Trigonometry

You draw a single ray and label it twice: once for the incoming ray and again for the outgoing ray.

Increase the amplitude and the frequency

Please don't change the subject to the weather or your journey home yesterday. I do not want you going off in tangents when we are discussing trigonometric ratios such as tangents.

You have not indicated which side the angle is opposite of. Can us law of cosines then by calling sides c and a.

b^2 = a^2 + c^2 - 2(a)(c) cos(B)

I would arbitrarily have to assign values you have not given me.

The trigonometric functions: sine, cosine, tan as well as their reciprocals are periodic so that their value repeats every 2Ï€ radians. As a result, the inverse operations are functions only over restricted domains, as follows.

Sine : −π/2 ≤ y ≤ π/2

Cosine: 0 ≤ y ≤ π

Tangent: −π/2 < y < π/2

The principal value is the angle that lies within the relevant range. All other solutions are non-principal.

Similarly, the principal square root of a positive real number is the positive square root. The negative root is not the principal root.

cosecant = 1/sine

csc 90 deg = 1/(sin 90 deg)

= 1/1

= 1

False.

Electromagnetic radiation is propogated as sine waves.

y=|x|/4

The range is [0 , ∞ )

jomar

In the complex field, the two numbers are the same.

If you restrict yourself to real solutions, the relationship is as follows:

A polynomial of degree p has p-2k real solutions where k is an integer such that p-2k is non-negative. [There will be 2k pairs of complex conjugate roots.]

sin (theta) = [13* sin (32o)]/8

= 13*0.529919264/8

= 0.861118804

[theta] = sin-1 (0.861118804)

[theta] = 59.44o

Let: o = opposite

h = hypotenuse

a = adjacent

sin = o/h; tan = o/a

Therefore, sin/tan = (o/h)/(o/a)

= (o/h)*(a/o)

= a/h

= cos

If the angle between the ladder and the ground is 60 deg, and you know the angle between the ground and the wall is 90 deg, then you have a 30-60-90 degree triangle, which is a common triangle. You should memorize this one. The commonest sides of this right triangle are 4-5-6, with the longest side being the hypoteneuse, in this case the ladder leaning from the ground to the wall. The wall is 4m high, the base of the ladder would be 5m out from the wall, and the length of the ladder is 6m.

No.

All of the angles in a triangle add up to 180, so if one angel was 34, and another angle was 67, than the mystery angle would be 79. I got that because 180-(a+b)=c. 34 is a, and 67 is b.

0.5*0.24*72 = 5.88 units of distance per second.

The answer depends on what information you have about theta!

The sine of 52.5 degrees equals 0.79335334029124. Hope I helped!

If the hypotenuse and one leg of a right angled triangle are congruent to the hypotenuse and leg of another right angled triangle, then the two triangles are congruent.

cos (3pi x/3) = 0.5

The number of answers depends on the range of angles. I will solve this question using the range 0<x<2 pi. Draw a unit circle, you will see that the quadrants where cosine gives a positive value are the first and the fourth quadrants.

The Angles whose cosines give + 0.5 are: pi / 3 and 5pi / 3. (Note that 7 pi/3 and 11 pi/3 are also possible solutions but they lie out of the range 0<x<2pi.)

3pi x/3 = pi/3 OR 3pi x/3 = 5pi/3

Here you can solve the equations to give x = 1/3; 5/3

If you meant "Pythagorean Theorem" , the uses are almost infinite.

It is associated with finding the length of the "hypotenuse" of any right-angled triangle, given that the other two sides are known. However, a modified version of the Pythagorean Theorem allows us to find the length of any one side of any triangle, given that we know the other two sides, and the angle between them.

In physics, many calculations are based on the Pythagorean Theorem.

For Example,

The use of Trigonometric Parallax allows us to calculate the distance to relatively near stars.It involves the usage the Sun, Earth and the star in question as vertices of the right-angled triangle.

He is known for his significant advances in trigonometry.

None. A wheel, be definition, is circular and therefore has no vertices!