The value of 2(sin6735∘+cos6735∘) - 3 (sin4735∘+cos4735∘) + 1 is
Given expression
2(sin6735∘+cos6735∘) - 3 (sin4735∘+cos4735∘) + 1
Sin (735∘) = sin (2 × 360∘ + 15∘)
= sin 15∘
= sin (45∘-30∘) = sin45∘cos30∘ - cos45∘ sin30∘
= 1√2 × √32 - 1√2 × 12 = √3−12√2
Similarly cos 735∘ = cos (2 × 360∘ + 15∘) = cos15∘
Cos (45∘-30∘) = cos45∘cos30∘ + sin45∘.sin30∘
= 1√2 × √32 + 1√2 × 12
= √3+12√2
Substituting sin 735∘ & cos 735∘ is the given expression.
2.[(√3−12√2)6+(√3+12√2)6]−3[(√3−12√2)4+(√3+12√2)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