Happyface's Methods Questions

[Inside Out]

For pi/2<a<pi with sina=cosb where 0<b<pi/2, find a in terms of b

(pg 206, q. 5e of essential maths bk)





I got a=pi/2 - b
Answer a=pi/2 + b

[xZero]

since a is in the second quadrant, sin(a) = -cos(pi/2 -a)

-cos(pi/2 - a) = cos(b)

cos(-(pi/2-a)) = cos(b)  (note that cosine is a even function)

-(pi/2 - a) = b

-pi/2 + a = b

a = b + pi/2

[evaever]

cosb=sin(pi/2-b) or sin(pi/2+b)
given a is in the second q, .: a=pi/2+b

[jane1234]

xZero's method is fine... but here's a different way you could do it...

Remember that a is in the second quadrant.
So a=pi/2 - b is only the angle if a was in the first quadrant (as that is what b is in).
To make any first-quadrant angle (sometimes called reference angle) into second quadrant just take away the angle from pi.
So pi - (pi/2 - b) = pi/2 + b ... which is the answer
Hope that made sense!!

[Inside Out]

i dont get it....

sin(a) = -cos(pi/2 -a)
why is there a negative in front of cos(pi/2-a)

[jane1234]

i dont get it....

sin(a) = -cos(pi/2 -a)
why is there a negative in front of cos(pi/2-a)

Because cos is negative in the second quadrant. So cos(pi/2-a) would be a negative number (as a is in the second quadrant) so another negative needs to be added to make it a positive (so it equals sin(a))