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mr_godi [17]
2 years ago
12

Prove the identity: 2sin(a+b)sin(a-b) = cos(2b)-cos(2a)

Mathematics
2 answers:
Aleonysh [2.5K]2 years ago
4 0

2sin(a+b)sin(a-b)=cos(2b)-cos(2a) \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{let's do this part first, we'll add the "2" later}}{sin(a+b)sin(a-b)} \\\\\\ \underset{\textit{difference of squares}}{[sin(a)cos(b)+cos(a)sin(b)][sin(a)cos(b)-cos(a)sin(b)]} \\\\\\ sin^2(a)cos^2(b)-cos^2(a)sin^2(b) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{\textit{we know that}}{cos(2\theta )=cos^2(\theta)-sin^2(\theta)}\implies sin^2(\theta)=cos^2(\theta)-cos(2\theta ) \\\\[-0.35em] \rule{34em}{0.25pt}

[cos^2(a)-cos(2a)]cos^2(b)-cos^2(a)[cos^2(b)-cos(2b)] \\\\\\ \underline{cos^2(a)cos^2(b)}-cos(2a)cos^2(b)\underline{-cos^2(a)cos^2(b)}+cos^2(a)cos(2b) \\\\\\ cos^2(a)cos(2b)-cos(2a)cos^2(b) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{\textit{we also know that}}{cos(2\theta)=2cos^2(\theta)-1}\implies \cfrac{cos(2\theta)+1}{2}=cos^2(\theta) \\\\[-0.35em] \rule{34em}{0.25pt}

\left[ \cfrac{cos(2a)+1}{2} \right]cos(2b)-cos(2a)\left[ \cfrac{cos(2b)+1}{2} \right] \\\\\\ \stackrel{\textit{now let's bring back the "2"}}{2\left( \left[ \cfrac{cos(2a)+1}{2} \right]cos(2b)-cos(2a)\left[ \cfrac{cos(2b)+1}{2} \right] \right)} \\\\\\\ [cos(2a)+1]cos(2b)-cos(2a)[cos(2b)+1] \\\\\\ \underline{cos(2a)cos(2b)}+cos(2b)-\underline{cos(2a)cos(2b)}-cos(2a)\implies cos(2b)-cos(2a)

Alex777 [14]2 years ago
3 0

Answer:

see below

Step-by-step explanation:

see attached for my workings, step-by-step and the trig identities I used.

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Choose what the expressions below best represent within the context of the word problem. The tens digit of a number is twice the
Usimov [2.4K]
<span>2x + x = 12
=> x =12/3 =4
so, original number is 84.

</span>
6 0
3 years ago
Graph f(x) =-2x^+16x-30 by factoring to find the solutions, then find the coordinates of the vertex, and the axis of symmetry. I
fredd [130]

Answer:

<em>Observe attached image</em>

<em>Function zeros:</em>

(3, 0), (5, 0)

<em>Vertex:</em>

(4, 2)

<em>Axis of symmetry:</em>

<em>x =4</em>

Step-by-step explanation:

<u>First factorize the function</u>

f (x) = -2x ^ 2 + 16x-30

<em>Take -2 as a common factor.</em>

-2(x ^ 2 -8x +15)

<em>Now factor the expression x ^ 2 -8x +15</em>

You must find two numbers that when you add them, obtain the result -8 and multiplying those numbers results in 15.

These numbers are -5 and -3

Then we can factor the expression in the following way:

f (x) = -2(x-5)(x-3)

<em><u>The quadratic function cuts the x-axis at </u></em><em>x = 3 and at x = 5.</em>

Now we find the coordinates of the vertex.

For a function of the form ax ^ 2 + bx + c the x coordinate of its vertex is:

x = \frac{-b}{2a}

In the function f (x) = -2x ^ 2 + 16x-30

a = -2\\b = 16\\c = 30

<u>Then the vertice is:</u>

x = \frac{-16}{2(-2)}\\\\x = 4

The y coordinate of the symmetry axis is

y = f (4) = -2 (4) ^ 2 +16 (4) -30\\\\y = 2

The axis of symmetry is a vertical line that cuts the parabola in two equal halves. This axis of symmetry always passes through the vertex.

<u>Then the axis of symmetry is the line</u>

x = 4

<u>The solutions and the vertice written as ordered pairs are:</u>

<em>Function zeros:</em>

(3, 0), (5, 0)

<em>Vertex:</em>

(4, 2)

6 0
3 years ago
Pls solve question 4b. Verrry urgent . Ok tyy
kodGreya [7K]

Answer:

y = 3\left(x-\dfrac{3}{2}\right)^2+\dfrac{41}{4}

Step-by-step explanation:

Given equation:

y = 3x^2 - 9x + 17

Factor out 3 from the first 2 terms:

y = 3(x^2 - 3x) + 17

Divide the coefficient of x by 2 and square it:  (-3 ÷ 2)² = 9/4

Add this inside the parentheses and subtract the distributed value of it outside the parentheses:

y = 3\left(x^2 - 3x+\dfrac{9}{4}\right) + 17-\dfrac{27}{4}

Factor the parentheses and combine the constants:

y = 3\left(x-\dfrac{3}{2}\right)^2+\dfrac{41}{4}

8 0
1 year ago
Help me on how to do this
Mashcka [7]

Answer:

13 quarters ($3.25) and 12 dimes ($1.2) is 25 together ($4.45). Thats the closest I got without going over 25 dimes and quarters. Sorry

7 0
3 years ago
C
NNADVOKAT [17]

Answer:

Cells come from pre-existing cells.

5 0
2 years ago
Read 2 more answers
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