Trigonometry

When none of the angles are known, and using Pythagoras, the triangle is known not to be right angled.

cosecant(x) = 1/sin(x)

For any angle in the triangle that is not a right angle, the opposite side is the side does not touch the angle and the adjacent side touches it and is perpendicular to the opposite side. The third side is the longest side and is the hypotenuse

cot(15)=1/tan(15)

Let us find tan(15)

tan(15)=tan(45-30)

tan(a-b) = (tan(a)-tan(b))/(1+tan(a)tan(b))

tan(45-30)= (tan(45)-tan(30))/(1+tan(45)tan(30))

substitute tan(45)=1 and tan(30)=1/√3 into the equation.

tan(45-30) = (1- 1/√3) / (1+1/√3)

=(√3-1)/(√3+1)

The exact value of cot(15) is the reciprocal of the above which is:

(√3+1) /(√3-1)

I was wondering if you knew but I think sophomore or junior. I was two grade levels ahead in math so I don't know for sure.

It is trigonometry.

It is approx 0.9725

This may not be the most efficient method but ...

Let the three angle be A, B and C.

Then note that A + B + C = 20+32+38 = 90

so that C = 90-A+B.

Therefore,

sin(C) = sin[(90-(A+B) = cos(A+B)

and cos(C) = cos[(90-(A+B) = sin(A+B).

So that tan(C) = sin(C)/cos(C) = cos(A+B) / sin(A+B) = cot(A+B)

Now, tan(A+B) = [tan(A)+tan(B)] / [1- tan(A)*tan(B)]

so cot(A+B) = [1- tan(A)*tan(B)] / [tan(A)+tan(B)]

The given expressin is

tan(A)*tan(B) + tan(B)*tan(C) + tan(C)*tan(A)

= tan(A)*tan(B) + [tan(B) + tan(A)]*cot(A+B)

substituting for cot(A+B) gives

= tan(A)*tan(B) + [tan(B) + tan(A)]*[1- tan(A)*tan(B)]/[tan(A)+tan(B)]

cancelling [tan(B) + tan(A)] and [tan(A) + tan(B)], which are equal, in the second expression.

= tan(A)*tan(B) + [1- tan(A)*tan(B)]

= 1

2.41

(40000000 x .00002)=800

-2(cot2theta)

The cosine of 10 degrees is 0.98480775301221. Hope I helped!

cosine = adjacent/hypotenuse

Period is how long it takes for the sine and cosine functions to restart repeating themselves. Both have a period of 2pi (360 degrees).

The cosine of 70 degrees is 0.34202014332567.

To find the exact value of tan 105°.

First, of all, we note that

sin 105° = cos 15°; and

cos 105° = -sin 15°.

Thus, tan 105° = -cot 15° = -1 / tan 15°.

Using the formula

tan(α - β) = (tan α - tan β) / (1 + tan α tan β);

and using, also, the familiar values

tan 45° = 1, and

tan 30° = ½ / (½√3) = 1/√3 = ⅓√3;

we have,

tan 15° = (1 - ⅓√3) / (1 + ⅓√3);

whence,

cot 15° = (1 + ⅓√3) / (1 - ⅓√3)

= (√3 + 1) / (√3 - 1) {multiplying through by √3}

= (√3 + 1)2 / (√3 + 1)(√3 - 1)

= (3 + 2√3 + 1) / (3 - 1)

= (4 + 2√3) / 2

= 2 + √3.

Therefore,

tan 105° = -cot 15° = -2 - √3,

which is the result we sought.

We are asked the exact value of tan 105°, which we gave above.

We can test the above result to 9 decimal places, say, by means of a calculator:

-2 - √3 = -3.732050808; and

tan 105° = -3.732050808;

thus indicating that we have probably got the right result.

sin(theta) = 15/17, cosec(theta) = 17/15

cos(theta) = -8/17, sec(theta) = -17/8

cotan(theta) = -8/15

theta = 2.0608 radians.

Just as at 90 degrees, it goes to infinity. That is because tangent of angle is opposite side over adjacent side. The oppsoite side gets bigger and bigger after 45 dgerees ( and in your case then after 225 degrees) until it grows beyond all bounds at 270 degrees

-1

sin2x / (1-cos x) = (1-cos2x) / (1-cos x) = (1-cos x)(1+cos x) / (1-cos x) = (1+cos x)

sin2x=1-cos2x as sin2x+cos2x=1

1-cos2x = (1-cos x)(1+cos x) as a2-b2=(a-b)(a+b)

The radical answer is sqrt(3)/2.

(0.86602540378443864676372317075294)

It is a simple trigonometric equation. However, without information on whether the angles are measures in degrees or radians, and with no domain for theta, the equation cannot be solved.

All you have to do is to select the values you want to have compared with Excel. Then click on Insert on the top, then Scatter, then Scatter with only markers. After this, you can give name to the axises, but then in the Layout section you have to select the regression type of the trendline as well.

A function f(x) is even if:

f(x) = f(-x)

In layman's terms this property simply means that any real number in the domain and it's opposite will yield the same function value in the range.

To simplify this down even further, an even function, when graphed will appear to be symetric about the y-axis (assuming that you use the standard Cartesian coordinate plane).

In the case of trig functions, you would have to test whether the even function property holds true for each. We will the test points π, π/2, or π/4. NOTE: The # signs are present next to the functions that are even:

1. Sine: f(x) = sin(x)

-> sin(π/2) = 1, but sin(-π/2) = -1. Since 1 does not equal -1, sine is NOT an even function.

2. #Cosine: f(x) = cos(x)

-> cos(π) = -1 = cos(-π). Since both are equal, cosine IS an even function.

3. Tangent: f(x) = tan(x)

-> tan(π/4) = 1, but tan(-π/4) = -1. Therefore, tangent is NOT an even function.

4. Cosecant: f(x) = csc(x)

-> csc(π/2) = 1, but csc(-π/2) = -1. Therefore, cosecant is NOT an even function.

5. #Secant: f(x) = sec(x)

-> sec(π) = -1 = sec(-π). Since the secant function has asymptotes, it IS an even function provided that x does not equal π(2n+1)/2, where n may be all integers.

6. Cotangent: f(x) = cot(x)

-> cot(π/4) = 1, but cot(-π/2) = -1. Therefore cotangent is NOT even.

f(x) = 1/x except where x = 0.