ahh, test 2mo,, plz help me on this?

Question

hi,, thanks for helping..
so my question involes sine, cosine and tangent.

using a calucaltor how would i find .. lets say sinx = 45 int he interval 0 < x < 560

i cant rember how to do it using that stupid postive quadrent thing,, so do you know the forumla..

like for sin it is 180+x or 360-x (i cant rember this)

so my question... what is the forumula for sin, cos, tan...

and if your fealing relaly darey... can you explain how to do it using the quadrents,, so i dont have to rember all the forumlas...

thanks alot.

wow my spelling sucks when I type in a rush

ok stupid me.. say sinx = 0.5 for my question

JimBurnel... thank you very much for your time/effort.

Unfortunately, this does not solve my dilemma.
What I don’t know/understand is:

Well, lets take sin x = 0.5

Using a calculator I am amble to find out that a = arcsin 0.5 = 30

But my problem, I know that there are also other values.
Doing some more research I know think that it is 180 – a
Or … 180 – 30 = 150

But again... these are only 2 values… how do you calculate all the values…
Also,, for cosine and tangent.

Thanks a lot for help so far, and would really appreciate if you could further assist me with this.

sk

arcsin x = θ + k360° and (180° - θ) + k360°
arccos x = θ + k360° and -θ + k360°
arctan x = θ + k180°

Tats perfect.

I cant thank you enough for helping me with this.

Thank you alot!!!!!!!!!!! This really saved me for tomorrow. I think i also understand how it is done using the graphs.

hmm, but can you explain how we do it (using the graphs) of sin x = -0.5
so a = -30

All the best.


Thx agian

Answer 1

Eh, is this any more helpful?

arcsin x = θ + k360° and (180° - θ) + k360°
arccos x = θ + k360° and -θ + k360°
arctan x = θ + k180°

where k is any integer.

So for example

arccos 0.5 = 60° + k360° and -60° + k360°

or {..., -420°,-300°, -60°, 60°, 300°, 420°, ....}

The reason for these formulas (or at least the sin one) is explained below.

----

arcsin -0.5 = -30° + k360° or (180° - -30°) + k360°

= -30° + k360° or 210° + k360°
...
----

So you're trying to find the angle θ such that:

sin θ = 0.5

If you remember in a 30-60-90 triangle, the shorter side is 1/2 the length of the hypotenuse. Also the shorter side of a 30-60-90 triangle is the side OPPOSITE the 30° angle.

So, sin(30°) = 0.5, and 30° is one of the angles you're looking for.

But also remember that sin is positive in the second quadrant!

So picture drawing a 30° angle in the first quadrant and then "flipping" it around the y-axis. Now, instead of being 30° more than the positive x-axis, it's 30° less than the NEGATIVE x-axis.

The negative x-axis has angle 180°, because it's halfway around the circle.

So the other angle that has a sin of 0.5 is 180° - 30° = 150°.

And then, since you can add or subtract 360° as many times as you want, you can say that the answers are:

30° + k x 360° and 150° + k x 360°, where k is any integer.

Does that help?

If not, read below.

-----

Here's a quick trig refresher.

FUCNTIONS: SOH-CAH-TOA:

SOH: Sine is Opposite over Hypotenuse
sin θ = opposite/hypotenuse.

sin is related to y: wherever y is positive, sin is positive; wherever y is negative, sin is negative.

CAH: Cosine is Adjacent over Hypotenuse
cos θ = adjacent/hypotenuse.

cos is related to x: wherever x is positive, cos is positive; wherever x is negative, cos is negative.

TOA: Tangent is Opposite over Adjacent
tan θ = opposite/adjacent.

tan is related to both y and x: wherever y and x have the same sign, tan is positive; wherever y and x have different signs, tan is negative.

ANGLES:

If you have any angle, you can get the same angle by adding or subtracting 360° or 2π radians any number of times. (Since 2π radians equals 360°, you can convert back and forth easily.)

You can remember this by remembering that there are 360° in a circle, and if you go all the way around in a circle, you end up where you started.

QUADRANTS: Alabama State Teachers College!

Quadrant 1 - Alabama: ALL are positive
The first quadrant is where x and y are both positive, the upper right quadrant. It corresponds to angles between 0° and 90°, or 0 to π/2 radians. Since x and y are both positive, sin, cos, and tan are all positive in the first quadrant.

Quadrant 2 - State: SIN is positive
The second quadrant is where x is negative and y is positive, the upper left quadrant. It corresponds to angles between 90° and 180°, or π/2 to π radians. Since x is negative and y is positive, only sin is positive in the second quadrant. cos and tan are negative.

Quadrant 3 - Teachers: TAN is positive
The third quadrant is where x and y are both negative, the lower right quadrant. It corresponds to angles between 180° and 270°, or π to 3π/2 radians. Since both x and y are negative in the third quadrant, only tan is positive; sin and cos are both negative.

Quadrant 4 - College: COS is positive
The fourth quadrant is where x is positive and y is negative, the lower right quadrant. It corresponds to angles between 180° and 360°, or 3π/2 to 2π radians. Since x is positive and y is negative, only cos is positive in the fourth quadrant. sin and tan are negative.

Answer 2

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