Trigonometry

We stat with the law of cosines, which we can assume to be true:

* c2 = a2 + b2 - 2ab*cos(C) Then rearrange it:

* cos(C) = a2 + b2 - c2/2ab Use the identity sin(x)=SQRT(1-cos(x))

* sin(C) = SQRT( 1 - (a2 + b2 - c2/2ab)2) Use the operator A = 1/2ab*sin(C) where A is area. Also, set one equal to 4a2b2 and factor it out.

* 2A/ab = SQRT(4a2b2 - (a2 + b2 - c2)2)/2ab ab's cancel, and the term inside the square root is the difference of two squares.

* A = 1/4*SQRT((2ab - (a2 + b2 - c2))(2ab + (a2 + b2 - c2))) when the two groups are simplified, the can be factored in binomial squares.

* A = 1/4*SQRT((c2 - (a - b)2)((a + b)2 - c2) Once again, we have differences of squares.

* A = 1/4*SQRT((c - (a - b))(c + (a - b))((a + b) - c)((a + b + c)) Simplify.

* A = 1/4*SQRT((c + b - a)(c + a - b)(a + b - c)(a + b +c)) Here comes the tricky part. We have four parts here. Three have two terms positive and one negative. Having a + b + c is like having the P. If we have a + b - c, that is like saying P - 2c, right? So to make it even easier, we can call s, the semi-perimeter, P/2. Then we can say a + b - c is 2s - 2c, or 2(s - c). We can apply that to all parts except the last one, which is just 2s.

* A = 1/4*SQRT(2(s - a)*2(s - b)*2(s - c)*2s) The two's can multiply together to 16 and come out of the root, canceling with the 1/4. we are left with good old Heron's formula.

* A = SQRT(s(s - a)(s - b)(s - c))

It's possible that either the angles or sides are labeled according to length or size.

Let's look at right triangles for a moment. In any right triangle, the hypotenuse is the side opposite the right angle. There exist three ratios (and their inverses) as regards the length of the sides of the right triangle. These are opposite/hypotenuse (called the sine function), adjacent/hypotenuse (called the cosine function), and opposite/adjacent (called the tangent function). The inverse of the sine is the cosecant, the inverse of the cosine is the secant, and the inverse of the tangent is the cotangent. The abbreviations for these functions are, sin, cos, tan, csc, sec and cot, respectively.

What is underneath this idea is that for any (every!) right triangle, there is a fundamental relationship or ratio between the lengths of the sides for all triangles with the same angles. For instance, if we have a triangle with interior angles of 30 and 60 degrees (in addition to the right angle), regardless of what size it is, the ratio of the lengths of the sides is always the same. And the trigonometric functions express the ratios of the lengths of the sides.

well in order to get sine b you will have to got to your calculator and reverse the equation ... in other words on the calculator you will see sin-1 you will hit that and in the parenthesis you put .96 .so it should look like this sin-1(.96) and you qet your answer .!

Can you use trigonomic functions in real life situations? It's not like you carry a calculator with you everywhere...

Very unlikely unless you have a job that requires trig skills.

12.0 if you want 3 sig-figs, or 12 if you only want 2.

There are four quadrants. They are represented by Roman numerals : I(one), II(two), III(three), IV(four). The first quadrant contains all positive points , (+x, +y) The second quadrant contains negative x's and positive y's , (-x, +y)

The third quadrant is all negative , (-x, -y)

The fourth quadrant has negative y's and positive x's , (+x, -y)

Placing a question mark at the end of a phrase does not make it a sensible question. Try to use a whole sentence to describe what it is that you want answered.

Since theta is in the second quadrant, sin(theta) is positive.

sin2(theta) = 1 - cos2(theta) = 0.803

So sin(theta) = +sqrt(0.803) = 0.896.

x + y = 28 x*y = 7 x=7/y replacing X in eq 1 7/y+y=28 y^2 - 28y + 7 = 0 using above solutions find the reciprocal and sum -- Dhruv

An equilateral triangle has 3 equal sides and 3 equal 60 interior angles that add up to 180 degrees

Whenever architecture involves the use of lines that are not on the x or y axis, it will involve trigonometry to calculate the length of lines and the angles they make from one another.
One example is calculating roof pitch.

8pi/9

Think of tangent as sin divided by cos or sin/cos. Now cos 90 degrees is 0 so tan90 would be 1/0 which is not define since you are now allowed to divide by 0. It's best to visualize what Tan(90) means: Take a typical right angle triangle: From the angle in question (A), you have the adjacent leg (The leg extending from angle A to the right angle opposite), the opposite leg (the leg directly opposite the angle A, of course) and the hypotenuse (the leg intersecting the aforementioned two legs) Tan(A) is the ratio of the opposite leg and the adjacent leg. To solve for Tan (A): Tan (A) = opposite/adjacent. As angle (A) increases, the length of the opposite side also increases. So what happens at A=90? Well, you no longer have a triangle! The hypotenuse (the leg that intersects the opposite side from the adjacent side) NEVER INTERSECTS. Therefore, no longer being a triangle, you can no longer define the ratio.

If none of the letters were repeated, the answer would be 7!. However, the letters M & E are repeated. So, we need to divide the 7! by 2! twice to account for the two letters repeated twice. The solution is: 7! / (2! * 2!). This written out is: (7*6*5*4*3*2*1)/(2*1*2*1) or 1260.

The circumference of a circle C is 2Pixr So solving for r we have C/2Pi=r

If sin θ = tan θ, that means cos θ is 1 (since tan θ = (sin θ)/(cos θ))

(Usually in and equation a/b=a, b doesn't have to be 1 when a is 0, but cos θ = 1 if and only if sin θ = 0)

The angles that satisfy cos θ = 1 is 2n(pi) (or 360n in degrees)

When n is an integer.

But if sin θ = tan θ = θ, the only answer is θ = 0.

Because sin 0 is 0 and cos 0 is 1 and tan 0 is 0

The only answer would be when θ = 0.

Cosine squared theta = 1 + Sine squared theta

If x2 + y2 = 1, then the point (x,y) is a point on the unit circle.

A Quadrantal angle is an angle that is not in Quadrant I. Consider angle 120. You want to find cos(120) . 120 lies in quadrant II. Also, 120=180-60. So, it is enough to find cos(60) and put the proper sign. cos(60)=1/2. Cosine is negative in quadrant II, Therefore, cos(120) = -1/2.

It is 3 minus 2i

Acute angles (less than 90 degrees)

Obtuse angles (greater than 90 degrees)

Right angles ( equal to 90 degrees)

The trigonometric functions are sine, cosine and tangent along with their reciprocals and the inverses. Whether the angle is acute or obtuse (or reflex) makes no difference).

The related link shows a set of complex numbers that depict the Electro-Magnetic

fields around two wires. The formula is (z-1)/(z+1)