

Therefore, sin²A + sin²B + sin²C = 2.

Given to find the value for,

cos² 76^o + cos² 16^o – cos 76^o cos 16^o

In the above expression consider cos 76^o cos 16^o

[By using the trigonometric sum formula, we can say that,

cos(C+D) + cos ( C-D) = 2 cos C cos D]

Now multiply and divide this with 2, we get

Consider the full expression,

Multiplying and dividing the terms cos² 76° + cos² 16° with 2

[ by using the formula, cos 2θ = 2cos²θ – 1 � 2cos²θ = cos 2θ +1 ]

[ by using the formula, cos A + cos B ]

Hence, cos² 76^o + cos² 16^o – cos 76^o cos 16^o

Given,

We should find the value for

[by using the formulae, sin²θ + cos²θ = 1 and sin2θ = 2 sinθcosθ]

As already mentioned in the question, , so x lies in the 1^st quadrant and both sine and cosine functions are positive.

Therefore, sin x + cos x

Given expression is

[by using sin2θ = 2 sinθ cosθ � cos θ ]

Hence

Given,

To find the value for tan 2A,

Consider

[ by using the formula for ]

[by substituting the value of tan A as given in the problem]

= tan B

Therefore, tan 2A = tan B

Given, sin x + cos x = a

We need to find the value of the expression,

sin⁶ x + cos⁶ x = (sin² x)³ + (cos² x)³

= (sin² x + cos² x)³ – 3 sin² x cos² x (sin² x + cos² x)

[ by using the formula a³ + b³ = (a+b)³ – 3ab(a+b)]

= (1)³ – 3 sin² x cos² x (1)

[ by using the formula sin² x + cos² x = 1]

[ by using the formula sin² x + cos² x = 1]

Hence sin⁶ x + cos⁶ x

Given, sin x + cos x = a

To find the value of |sin x – cos x|

Consider square of |sin x – cos x|

|sin x – cos x|² = |sin x|² + |cos x|² – 2|sin x| |cos x|

[using the formula (a + b)²= a² + b² +2 ab]

|sin x – cos x|² = |sin x|² + |cos x|² – 2|sin x| |cos x|

= (sin² x + cos² x)–[(sin x + cos x)² –sin² x –cos² x]

= (sin² x + cos² x)–[a² – (sin² x + cos² x) ]

[using the formula sin² x + cos² x = 1]

= 1 – a² + 1

= 2 – a²

|sin x – cos x|² = 2 – a²

Taking square root on both sides.

Hence

Given expression,

[by rearranging terms]

[using the formula sin2θ = 2sinθcosθ]

= sin x

Hence

Given expression is

[using ]

[using cos2θ = 1 – 2 sin²θ ]

[using sin2θ = 2sinθcosθ ]

[using ]

Given expression,

Multiply and divide the expression with

[using the formula sin2θ =2 sin θ cos θ]

Multiply and divide the expression with 2

[using the formula sin2θ =2 sin θ cos θ]

Multiply and divide the expression with 2

[using the formula sin2θ =2 sin θ cos θ]

Multiply and divide the expression with 2

[using the formula sin2θ =2 sin θ cos θ]

Multiply and divide the expression with 2

[using the formula sin2θ =2 sin θ cos θ]

Multiply and divide the expression with 2

As

Hence answer is option D.

Given cos 2x + 2 cos x = 1, we need to find the expression,

(2 – cos² x) sin² x

Consider cos 2x + 2 cos x = 1

2cos² x – 1 + 2 cos x -1 = 0

2cos² x + 2cos x – 2 = 0

cos² x + cos x = 1 -------- (1)

Now consider the expression

(2 – cos² x) sin² x = (2 – cos² x)(1-cos²x)

= {2 – (1 – cos x)} { 1- ( 1 – cos x)}

[from equation (1) cos² x = 1 - cos x]

= (1+ cos x) ( cos x)

= cos x + cos² x

[from equation (1) cos² x + cos x = 1]

= 1

Hence (2 – cos² x) sin² x = 1, so option A is the answer.

Given expression is cot x – 2 cot 2x for all real values of x

Consider

[ using and ]

= tan x

Therefore cot x – 2 cot 2x = tan x.

Option B is the answer.

Given expression is

Now

Multiplying and dividing the whole expression with

[using sin 2x = 2 sin x cos x formula]

[using cos 3x = 4cos³x – 3 cos x formula]

[using ]

= 0

Therefore

The answer is option A.

Given ABC is a triangle, so ∠ A + ∠ B + ∠ C = 180°

Now applying tan on both sides

tan (A+B +C) = tan (180°)

tan (A + B + C) = 0 ----- (1)

Also given tan A + tan B + tan C = 0 ------ (2)

As per the formula of tan (A+B+C)

Now,

[from equation (1) ]

[from equation (2) ]

By cross multiplying

-tan A tan B tan C = 0

tan A tan B tan C = 0

therefore

Hence cot A cot B cot C = 0

The answer is option D.

