cos(105°)+sin(105°)=cos45°
Answers
Answer:
Taking LHS:
sin 105° + cos 105°
cos (90° - 105°) + cos 105°
cos 15° + cos 105°
cos C + cos D = 2(cos(C+D)/2)(cos(C-D)/2)
2cos(120°/2)cos(90°/2)
2cos60°cos45°
2(1/2)cos45°
cos45°
= RHS
I could also have converted to sin:
sin 105° + sin (90° – 105°)
sin 105° - sin 15°
sin C - sin D = 2(cos(C+D)/2)(sin(C-D)/2)
2cos(120°/2)sin(90°/2)
2cos60°sin45°
2(1/2)sin45°
sin 45°
= cos 45°
= RHS
- Route 2: Trigonometric Identities
We have:
sin 105° + cos 105° = cos 45°
Let me multiply both sides by 1/√2
Why? You'll see shortly.
(1/√2)(sin 105° + cos 105°) = (1/√2)cos 45°
You must know that (1/√2) = cos 45° = sin 45°
That gives me
(1/√2)sin 105° + (1/√2)cos 105° = (1/√2)cos45°
At this point, I can do two simple things:
- Write that as:
sin45°sin105° + cos45°cos105° = cos^2 45°
sinAcosB + cosAsinB = sin(A+B)
cos (105° - 45°) = cos^2 45°
cos 60° = cos^2 45°
1/2 = 1/2
- Write that as:
cos45°sin105° + sin45°cos105° = sin45°cos45°
2sinAcosA = sin(A+A) || sinAcosA = (1/2)sin(A+A)
sin(105° + 45°) = (1/2)sin(45° + 45°)
sin 150° = (1/2)sin 90°
sin (180° - 30°) = (1/2)
sin 30° = 1/2
1/2 = 1/2