Question

If cos a = 12/13 then find the value of 1/(1 + sin^25 a) + 1/( 1+ cos^25 a)

Answer 1

Given:

Given the value of cosA = 12/13

To Find:

The expression :

1/(1+sin²5A) + 1/(1+cos²5A)

Solution:

We should first find the value of sin5A .

Using the following formula for sin(A+B) ,

• sin5A = sin (2A + 3A)
• sin (2A + 3A) = sin2Acos3A + cos2Asin3A

Given cos A = 12/13, Therefore

• sin A = 5/13

We know the value of sin2A = 2sinAcosA

• sin2A = 2 x 5 x 12 / 13 x 13 = sin2A = 120/169
• cos2A = √1-sin²2A = √1 - 120²/169² =√ (169-120)(120+169)/169
• cos2A = √49x289/169 = 7x17/169 = 119/169

Also, sin3A = sin(2A+A) = sin2AcosA + cos2AsinA

• sin3A = 120 x 12/169 x 13 + 119 x 5/169 x 13
• sin3A  =(1440 + 595 )/13³
• sin3A = 2035/2197
• cos3A = √1-sin²3A = √(2197-2035)(2197+2035)/2197
• cos3A = 828/13³

Now we need to  find sin5A = sin3Acos2A + sin2Acos3A = 2035x119/$13^{5}$ + 120x828/$13^{5}$

• sin5A = (242165 + 99360)/371293  =341525/371293 = 0.9198

• cos5A = √1- sin²5A = √(371293-341525)(371293+341525)/371293

• cos5A = 145668/371293 = 0.39

Therefore sin10A = 2sin5Acos5A

• sin5Acos5A = 0.36

Now

• 1/(1+sin²5A) + 1/(1+cos²5A) = (1+ cos²5A + 1 + sin²5A)/(1+sin²5A)(1+cos²5A) =>
• 3/(1 + sin²5A + cos²5A + (sin5Acos5A)²
• ==>
• 3/(2+0.1296)
• 3/2.1296
• The answer = 1.41

Therefore the value of the expression 1/(1+sin²5A) + 1/(1+cos²5A) is equal to 1.41

Answer 2

1/(1 + sin² 5a) + 1/( 1 + cos² 5a) = 1.41

Step-by-step explanation:

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

On applying sin(A + B) = (sin A × cos B) + (cos A × sin B)

sin (2a + 3a) = (sin 2a × cos 3a) + (cos 2a × sin 3a)

From question, cos a = 12/13 whereas sin a = 5/13

sin 2a = 2 sin a × cos a

sin 2a = 2 × 12/13 × 5/13

∴ sin 2a = 120/169

cos 2a = √(1 - sin² 2a) = √(1 - 120²/169²) =√((169-120)(120+169))/169

cos 2a = √(49 × 289)/169 = (7 × 17)/169

∴ cos 2a = 119/169

On applying sin 3A = sin(2A + A) = (sin2A × cosA) + (cos2A × sinA )

sin 3a = (120/169 × 12/13) + (119/169 × 5/13 )

sin 3a  =(1440 + 595)/13³

∴ sin 3a = 2035/2197

cos 3a = √1 - sin² 3a = √(1 - 2035²/2197²)  = √(2197-2035)(2197+2035)/2197

∴ cos 3a = 828/13³

On applying sin 5A = (sin 3A × cos 2A) + (sin 2A × cos 3A)

sin 5a = (2035 × 119/13⁵) + (120 × 828/13⁵)

sin 5a = (242165 + 99360)/371293  = 341525/371293

∴ sin 5a = 0.9198

cos 5a = √(1- sin² 5a) = √(371293 - 341525)(371293 + 341525)/371293

cos 5a = 145668/371293

∴ cos 5a = 0.39

Therefore sin 10A = 2 sin 5A × cos 5A

sin 5A × cos 5A = 0.36

Now,

1/(1 + sin²5A) + 1/(1 + cos²5A) = (1 + cos² 5A + 1 + sin² 5A)/(1 + sin² 5A)(1 + cos² 5A)

1/(1 + sin²5A) + 1/(1 + cos²5A) = 3/(1 + sin²5A + cos²5A + (sin5Acos5A)²

1/(1 + sin²5A) + 1/(1 + cos²5A) = 3/(2+0.1296)

∴ 1/(1 + sin²5A) + 1/(1 + cos²5A) = 3/2.1296 = 1.41