Question

prove that cos 10°.cos 50°.cos 60° .cos 70° =​

Answer 1

Ans---> √3 / 16

To find---> Cos10° Cos50° Cos60° Cos70°

Solution---> We have an identity,

2CosA CosB = Cos( A + B ) Cos( A - B )

Cos10° Cos50° Cos60° Cos70°

Multiplying and dividing by 2, we get,

= ( 2 Cos70° Cos 50° ) Cos10° ( 1/2 ) / 2

= {cos( 70° + 50° ) + Cos( 70° - 50° )}Cos10° / 4

= ( Cos 120° + Cos 20° ) Cos10° / 4

= (1/4) ( Cos120° Cos10° + Cos20° Cos10° )

= ( 1 / 8 ) (2Cos120° Cos10° + 2 Cos20° Cos10° )

Putting Cos120° = -1/2 in it we get

= ( 1/8 ){2(-1/2) Cos10°+Cos(20° + 10°)+Cos(20°-10°)}

=(1/8) (- Cos 10° + Cos30° + Cos 10° )

- Cos10° and Cos10° cancel out each other

= ( 1 / 8 ) Cos30°

Putting Cos30° = √3 / 2 in it

= ( 1 / 8 ) ( √3 / 2 )

= √3 / 16

Answer 2

Good Night!

Step-by-step explanation:

2CosA CosB = Cos( A + B ) Cos( A - B )

Cos10° Cos50° Cos60° Cos70°

Multiplying and dividing by 2, we get,

= ( 2 Cos70° Cos 50° ) Cos10° ( 1/2 ) / 2

= {cos( 70° + 50° ) + Cos( 70° - 50° )}Cos10° / 4

= ( Cos 120° + Cos 20° ) Cos10° / 4

= (1/4) ( Cos120° Cos10° + Cos20° Cos10° )

= ( 1 / 8 ) (2Cos120° Cos10° + 2 Cos20° Cos10° )

Putting Cos120° = -1/2 in it we get

= ( 1/8 ){2(-1/2) Cos10°+Cos(20° + 10°)+Cos(20°-10°)}

=(1/8) (- Cos 10° + Cos30° + Cos 10° )

- Cos10° and Cos10° cancel out each other

= ( 1 / 8 ) Cos30°

Putting Cos30° = √3 / 2 in it

= ( 1 / 8 ) ( √3 / 2 )

= √3 / 16