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Pani-rosa [81]
3 years ago
9

Prove the identity (cosx+cosy)^2+(sinx-siny)^2= 2+2cos(x+y)

Mathematics
2 answers:
muminat3 years ago
8 0
TRIGONOMETRIC \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: RESOLUTIONS \\ \\ \\\\Given \: expression \: - \\ \\ \\ { (\cos(x) + \cos(y)} )^{2} + ( {( \sin(x ) - \sin(y) )}^{2} \\ \\ = { \cos(x) }^{2} + { \cos(y) }^{2} + 2 \cos(x) \cos(y) \\ \: \: \:+\: \: { \sin(x) }^{2} + { \sin(y) }^{2} - 2 \sin(x) \sin(y) \\ \\ = { \cos(x) }^{2} \: + { \sin(x) }^{2} + \\ \: \: \: \: \: { \cos(y) }^{2} + { \sin(x) }^{2} \: + \\ \: \: \: \: \: 2( \cos(x) \cos(y) - \sin(x) \sin(y) ) \\ \\ = \: 1 + 1 + 2( \cos(x + y) ) \\ \\ Using \:Trigonometric \: Identities \: - \\ \\ { \sin(q) }^{2} + { \cos(q) }^{2} = \: 1 \: \\ \cos(x) \cos(y) - \sin(x) \sin(y) = \cos(x + y) \\ \\ \\ \\ { (\cos(x) + \cos(y)} )^{2} + ( {( \sin(x ) - \sin(y) )}^{2} \\ = 2 + 2( \cos(x + y) ) \: \: \: \: \: \: \: \: \: \: \: \: \: \: Ans.\\ \\
astra-53 [7]3 years ago
4 0
(cos(x) + cos(y))^2 + (sin(x) - sin(y))^2 Remove the brackets

cos^2(x)
+ cos^2(y) + 2cos(x)*cos(y) + sin^2(x) - 2(sin(x)*sin(y) + sin^2(y) Combine these two in bold to make 1 because sin^2(x) + cos^2(x) = 1

1 + cos^2(y) + 2cos(x)*cos(y) - 2*sin(x)*cos(y) + sin^2(y) 
These two in bold also make 1

2 + 2cos(x)*cos(y) - 2*sin(X)*sin(y) Bring out a common factor of 2
2 +2(cos(x)*cos(y) -  sin(x)*sin(y) )

but cos(x+y ) = cos(x)*cos(y) - sin(x)*sin(y)

2 + 2* cos(x + y) is your final answer. 

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How do you solve z= - 2/3a, for a
gogolik [260]
Z = -2/3a
divide by -2/3 on both sides
a = z divided by -2/3
a = -3/2z
6 0
4 years ago
Read 2 more answers
Simplify -6i(8-6i)(-8-8i)
padilas [110]

Answer:

-96 + 672\,i

Step-by-step explanation:

This is a product of complex numbers, so we have in mind not only the general rules for multiplying binomials, but also the properties associated with the powers of the imaginary unit "i", in particular i^2=-1

We start by making the first product indicated which is that of a pure imaginary number (-6i) times the complex number (8-6i). We use distributive property and obtain the new complex number that results from this product:

-6\,i\,(8-6\,i)= (-6\,i)\,* 8 \, -\,6\,i\,(-6\,i)=-48\,i+36\,i^2=-48\,i+36\,(-1)=-36-48\,i

Now we make the second multiplication indicated (using distributive property as one does with the product of binomials), and combine like terms at the end:

(-36-48\,i)\,(-8-8\.i)=(-36)\.(-8)+(-36)(-8\,i)+(-48\,i)\,(-8)+(-48\,i)(-8\,i)=\\=288+288\,i+384\,i+384\,i^2=288+288\,i+384\,i+384\,(-1)=\\=288-384+288\,i+384\,i=-96+672\,i

5 0
3 years ago
A driver noted that he has traveled 8 miles in the last 10 minutes. What is his average speed in miles per hour?
Charra [1.4K]

Answer:

48 miles per hour.

Step-by-step explanation:

There are 60 minutes in an hour

Divide 60 by 10= 6

Multiply 8 by 6=48

7 0
4 years ago
Where does the graph of y= sin x from x = 0 to x = 2n1 start?
stepladder [879]

Answer:

Step-by-step explanation:

when x=0,sin x=0

when x=30° or π/6

sinx=1/2=0.5

when x=45° or π/4

sin x=√2/2≈1.414/2≈0.707

when x=60° or π/3

sinx=√3/2≈1.732/2≈0.866

when x=90° or π/2

sin x=1

when x=120° 2π/3

sin 120=sin (180-60)=sin 60 ≈0.866

when x=135° or 3π/4

sin 135=sin (180-45)=sin 45≈0.707

when x=150° or 5π/6

sin 150=sin (180-30)=sin 30=0.5

when x=180 or π

sin 180=sin(180-0)=sin 0=0

when x=210° or 7π/6

sin 210=sin (180+30)=-sin 30=-0.5

when x=225° or 5π/4

sin x=sin (180+45)=-sin 45≈-0.707

when x=240° or 4π/3

sin 240=sin (180+60)=-sin 60≈ -0.866

when x=270° or 3π/2

sin 270=sin (180+90)=-sin 90=-1

when x=300° or  5π/3

sin 300=sin(360-60)=-sin 60≈ -0.866

when x=315°or 7π/4

sin 315=sin (360-45)= -sin 45≈ -0.707

when x=330° or 13π/6

sin 330=sin (360-30)= -sin 30 =-0.5

when x=360 or 2π

sin (360)=sin (360+0)=sin 0=0

plot all these points on the graph,you get the reqd. graph.

you see it starts on (0,0)

6 0
4 years ago
Optimization problem:
GuDViN [60]
Cal problem!

given production 
P(x)=75x^2-0.2x^4

To find relative extrema, we need to find P'(x) and solve for P'(x)=0.

P'(x)=150x-0.8x^3    [by the power rule]

Setting P'(x)=0 and solve for extrema.
150x-0.8x^3=0  =>
x(150-0.8x^2)=0 =>
0.8x(187.5-x^2)=0
0.8x(5sqrt(15/2)-x)(5sqrt(15/2)+x)=0
=>
x={0,+5sqrt(15/2), -5sqrt(15/2)}   by the zero product rule.
[note: eqation P'(x)=0 can also be solved by the quadratic formula]

Reject negative root because we cannot hire negative persons.

So possible extrema are x={0,5sqrt(15/2)}

To find out which are relative maxima, we use the second derivative test.  Calculate P"(x), again by the power rule, 
P"(x)=-1.6x
For a relative maximum, P"(x)<0, so
P"(0)=0  which is not <0  [in fact, it is an inflection point]
P"(5sqrt(15/2))=-8sqrt(15/2) < 0, therefore x=5sqrt(15/2) is a relative maximum.

However, 5sqrt(15/2)=13.693 persons, which is impossible, so we hire either 13 or 14, but which one?

Let's go back to P(x) and find that
P(13)=6962.8
P(14)=7016.8

So we say that assigning 14 employees will give a maximum output.
8 0
3 years ago
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