Given and

Consider the equation

Now take the LHS of the equation,

cos 3x = 4cos³ x – 3cos x

[using the formula for cos 3x = 4cos³ x – 3cos x]

From the question we know,

Substituting the known cos x values in the cos 3x expansion,

------ (1)

If we compare the RHS of the cos3x equation with the now derived equation (1) we get,

From the here we can clearly say that

Hence the answer is option B.

Given, 2 tan α = 3 tan β

From here we get, ------ (1)

Now consider tan (α - β),

The expansion of tan (α - β) is given by

As we already know the value of tan α from equation (1), we have,

[ by using ]

Multiplying and dividing the equation with 2

[using sin2θ = 2 sinθ cosθ]

In the denominator adding and subtracting 1

[using cos2θ = 2cos²θ – 1]

Hence, in the question the answer matches with option A.

Given,

As there are 2 option in terms of tan 2A, let us consider tan 2A

[by using the formula for ]

[by substituting the value of tan A as given in the problem]

= tan B

Therefore, tan 2A = tan B

Hence the option B is the correct answer.

Given, sin α + sin β = a and cos α – cos β = b, then the value of

Consider sin α + sin β = a

As per the expansion of

Now , ----- (1)

Similarly, cos α - cos β = b

As per the expansion of

Now ------ (2)

By dividing equation (1) with (2) we get,

[As ]

Therefore the answer is option B.

Given to find the value of

We will solve the expression in two parts,

Now solving 1^st term

If we multiply and divide the term by 2, we get,

[using the formula for and ]

----- (1)

Solving the 2^nd term

[using the formula for ]

1 – 2 tan x cot 2x = 1 – ( 1 – tan²x)

= 1 – 1 + tan²x

1 – 2 tan x cot 2x = tan² x ----- (2)

Now by combining (1) and (2) we get,

Hence the answer is option D.

Given to find the value of the expression

(as sine is positive in 2^nd quadrant)

(as cosine is positive in 1^st quadrant)

There for

Hence the answer is option D.

Given to find the value of

The angles can be modified as and

Using the identity , we have

[using the identity cos²θ + sin²θ = 1]

= 1 + 1 = 2

Hence the answer is option B.

Given 5 sin α = 3 sin (α + 2 β) ≠ 0, then the value of tan (α + β) is

Consider the given equation,

5 sin α = 3 sin (α + 2 β)

By applying componendo and dividendo

We get

[ using and sum of angles ]

This clearly shows, tan (α + β) = 4 tan β

Hence the answer is option C.

Given expression is 2 cos x – cos 3x – cos 5x – 16 cos³ x sin² x

Consider the expression

2 cos x – cos 3x – cos 5x – 16 cos³ x sin²

= 2 cos x – (cos 5x + cos 3x) – 16 cos³ x sin² x

[using the sum of angles ]

= 2 cos x – [2 cos 4x cos x] - 16cos³ x sin² x

= 2 cos x (1 – cos 4x ) - 16cos³ x sin² x

[ using the property cos 2θ = 1- 2 sin² θ ]

= 2 cos x [ 1 – (1 – 2 sin² 2x)] -16cos³ x sin² x

= 2 cos x [2 sin² 2x] -16cos³ x sin² x

= 4cos x [2sin x cos x]² -16cos³ x sin² x

[ using sin 2θ = 2 sin θ cos θ ]

= 4 × 4 (cos x sin² x cos² x) -16cos³ x sin² x

= 16cos³ x sin² x -16cos³ x sin² x

= 0

Hence cos x – cos 3x – cos 5x – 16 cos³ x sin² x = 0

The answer is option C.

Given A = 2 sin² x – cos 2x

[ using cos 2x = 1 – 2 sin² x ]

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

= 2 sin² x -1 + 2 sin² x]

= 4 sin² x – 1

Now A = 2 sin² x – cos 2x = 4 sin² x – 1

As we know sin x lies between -1 and 1

-1 ≤ sin x ≤ 1

0 ≤ sin²x ≤ 1

Multiplying the inequality by 4

0 ≤ 4 sin² x ≤ 4

Subtracting 1 from the inequality

-1 ≤ (4 sin² x – 1) ≤ 3

From the above inequation, we can say that

A = (4 sin² x – 1) belongs to the closed interval [-1,3]

Hence the answer is A.

Given expression is

Consider

[using the formulae cos 3x = 4 cos³ x – 3 cos x and

cos 2x = 2cos²x -1]

= cos x

Therefore

Hence the answer is option A.

Given equation is

Let us consider LHS

[ using the formulae and ]

[ the value of tan 45° = 1 ]

[using the formulae cos 2x = cos² x – sin² x and cos² x + sin² x = 1]

= 2 sec 2x

Now comparing with the LHS with RHS

From here we can clearly say that the answer is option D.