CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The value of 2(sin6735+cos6735) - 3 (sin4735+cos4735) + 1 is __.

Open in App
Solution

Given expression

2(sin6735+cos6735) - 3 (sin4735+cos4735) + 1

Sin (735) = sin (2 × 360 + 15)

= sin 15

= sin (45-30) = sin45cos30 - cos45 sin30

= 12 × 32 - 12 × 12 = 3122

Similarly cos 735 = cos (2 × 360 + 15) = cos15

Cos (45-30) = cos45cos30 + sin45.sin30

= 12 × 32 + 12 × 12

= 3+122

Substituting sin 735 & cos 735 is the given expression.

2.[(3122)6+(3+122)6]3[(3122)4+(3+122)4]+1

Can we further simplify the given expression?

We might be able to simplify. But It's going to be time consuming. It doesn't look like an effective method to solve the given expression.

Just look at the expression it contains the terms of sin6x + cos6x

If we use the identity a3 + b3 = (a+b)(a2-ab+b3)

Where a & b are sin2735 & cos2735 respectively. Using the identity sin2x + cos2x = 1 we can easily simplify the expression.

2[(sin2735)3+ (cos2735)3]- 3(sin4735 + cos4735) + 1

a3 + b3 = (a+b) (a2 + b2 - ab)

2[(sin2735+ cos2735) (sin4735 + cos4735 - sin2735 cos2735)] - 3(sin4735+ cos4735) + 1

Using identity sin2x + cos2x = 1

2(sin4735 + cos4735)- 2sin2735 cos2735 - 3(sin4735 + cos4735) + 1

= -sin4735 - cos4735 - 2sin2735 cos2735 + 1

=-[(sin2735)2+ (cos2735)2 + 2(sin2735) (cos2735)]+1

= - (sin2735+cos2735)2+ 1

= -1+1 = 0


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
General Solutions
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon