Question on basic trigometric equation.. pls help me!!?

Question

pls include proper steps and working for the questions. extra explaination on steps would be highly appreciated. Help needed urgently.. Thks for helping

1. Find all the angles between -360 degree and 180 degree such
that

(a) sin x = - 0.5
(b) cos x = (√3) / 2
(c) tan (-x) + 1 = 0
(d) (√2) sin (90 degree -x) +1 = 0


for your information, here are the ans.. pls give me the steps in achieving this ans.
(a) -150, -30
(b) -330, -30, 30
(c) -315, -135, 45
(d) -225, -135, 135
*** ALL ANSWER ARE IN DEGREE!!***

Answer 1

Let's go one by one:

(a) sin x = -0.5 → x = 2kπ + Arc sin (-0.5) or x = 2kπ + [π – Arc sin (-0.5)]
Arc sin (-0.5) is the angle like θ which is between -90 and 90 degree (between -π/2 and π/2). So, Arc sin (-0.5) is -30 degrees or -π/6. So x = 2kπ + Arc sin (-0.5) = 2kπ + (-π/6) = 2kπ - π/6 or x = 2kπ + [π – Arc sin (-0.5)] = 2kπ + [π - (-π/6)] = 2kπ + [π + π/6)] = 2kπ + 7π/6
You want the answers between -360 and 180, so we start we k = -1.
k = -1 → x = -2π - π/6 = -13π/6 = -390 degree this is not in the range. Now we put k = -1 in the second formula. k = -1 → x = -2π + 7π/6 = -5π/6 = -150 degree. It's in the range so its one of the answers: x = -150.
Now let try k = 0 → x = 0 - π/6 = -π/6 = -30 degree. It's in the range too so it's one other answer: x = -30. Now we put k = 0 in the second formula. k = 0 → x = 0 + 7π/6 = 7π/6 = 210 degree and its out of the range. So the answers were x=-150 and x=-30.

(b) cos x = (√¯3)/2 → x = 2kπ ± Arc cos (√¯3)/2
Arc cos (√¯3)/2 is an angle like θ which is between 0 and 180 degree and cos θ = (√¯3)/2
So, Arc cos (√¯3)/2 is +30 degree or π/6.
x = 2kπ ± Arc cos (√¯3)/2 → x = 2kπ ± π/6
k = -1 → x = -2π ± π/6 → x = -13π/6 = -390 degree or x = -11π/6 = -330 degree. Just the second one is in the range.
K = 0 → x = 0 ± π/6 → x = π/6 = 30 degree or x = -π/6 = -30 degree. Both of these answers are in the range. Therefore, x = -330, x = -30 or x = 30 degree.

(c) tan (-x) + 1 = 0
Note: tan (-x) = -tan x. Thus, -tan x + 1 = 0 → tan x = 1 → x = kπ + Arc tan x = kπ + Arc tan 1 = kπ + π/4.
K = -2 → x = -2π + π/4 = -7π/4 = -315 degree. Its in the range.
K = -1 → x = -π + π/4 = -3π/4 = -135 degree. Its in the range too.
K = 0 → x = 0 + π/4 = π/4 = 45 degree. Its in the range too.
K = 1 → x = π + π/4 = 5π/4 = 225 degree. Its not in the range.
So the answers are x = -315, x = -135 or x = 45 degree.

(d) √¯2 sin (90 degree - x) + 1 = 0
Note: sin (90 degree - x) = cos x. so √¯2 cos x + 1 = 0 → cos x = -1/√¯2 = -(√¯2)/2 → x = 2kπ ± arcos[-(√¯2)/2] = 2kπ ± 3π/4
K = -1 → x = -2π ± 3π/4 → x = -7π/4 = -495 degree or x = -π/4 = -45 degree.
Between these answers just -45 degree is in the range.
K = 0 → x = 0 ± 3π/4 → x = -3π/4 = -135 degree or x = 3π/4 = 135 degree.
Both of them are in the range.
K = 1 → x = 2π ± 3π/4 → x = 5π/4 = 225 degree or x = 11π/4 = 495 degree.
No one is in the range.
So there are three answers: x = -135, x = -45 or x = 135 degree.

Have fun.

Answer 2

You already probably know that sin 30 = .5. In a circle with radius 1, sinx is equal to the y-coordinate where the radius intersects the circle. Each quadrant of the x-y graph has an angle where the sine is positive or negative .5 because the graph is symmetric. So for (a), first figure out which quadrants have a negative sine or y-coordinate. Those quadrants will be the third and fourth. Use 30 degrees as your reference. Since you're going in the negative direction, the 4th quadrant angle will be -30. The third quadrant angle will be directly accross from the -30 (symmetry again) which puts it at -150.

For cos, think of the x-coordinate instead of y. Tan is sin/cos, so you have to see what the combination of two numbers does. For example, tangent is negative in quadrants 2 and 4 because the signs are different. So for (c), move the 1 to the other side and ask which basic angle has a sin and cos that would divide to -1.

I know that this doesn't cover everything, but I hope it's a good start.

Answer 3

First of all I can give you the signs for each quadrant.

First Quadrant: Sin = + Cos = + Tan = +
Second quadrant: Sin = + Cos = - Tan = -
Third Quadrant: Sin = - Cos = - Tan = +
Fourth Quadrant: Sin = - Cos = + Tan = -

This information is helpful when you are dealing with angles greater than 360 degrees.

Inverse sine, Cos, and Tangents may be found with a Slide rule or from Trig Tables.

Without seeming to do your work for you, Which would do you no good. I am sending you the only tools you will need to verify the answers.

Good Luck and happy studying

Answer 4

You need to consider the periods of the sin and cosine functions, both are 2pi radians (i.e 360 degrees)

if sin x = y
x = n.pi + (-1)^n . arcsin y
where n is an integer

if cos x = y
x= 2n.pi + arccos y
where n is an integer
For the above radian measure is necessary, you can easily convert as pi radians = 180 degrees
Note that the arcsin and arccos functions have a limited domain such that they can be unique functions.

Also sin 30 = 1/2
cos 30 = V3/2
tan 45 = 1
sin 45 = cos 45 = 1/V2

are principle angles.
Note also that sin (-x) = -sin x
cos (-x) = cos x
tan (-x) = -tan x

Answer 5

sinx=-.5=-1/2 draw triangle with opp=-1 and hyp=2
one is in Q3 other in Q4 it is a 30 deg triangle so ans. are -180+30=-150 and 0-30=-30

let R(X) mean square root of x. draw triangle with adj=R(3) and hyp=2 it is a 30 deg triangle in Q1 it is 30 in Q4 it is -30 and -360+30=-330

tan(-x)+1=0 =>tan(-x)=-1 =>-tan(x)=-1 =>tan(x)=1 draw triangle with opp=adj=1 in their respective quadrants reason answer out as i did first 2

subtract 1 divide by R(2) draw appropriate triangles in Q's
keeping in mind hyp>0 always. find angles then set each=90-x solve for x and find angles so they are between -360 and 180 as required