sin(90+x)=cosx
therefore substitute sin(90+x)=cosx
and then solve the trigonometric equation cosx = 2 cos^2 (x) which is easy...
Hay,
I started with co-functions a little while back, I got the concept and understand the A + B = 90^{0},
But i got this one that threw me a curve-ball,
sin (90^{0} + x) = 2 cos^{2}x
I spoke to my lecturer about this one and he said just to treat the second part as a normal square,
Where is what i tried...
sin (90^{0} + x) = 2[sin(90^{0}-x)^{2}]
sin (90^{0} + x) = 2[(90^{0}-x)(90^{0}-x)]
90^{0}-x = 2[810^{0} - 180^{0}x + x^{2}]
90^{0}-x = 1620^{0} - 360^{0}x + 2x
2x^{2} - 361^{0}x + 1530^{0} = 0
Using quadratic formula:
.'. x = 3 + "root 12231" /4 or .'. x = 3 -"root 12231" /4
.'. x = 28.398 or .'. x = -27.145
Did i miss something here or look at it from a wrong angle?
Thanks Minoanman,
I didn't know you where allowed to substitute,
I thought the point of the exercise was to determine x with only using co-functions,
sin (90^{0} + x) = 2 cos^{2}x
cosx = 2cos^{2}x
divide both side with cosx
.'. 1 = 2 cosx
.'. cosx = 1/2
.'. x = 60^{0}; 90^{0}; 270^{0}; 300^{0}
I must say i'm not sure where they got the 90 and 270 degree from,
I just copied it from the text book,
Doesn't 90 and 270 degree come from the fact the cos 90 and cos 270 = 0 and -0,
So if i put in 0 on both sides of the equation ( cosx = 2cos^{2}x )
Then i sould be ok???
Since you say you understand how "co-functions" are related you could do it this way: sin(90+ x)= sin(90- (-x))= cos(-x). And since cosine is an even function, cos(-x)= cos(x), so
sin(90+ x)= 2cos^2(x) is the same as cos(x)= 2 cos^2(x). cos(x)= 0 is one solution which gives for any integer n. If cos(x) is not 0, we can divide both sides by cos(x) giving 1= 2cos(x) or cos(x)= 1/2. From that, [tex]x= \pi/3+ 2n\pi[tex] and .