Cos 15degrees
the answer is: squareroot6+squareroot2/4
can you please explain how they got the answer?? thankss
10 points :)
the answer is: squareroot6+squareroot2/4
can you please explain how they got the answer?? thankss
10 points :)
-
In order to use a sum or difference identity to evaluate cos(15°), we need to write 15° as a sum or difference of two special angles. The special angles in Quadrant I are:
0°, 30°, 45°, 60°, and 90°.
Obviously, we cannot pick two of these that sum to 15; however, we can pick two whose difference is 15° (45° and 30°). This suggests that we should write:
15° = 45° - 30°.
So:
cos(15°) = cos(45° - 30°)
= cos(45°)cos(30°) + sin(45°)sin(30°), by the cosine subtraction formula
= (√2/2)(√3/2) + (√2/2)(1/2), from the unit circle
= √6/4 + √2/4
= (√6 + √2)/4, as required.
I hope this helps!
0°, 30°, 45°, 60°, and 90°.
Obviously, we cannot pick two of these that sum to 15; however, we can pick two whose difference is 15° (45° and 30°). This suggests that we should write:
15° = 45° - 30°.
So:
cos(15°) = cos(45° - 30°)
= cos(45°)cos(30°) + sin(45°)sin(30°), by the cosine subtraction formula
= (√2/2)(√3/2) + (√2/2)(1/2), from the unit circle
= √6/4 + √2/4
= (√6 + √2)/4, as required.
I hope this helps!
-
cos15
= cos(45-30)
= cos45 cos30 + sin45 sin30
= (sqrt(2)/2)(sqrt(3)/2) + (sqrt(2)/2)(1/2)
= (sqrt(6) + sqrt(2))/4
= cos(45-30)
= cos45 cos30 + sin45 sin30
= (sqrt(2)/2)(sqrt(3)/2) + (sqrt(2)/2)(1/2)
= (sqrt(6) + sqrt(2))/4