If un = cosnθ + sinnθ, then prove that 2u6 – 3u4 + 1 = 0

Trigonometry is a discipline of mathematics that studies the relationships between the lengths of the sides and angles of a right-angled triangle. Trigonometric functions, also known as goniometric functions, angle functions, or circular functions, are functions that establish the relationship between an angle to the ratio of two of the sides of a right-angled triangle. The six main trigonometric functions are sine, cosine, tangent, cotangent, secant, and cosecant.

Trigonometric angles are the Angles defined by the ratios of trigonometric functions. Trigonometric angles represent trigonometric functions. The value of the angle can be anywhere between 0-360°.

Right-angled Triangle

As given in the above figure in a right-angled triangle:

• Hypotenuse: The side opposite to the right angle is the hypotenuse, It is the longest side in a right-angled triangle and opposite to the 90° angle.
• Base: The side on which angle C lies is known as the base.
• Perpendicular: It is the side opposite to angle C in consideration.

Trigonometric Functions

Trigonometry has 6 basic trigonometric functions, they are sine, cosine, tangent, cosecant, secant, and cotangent. Now let’s look into the trigonometric functions. The six trigonometric functions are as follows,

1. sine: the ratio of perpendicular and hypotenuse is defined as sine and It is represented as sin θ
2. cosine: the ratio of base and hypotenuse is defined as cosine  and it is represented as cos θ
3. tangent: the ratio of sine and cosine of an angle is defined as tangent. Thus the definition of tangent comes out to be the ratio of perpendicular and base and is represented as tan θ
4. cosecant: It is the reciprocal of sin θ and is represented as cosec θ.
5. secant: It is the reciprocal of cos θ and is represented as sec θ.
6. cotangent: It is the reciprocal of tan θ and is represented as cot θ.

According to the above image, Trigonometric Ratios are

1. sin θ = Perpendicular / Hypotenuse = AB/AC
2. cosine θ = Base / Hypotenuse = BC / AC
3. tangent θ = Perpendicular / Base = AB / BC
4. cosecant θ = Hypotenuse / Perpendicular = AC/AB
5. secant θ = Hypotenuse / Base = AC/BC
6. cotangent θ = Base / Perpendicular = BC/AB

Reciprocal Identities

sin θ = 1/ cosec θ                     OR                 cosec θ = 1/ sin θ

cos θ = 1/ sec θ                       OR                sec θ = 1 / cos θ

cot θ = 1 / tan θ                     OR                tan θ = 1 / cot θ

cot θ = Cos θ / sin θ               OR               tan θ = sin θ / cos θ

tan θ.cot θ = 1

Values of Trigonometric Ratios

	0°	30°	45°	60°	90°
Sin θ	0	1/2	1/√2	√3/2	1
Cos θ	1	√3/2	1/√2	1/2	0
Tan θ	0	1√3	1	√3	NOT DEFINED
Sec θ	NOT DEFINED	2	√2	2/√3	1
Cosec θ	1	2/√3	√2	2	NOT DEFINED
Cot θ	NOT DEFINED	√3	1	1/√3	0

Trigonometric Identities of Complementary and Supplementary Angles

• Complementary Angles: Pair of angles whose sum is equal to 90°.

Identities of Complementary angles are:

sin (90° – θ) = cos θ

cos (90° – θ) = sin θ

tan (90° – θ) = cot θ

cot (90° – θ) = tan θ

sec (90° – θ) = cosec θ

cosec (90° – θ) = sec θ

• Supplementary Angles: Pair of angles whose sum is equal to 180°.

Identities of supplementary angles are:

sin (180° – θ) = sin θ

cos (180° – θ) = – cos θ

tan (180° – θ) = – tan θ

cot (180° – θ) = – cot θ

sec (180° – θ) = – sec θ

cosec (180° – θ) = – cosec θ

Question: If u_n = cosⁿ θ + sinⁿ θ then prove that, 2u₆ – 3u₄ + 1 = 0.

