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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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den301095 [7]

Answer:

15.9

Step-by-step explanation:

4 1/4 x 3 x 1 1/4 = 15.93

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5 0
3 years ago
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Bumek [7]

Answer:

58

Step-by-step explanation:

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6 0
2 years ago
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H(t) = −t2 + t + 1 h(x + 1)
Harman [31]

The value of the function h(x + 1) is -x^2 - x + 1

<h3>How to evaluate the function?</h3>

The equation of the function is given as:

h(t) =-t^2 + t + 1

The function is given as:

h(x + 1)

This means that t = x + 1

So, we substitute t = x + 1 in the equation h(t) =-t^2 + t + 1

h(x + 1) =-(x + 1)^2 + (x + 1) + 1

Evaluate the exponent

h(x + 1) =-(x^2 + 2x + 1) + x + 1 + 1

Expand the brackets

h(x + 1) = -x^2 - 2x - 1 + x + 1 + 1

Evaluate the like terms

h(x + 1) = -x^2 - x + 1

Hence, the value of the function h(x + 1) is -x^2 - x + 1

Read more about functions at:

brainly.com/question/1415456

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<u>Complete question</u>

Consider the following function definition, and calculate the value of the function

h(t) = −t2 + t + 1 h(x + 1)

3 0
1 year ago
Help me asap!!!!!!!????
Elina [12.6K]

Answer:

5*10^-4

Step-by-step explanation:

10^-1=1

10^-2=0.1

10^-3=0.01

10^-4=0.001

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3 years ago
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Ilya [14]

Answer:

#3

Step-by-step explanation:

8 0
2 years ago
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