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# Which basic trigonometric identity is actually a statement of the Pythagorean Theorem?

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## Suggested Questions And Answer :

### Which basic trigonometric identity is actually a statement of the Pythagorean Theorem?

Sin^x + cos^2x = 1. (Opposite/hypotenuse)^2 + (adjacent/hypotenuse)^2 = 1, so opposite^2 + adjacent^2 = hypotenuse^2, which is Pythagoras' theorem.

### how do you answer this equation sec^2(pi/2-x)-1=cot^2x

Equations that are indented are NOT equal to the initial equation. They are just manipulations of various parts of the initial equation (essentially getting secant and cotangent in terms of cosine and sine). sec^2(pi/2 - x) -1 = cot^2(x) sec(x) = 1 / cos(x) so: sec^2(pi/2 - x) = 1 / cos^2(pi/2 - x) The given equation then becomes [1 / cos^2(pi/2 - x)] - 1 = cot^2(x) cot(x) = 1 / tan(x) tan(x) = sin(x) / cos(x) so: cot(x) = 1 / tan(x) = cos(x) / sin(x) cot^2(x) = [cos(x) / sin(x)]^ = cos^2(x) / sin^2(x) The equation becomes [1 / cos^2(pi/2 - x)] - 1 = cos^2(x) / sin^2(x) cos(pi/2 - x) = sin(x) so: cos^2(pi/2 - x) = sin^2(x) The equation becomes [1 / sin^2(x)] - 1 = cos^2(x) / sin^2(x) Multiply both sides by sin^2(x): [ [1 / sin^2(x)] - 1 ]*sin^2(x) = [cos^2(x) / sin^2(x)]*sin^2(x) 1 - sin^2(x) = cos^2(x) cos^2(x) + sin^2(x) = 1 This is a well-known trigonometric identity and should be enough to show that the initial equation is valid, but I'll prove this identity for the sake of completeness: The Pythagorean Theorem: A^2 +B^2 = C^2 sine is the ratio of the side opposite the angle to the hypotenuse. cosine is the ratio of the side adjacent to the angle and the hypotenuse. If we take a triangle with the right angle at the bottom left corner and put that corner at the origin on a graph, the side opposite the bottom right corner is along the y axis and the side adjacent to that angle is along the x axis. So we adjust our Pythagorean Theorem variables to: x^2 + y^2 = h^2 where h is the hypotenuse. (x^2) / (h^2) + (y^2) / (h^2) = (h^2) / (h^2) (x/h)^2 + (y/h)^2 = 1 ratio of x to h is: x/h = cos(angle) ratio of y to h is: y/h = sin(angle) plug back into equation (x/h)^2 + (y/h)^2 = 1 cos^2(angle) + sin^2(angle) = 1

### mathematical statistics

Mathematically, Bayes' theorem gives the relationship between the probabilities of A and B, P(A) and P(B), and the conditional probabilities of A given B and B given A, P(A|B) and P(B|A). P(A|B) = {P(B|A)*P(A)}/{P(B)} DERIVATION We can do it from set theory applied to conditional probabilities. P(A given B) = P(A and B)/P(B) Likewise, P(B given A) = P(A and B)/P(A) Rearrange to get the common term as follows: P(A and B) = P(A given B).P(B) = P(B given A).P(A) Divide the middle and right hand terms by P(B) on condition that it is not zero: P(A given B) = P(B given A).P(A)/P(B) That is a basic statement of Bayes' Theorem. Statistics Help - http://math.tutorvista.com/statistics.html

### given: cos theta = 3/5 theta is in Q4 FIND cos2theta, sin2theta,tan2theta

cos(2theta)=2cos^2(theta)-1; sin(2theta)=2sin(theta)cos(theta); tan(2theta)=sin(2theta)/cos(2theta). These are trigonometric identities, true for all values of theta. In Q4, draw a triangle ABC with the following coordinates for the vertices: B(0,0), A(3,0), C(0,-4). In this triangle, AB=3, BC=-4, AC=sqrt(3^2+4^2)=5 (hypotenuse) using Pythagoras' theorem. Angle BAC=theta. So cos(theta)=AB/AC=3/5; sin(theta)=BC/AC=-4/5, and sin(2theta)=2sin(theta)cos(theta)=2*3/5*(-4/5)=-24/25, cos(2theta)=2(3/5)^2-1=18/25-1=(18-25)/25=-7/25; tan(2theta)=-24/25÷(-7/25)=24/7. Draw a triangle DEF with E(0,0),  D(-7,0), F(0,-24) in Q3. Angle EDF=2theta, sin(2theta)=EF/DF=-24/25, cos(theta)=DE/DF=-7/25, tan(2theta)=EF/DE=-24/-7=24/7.

### simplify the following expression: cos^2 x + sin^2 x/ cot ^2x - csc^2x

cos^2 x + sin^2 x = 1 cot^2 x - csc^2 x = -1 So your answer is -1. To understand how I got these identities, look below. In a right triangle ABC with B=90 degrees, AB^2 + BC^2 = AC^2 (PYTHAGORAS THEOREM) Divide this by AC^2 and BC^2 and substitute the corresponding trigonometric ratios and you get the answer.

