Trig equations and inequality's for X

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Louise Johnson

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Jan 21, 2007
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Hello I am working through my trig and hopefully progressing a little bit. I have a couple of questions I could really use somebody to take a quick peek at.
Thank you
Louise

Solve each equation or inequality for X:

#1 Sin(x+pi/2)=1/2

I used the co-function identity of Sin(pi/2-x) = cos x
Answer Cos x =1/2 in radians = Pi/3 and 5Pi/3
I also put both those radians into the LHS and they worked. I guess I am just looking to see if I am on the right track.

#2 Sinx>Cos x where x is greater or equal to zero and less than 2Pi

The Hint included with this question: Sketch a graph and determine the values of x for which the curve y=sinx is above y= cosx

I graphed both Sin X and Cos X. From the hint I am thinking the answer is just erasing any part of the sin x graph that is underneath the cos x graph. Also because x is greater than 0 the sin wave is only drawn on the right or positive side of the graph.
My problem is that the original question states that I should be solving for something?

#3 If y=sin(bx+c)+d, how does each of the four constants affect the graph of this function

Answer a) represents Amplitude b) represents Period c) represents phase shift
d) represents verticle displacement
 
sin(x + pi/2) = sin(x)cos(pi/2) + cos(x)sin(pi/2) = sin(x)*0 + cos(x)*1 = cos(x)
cos(x) = 1/2 at pi/3 and 5pi/3


sin(x) > cos(x) is the same as sin(x) - cos(x) > 0 ... this would be true where the y-values on the unit circle are greater than the x-values ... pi/4 < x < 5pi/4.
In my mind, this is just a common sense/general knowledge of the unit circle type question rather than an inequality to "solve". Others may differ on this opinion.


y = a*sin(bx + c) + d
|a| = amplitude
2pi/|b| = period
c/b = phase shift (c/b > 0, shift left ... c/b < 0, shift right)
d = vertical shift (d > 0, shift up ... d < 0, shift down)
 
Thanks Skeeter!! I am pretty happy with getting something correct for once. The second question still seems alittle suspect... Anyways I am gonna move on and get some more Trig done :)
Thank you
Louise
 
#2 It should not be difficult to determine where sin(x) = cos(x). With that information, there is little difficult in the determination.

cos(x) is greater from x = 0
After they are equal, it switches and sin(x) is greater.
After they are equal again, it switches back and cos(x) is greater.
 
Hello tkhunny, I have read your explaination through a couple of times now and I see what you are saying. It has helped to draw out the graphs of both to see it for myself. I also understand what skeeter meant by using the unit circle as a reference and you can see where the Y coordinate is larger than the x coordinate.
In the end I graphed both the two overlapping waves and then skeeter's
sin(x)-cos(x)>0. All the question asks is solve the equation or inequality for x so hopefully I have satisfied that by now? Anyways the problem with distance education is by the time I find out I will have moved on to statistics or some other topic....
Thank you for taking the time to help
Sincerly
Louise
 
It could also be beneficial to figure this out:

\(\displaystyle \sin(x) - \cos(x) = \sqrt{2}*\sin(x-\frac{\pi}{4})\)

Then we need to care only about the function greater than zero.
 
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