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Sever21 [200]
3 years ago
9

How do you know if the question is asking for you to do Pythagoras or Trigonometry? Like in basic terms, what’s the difference??

?
Mathematics
1 answer:
mrs_skeptik [129]3 years ago
7 0

Pythagoras theorem only works for right triangle (90 degree)

Trigonometry (sin, cos, tan) works everywhere!

Hopefully I answered your question!

If you want to learn more about Trigonometry or Pythagoras theorem, send me a message and add me as friend on Brainly. I'm a mathematician and an astrophysicist

:) have a great day!

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Trig proofs with Pythagorean Identities.
lorasvet [3.4K]

To prove:

$\frac{1}{1-\cos x}-\frac{\cos x}{1+\cos x}=2 \cot ^{2} x+1

Solution:

$LHS = \frac{1}{1-\cos x}-\frac{\cos x}{1+\cos x}

Multiply first term by \frac{1+cos x}{1+cos x} and second term by \frac{1-cos x}{1-cos x}.

        $= \frac{1(1+\cos x)}{(1-\cos x)(1+\cos x)}-\frac{\cos x(1-\cos x)}{(1+\cos x)(1-\cos x)}

Using the identity: (a-b)(a+b)=(a^2-b^2)

        $= \frac{1+\cos x}{(1^2-\cos^2 x)}-\frac{\cos x-\cos^2 x}{(1^2-\cos^2 x)}

Denominators are same, you can subtract the fractions.

       $= \frac{1+\cos x-\cos x+\cos^2 x}{(1^2-\cos^2 x)}

Using the identity: 1-\cos ^{2}(x)=\sin ^{2}(x)

       $= \frac{1+\cos^2 x}{\sin^2x}

Using the identity: 1=\cos ^{2}(x)+\sin ^{2}(x)

       $=\frac{\cos ^{2}x+\cos ^{2}x+\sin ^{2}x}{\sin ^{2}x}

       $=\frac{\sin ^{2}x+2 \cos ^{2}x}{\sin ^{2}x} ------------ (1)

RHS=2 \cot ^{2} x+1

Using the identity: \cot (x)=\frac{\cos (x)}{\sin (x)}

        $=1+2\left(\frac{\cos x}{\sin x}\right)^{2}

       $=1+2\frac{\cos^{2} x}{\sin^{2} x}

       $=\frac{\sin^2 x + 2\cos^{2} x}{\sin^2 x} ------------ (2)

Equation (1) = Equation (2)

LHS = RHS

$\frac{1}{1-\cos x}-\frac{\cos x}{1+\cos x}=2 \cot ^{2} x+1

Hence proved.

5 0
3 years ago
Identify the area of the kite. HELP ASAP PLEASE!!
ExtremeBDS [4]

Answer:

2340 m^2

Step-by-step explanation:

The area of kite = multiply the lengths of the two diagonals and divide by 2

The top left triangle:

Using Pythagorean theorem:

c^2 = a^2 + b^2

so

b^2 = c^2 - a^2

b^2 = 51^2 - 24^2

b^2 = 2601 - 576

b^2 = 2025

b = 45

Top right triangle:

Using Pythagorean theorem:

c^2 = a^2 + b^2

so

b^2 = c^2 - a^2

b^2 = 53^2 - 45^2

b^2 = 2809 - 2025

b^2 = 784

b = 28

so  the two diagonals:

24 + 28 = 52 m and 45 + 45 = 90 m

Area of the kite:

A = 52 x 90 / 2

A = 2340 m^2

3 0
3 years ago
128.5 divided by 4.5
Ganezh [65]

Answer:

your answer is 28.5

7 0
3 years ago
Read 2 more answers
Find the ARE of the figure
Tresset [83]
First, break up the shapes into parts. The first one you see is 7 by 4 rectangle on the top.
A=l*w
A=7*4
A=28 for the rectangle, your not done
Now the triangle to the left
A=l*w/2
A=19*10/2
A=95
Now all you got left is the trapezoid, which area's formula is:
A=(b∧1+b∧2*h)/2
A=9+19*8/2
A=112.
Now you have the area for three of the figures, add them up:
112+95+28=235
So in total, the area of this figure is 235 cm^2
5 0
4 years ago
Choose the equation that represents the line that passes through the point (−1, 6) and has a slope of −3.
guapka [62]

Answer:

Step-by-step explanation:

The equation of a line is y=mx+b, where m is the slope and b is the y-intercept. y and x are the y and x values of any point respectively.

The information we have:

y = 6

x = -1

m = -3

6 = -3*-1 + b

6 = 3 + b

b = 3

So now we know that m = -3 and b = 3

y and x were 6 and -1 in this case but we need an equation that works for any point on the line so we can keep y and x just as it is (variables, not numbers).

The final equation is:

y = -3x + 3

4 0
3 years ago
Read 2 more answers
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