Solution:     

Here we have:

u_n  = cosⁿθ+sinⁿθ 

To prove : 2u₆ – 3u₄ + 1 = 0

Lets n = 6

So, u_n  = cosⁿθ+sinⁿθ 

u₆ = cos⁶θ+sin⁶θ 

u₆ = (cos²θ)³+ (sin²θ) ³

u₆ = (cos²θ +sin²θ)³− 3(cos ²θ)(sin ²θ)(cos ²θ+sin ²θ) 

u₆ =1−3cos²θ sin²θ                                                                                         ……. (1)                                                   { cos²θ +sin²θ = 1 }

Now let n = 4      

u₄ = cos⁴θ + sin⁴θ

u₄​  = (cos²θ)² + (sin²θ)²

u₄ = (cos²θ + sin²θ) ²−2(cos ²θ)(sin ²θ)                                                                                                                      { cos²θ +sin²θ = 1 }

u_4​  = 1− 2cos²θsin²θ                                                                                 ……. (2)                                                   

Now to prove 

2u₆ – 3u₄ + 1 = 0

Put values from equations (1) and (2).

So,

2u₆ – 3u₄ + 1 = 0

= 2(1−3cos ²θ sin ²θ) − 3(1−2cos ²θsin ²θ)+1

= 2 − 6cos²θ sin²θ − 3 + 6cos²θ sin²θ)+1

= 2 -3 +1 

= 0 

Therefore  2u₆ – 3u₄ + 1 = 0

Hence proved 

Similar Questions

Problem 1:  If cos θ + sin θ = √2 cos θ, prove that cos θ – sin θ = √2 sin θ?

Solution:

Here we have

cos θ + sin θ = √2 cos θ

Squaring both the sides

(cos θ + sin θ)² = (√2 cos θ)²

cos² θ + sin² θ + 2 sin θcos θ = 2 cos ²θ

cos² θ – sin² θ = 2 sin θcos θ

(cos θ + sin θ)(cos θ – sin θ) = 2 sin θcos θ

(cos θ – sin θ) = 2 sin θcos θ/(cos θ + sin θ)

cos θ – sin θ = 2 sin θcos θ/√2 cos θ               {cos θ + sin θ = √2 cos θ}

cos θ – sin θ = √2 sin θ

Hence proved

Problem 2: Prove that (cos θ sec θ)/cot θ = tan θ.

Solution:

Here we have   

cos θ sec θ / cot θ = tan θ

Therefore,

{ cos θ sec θ }/ cot θ = tan θ

By taking L.H.S

cos θ sec θ / cot θ

we can write cos θ sec θ as 1  

= (cos θ sec θ )/cot θ

= 1/cot θ                                                                                                                              {cos θ = 1/ sec θ   therefore Cos θ Sec θ  = 1}

= tan θ                                                                                                                                 {tan θ = 1 / cot θ }

Therefore LHS = RHS

{cos θ sec θ}/ cot θ = tan θ

Hence Proved

Problem 3 : prove {(cot A + cosec A – 1) / (cot A – cosec A + 1)} = {(1 + cos A) / sin A} ?

Solution: 

We have 

To prove : {(cot A + cosec A – 1) / (cot A – cosec A + 1)} = {(1 + cos A) / sin A}

First LHS = {(cot A + cosec A – 1) / (cot A – cosec A + 1)}

              = {(cot A + cosec A ) – ( cosec² A – cot²A ) } / {(cot A – cosec A) + 1}          { cosec² A – cot²A = 1 }

              =  {( cosec A + cot A ) –  ( cosec A + cot A )( cosec A – cot A ) } / (cot A – cosec A + 1 )

              =  {( cosec A + cot A ) [ 1 – ( cosec A – cot A ) ] } / (cot A – cosec A + 1 )

              =  {( cosec A + cot A ) ( cot A – cosec A  + 1 )} / (cot A – cosec A + 1 )

              =    cosec A + cot A

              =   (1 / sin A) + (cos A / sin A)

              =  (1 + cos A) / sin A

=  RHS
Hence Proved

Problem 4 : If sin θ + cos θ  = √2 , then prove tan θ  + cot θ  = 2 

Solution: 

We have sin θ + cos θ  = √2 

Squaring both sides

(sin θ + cos θ)²  = (√2 )²

sin² θ + cos² θ  + 2 sin θcos θ = 2 

1 + 2 sin θcos θ = 2                                                                                                                                             {sin² θ + cos² θ = 1}

2 sin θcos θ = 2 – 1

2 sin θcos θ = 1

2 sin θcos θ = sin² θ + cos² θ

Divide both sides by sin θcos θ

2 sin θcos θ / sin θcos θ = (sin² θ + cos² θ) / sin θcos θ

2 = (sin² θ / sin θcos θ) + (cos² θ / sin θcos θ)

2 = (sin θ / cos θ) + (cos θ / sin θ)

2 = tan θ  + cot θ

Hence proved, tan θ  + cot θ  = 2.