Question

"Question10
Evaluate, cos45° cos30° + sin45° sin30°
Chapter6, Trigonometry, Exercise -6 ,Page number 288"

Answer 1

HELLO DEAR,

we know that:-


cosAcosB + sinAsinB = cos(A - B)


NOW,

cos45° cos30° + sin45° sin30°

=> cos(45° - 30°)

=> cos15° = (√3 + 1)/2√2



another method,



cos45° = 1/√2 = sin45°

cos30° = √3/2 , sin30° = 1/2

NOW,

1/√2 * √3/2 + 1/√2 * 1/2

=> √3/2√2 + 1/2√2

=> (√3 + 1)/2√2



I HOPE ITS HELP DEAR,
THANKS

Answer 2

Hey there!

We know that , cos( A - B ) = cosA cos B + sinA sinB.

So, cos45° cos30° + sin45° sin30° is in the form of cosA cos B + sinA sinB. .

Now,

cos45° cos30° + sin45° sin30°

= cos ( 45 - 30 )

= cos 15

We know that, cos15° = √3+1/2√2

= √3 + 1 / 2√2

[ or ]

We know trigonometric ratios of special angles : cos45 = sin45 = 1/√2 , cos30 = √3/2 , sin30 = 1/2

= cos45° cos30° + sin45° sin30°

= 1/√2 ( √3/2 ) + (1/2 ) ( 1/√2 )

= √3/2√2 + 1/2√2

= √3 + 1 / 2√2

Therefore , cos45° cos30° + sin45° sin30° = √3+1/2√2

Hope helped!