### Triangle ABC is right angled at B and D is mid point of BC. Prove that :AC2=4AD2-3AB2

In △ABC, ∠B=90° By the Pythagorean theorem, AC²=AB²+BC² …Eq.1, In △ABD,  in the same manner, AD²=AB²+BD² …Eq.2    Here, BD=½BC   Thus Eq.2 is rewritten as follows: AD²=AB²+(½BC)² ⇔ 4AD²=4AB²+BC² ⇔ BC²=4AD²－4AB² …Eq.3    Plug Eq.3 into Eq.1 ⇒ AC²=AB²+(4AD²－4AB²) ⇔AC²=4AD²－3AB²                            Q.E.D.

### what kind of viva question can be asked by a teacher about the course of algebra by a masters students

Type of question:  "How can your subjects (algebra and calculus) best be explained progressively to children and young people to enable them to develop their understanding gradually to grasp the principles involved in these abstract subjects, so that they can demonstrate their understanding in applying it in a practical way to everyday problems?" "How can such learning be communicated so as to be a fun thing, rather than cold, formal and analytical, as is often taught?" (Children will often attempt to understand the subjects by learning by rote formulas and the like without any comprehension of how their understanding can be applied. This makes the subjects boring for them when, with the right teaching techniques, it could be fun! If children can be taught at every step of the way in terms of what they already fully understand, they will enthusiastically and gradually develop their understanding of more and more complex, abstract topics.) If you're actually looking for a problem to set students, there are many examples, but the ideal problem is one that gets students to apply their understanding of mathematics in a practical way. Combining geometry with algebra, is one way. Choosing a problem that tests students' understanding of the formula for solving quadratic equations, or trigonometric identities, or rules like the sine and cosine rules, or simultaneous equations, etc., without specifically stating what methods to apply in solving the problem, is a good way of discovering students' mathematical ability in practical application. Such problems could be presented as word problems and could be expressed concisely to appear simple, but nevertheless demanding on the intellect and comprehension.

### find the length of each boundary

find the length of each boundary Vista county is setting aside a large parcel of land to preserve it as open space. the county has hired Jane's surveying firm to survey the parcel, which is in the shape of a right triangle. the longer leg of the triangle measures 5 miles less than the square of the shorter leg, and the hypotenuse of the triangle measures 13 miles less than twice the square of the shorter leg. the length of each boundary is a whole number. find the length of each boundary. Make a the long side, b the short side and c the hypotenuse. a = b^2 - 5 b = b, obviously c = 2b^2 - 13 From the Pythagorean theorem, we have a^2 + b^2 = c^2 (b^2 - 5)^2 + b^2 = (2b^2 - 13)^2 (b^2 - 5) * (b^2 - 5) + b^2 = (2b^2 - 13) * (2b^2 - 13) b^4 - 5b^2 - 5b^2 + 25 + b^2 = 4b^4 - 26b^2 - 26b^2 + 169 b^4 - 9b^2 + 25 = 4b^4 - 52b^2 + 169 Multiply by -1. That way, when the terms on the right side are moved to the left side, the b^4 coefficient will be positive. -b^4 + 9b^2 - 25 = -4b^4 + 52b^2 - 169 -b^4 + 4b^4 + 9b^2 - 52b^2 - 25 + 169 = 0 3b^4 - 43b^2 + 144 = 0 We need to factor the left side. If we use 12 and 12 as the factors for 144, we don't get the required coefficient for the b^2 term. The factors 16 and 9 will work because we multiply the 9 by 3. (3b^2 - 16)(b^2 - 9) = 0 Using those factors we have (3b^2 -16) = 0 or (b^2 - 9) = 0 Solving for b in the first factor: (3b^2 -16) = 0 3b^2 = 16 b^2 = 16/3 = 5.33333 b = 2.3094 This doesn't satisfy the requirement that all lengths are whole numbers. Solving for b in the second factor: (b^2 - 9) = 0 b^2 = 9 b = 3 Actually, b = -3 is also valid, but the length cannot be negative. We have side b = 3 miles, a = b^2 - 5 = 9 - 5 = 4 miles, c = 2b^2 - 13 = 2(9) - 13 = 18 - 13 = 5 miles So, we have a simple 3-4-5 right triangle.

### What is the y-coordinate of the endpoint in quadrant III.

The point (1,6) is one endpoint of a line segement that has a length of square root of 80 units. The other endpoint is in quadrant III and has an x-coordinate of -3. We can draw a right triangle using these two points, and use the Pythagorean theorem to find what we are looking for. The length of the x leg is (1 - -3), which is 4. The length of the y leg is (6 - y). We are given the length of the hypotenuse, the square root of 80. The formula is x^2 + y^2 = c^2  (I used x and y in place of a and b) 4^2 + (6 - y)^2 = 80  (square root of 80, squared) 16 + 36 - 12y + y^2 = 80 Subtract 80 from both sides. 16 + 36 - 12y + y^2 - 80 = 80 - 80 y^2 - 12y - 80 + 36 + 16 = 0 y^2 - 12y - 28 = 0 Factoring, we get the following: (y - 14) * (y + 2) = 0 Because we are multiplying and getting a zero, one of the two factors has to be zero. y - 14 = 0 y = 14 Or... y + 2 = 0 y = -2 The problem statement says the endpoint we are interested in is in quadrant III, which means the y co-ordinate has to be negative, therefore y = -2 The point is (-3, -2)

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