Ask Singapore Homework?
Upload a photo of a Singapore homework and someone will email you the solution for free.
Question
secondary 4 | A Maths
2 Answers Below
Anyone can contribute an answer, even non-tutors.
need help with this qn, pls explain too
Date Posted: 3 years ago
Views: 352
See 2 Answers
done
{{ upvoteCount }} Upvotes
clear
{{ downvoteCount * -1 }} Downvotes
Q14i
Date Posted:
3 years ago
for the basic angle of sinx = +/- sqrt 1/2, all ASTC are supposed to be considered or is it only A and S?
why is 3pi/4 not rejected tho as the range is 0
why is 3pi/4 not rejected tho as the range is 0
In this particular question, the curve is defined for values of x from 0 to pi, so we only have to consider the first two quadrants.
3pi/4 radians, or equivalently 135 degrees, have a basic angle of pi/4, or equivalently 45 degrees. Clearly the sine of an angle in the second quadrant has a positive value. So, the sine of 3pi/4 has the same signage and value as the sine of pi/4.
From here, the angle 3pi/4 must follow the basic angle pi/4 in computing the value 1 / sqrt 2.
3pi/4 radians, or equivalently 135 degrees, have a basic angle of pi/4, or equivalently 45 degrees. Clearly the sine of an angle in the second quadrant has a positive value. So, the sine of 3pi/4 has the same signage and value as the sine of pi/4.
From here, the angle 3pi/4 must follow the basic angle pi/4 in computing the value 1 / sqrt 2.
sin pi/4 = 1 / sqrt 2 (or sqrt 2 / 2)
3pi/4 is an angle in the second quadrant with basic value pi/4, so
sin 3pi/4
= sin pi/4
= 1 / sqrt 2
3pi/4 is an angle in the second quadrant with basic value pi/4, so
sin 3pi/4
= sin pi/4
= 1 / sqrt 2
done
{{ upvoteCount }} Upvotes
clear
{{ downvoteCount * -1 }} Downvotes
Q14ii
The idea is largely the same as usual type of differentiation questions, just that trigo stuffs are included here.
The idea is largely the same as usual type of differentiation questions, just that trigo stuffs are included here.
Date Posted:
3 years ago
why is 6 included in differentiating (sinx) ^2cosx tho, is it not 6[(sinx)^2cosx]?
If you prefer to leave the 6 out entirely from the differentiation, you can do so. It's not a must for the 6 to follow the other terms during the differentiation.
360 Tutoring Program