Trigonometry> Prove that ~sin36° + sin72° + cos18° + co...

Prove that ~sin36° + sin72° + cos18° + cos54° = 5/16

Ayush Baghel , 6 Years ago

Grade 11

3 Answers

Arun

Last Activity: 6 Years ago

Dear Ayush

You have typed a wrong question.

There should be signs of multiplication instead of plus sign.

Arun

Last Activity: 6 Years ago

Sin36° sin72° sin108° sin144°

= sin36° sin72° sin(90° + 18°) sin(90° + 54°)

= sin36° sin72° cos18° cos 54°

= sin^2 36° sin^2 72° {since, cos18° = sin72° and cos54° = sin36°}

= { ( √(10 - 2√5 ) / 4 }2 [ { √(10 + 2√5) } / 4 ]2 {since, sin36° = ( √(10 - 2√5 ) / 4 and sin72° = { √(10 + 2√5) } / 4}

= [ (10 - 2√5 ) / 16 ] [ (10 + 2√5) } / 16 ]

= { (10)^2 - (2√5)^2 } / ( 16 * 16 ) {since, (A - B)(A + B) = A^2 - B^2 }

= (100 - 20) / 256

= 80 / 256

= 5 / 16

Approved

Siddharth Singh

Last Activity: 6 Years ago

Sin36° sin72° sin108° sin144°= sin36° sin72° sin(90° + 18°) sin(90° + 54°)= sin36° sin72° cos18° cos 54°= sin^2 36° sin^2 72° {since, cos18° = sin72° and cos54° = sin36°}= { ( √(10 - 2√5 ) / 4 }2 [ { √(10 + 2√5) } / 4 ]2 {since, sin36° = ( √(10 - 2√5 ) / 4 and sin72° = { √(10 + 2√5) } / 4}= [ (10 - 2√5 ) / 16 ] [ (10 + 2√5) } / 16 ]= { (10)^2 - (2√5)^2 } / ( 16 * 16 ) {since, (A - B)(A + B) = A^2 - B^2 }= (100 - 20) / 256= 80 / 256= 5 / 16Thanks