Prove the identity cos(3x)=cos^3(x)-3sin^2(x)cos(x), step by step.

1 answer:

8 0

Cos(A+B) = cosAcosB - sinAsinB
cos(3x) = cos(x+2x)
cos(x+2x)= cosxcos2x-sinxsin2x
cos2x = 2cos^2 (x) - 1
sin(2x) = 2sin(x)cos(x)
by putting the values we get
cos(x+2x)= cosx*cos2x-sinx*sin2x=cosx*(2cos^2 (x) - 1) -(sinx*2sin(x)cos(x))
<span>=cos^3(x)-3sin^2(x)cos(x)</span>
hope it helps

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An object launched vertically from a point that is 37.5 feet above ground level with an initial velocity of 48.6 feet per second

Elis [28]

Answer: 3.675 seconds

Step-by-step explanation:

Hi, when the object hits the ground, h=0:

h=−16t^2+48.6t+37.5

0=−16t^2+48.6t+37.5

We have to apply the quadratic formula:

For: ax2+ bx + c

x =[ -b ± √b²-4ac] /2a

Replacing with the values given:

a=-16 ; b=48.6; c=37.5

x =[ -(48.6) ± √(-48.6)²-4(-16)37.5] /2(-16)

x = [ -48.6 ± √ 4,761.96] /-32

x = [ -48.6 ± 69] /-32

Positive:

x = [ -48.6 + 69] /-32 = -0.6375

Negative:

x = [ -48.6 - 69] /-32 = 3.675 seconds (seconds can't be negative)

Feel free to ask for more if needed or if you did not understand something.

8 0

3 years ago

Can someone please help me with this problem??

Ostrovityanka [42]

Answer:

The slope is - $\frac{1}{3}$ and the y intercept is 2.

Step-by-step explanation:

So, we can see that the line moves down one unit every time it progresses to the right three times. Slope is defined by change in y/change in x. For this problem, the change in y is -1 and the change in x is +3, so the slope is $-\frac{1}{3}$ .

Now for the y intercept. The y intercept is the point where the line first passes through the y axis. In this case, it passes through 2. Your y intercept can be written in two ways; as a number (2) or as an ordered pair $(0,2)$ , although people usually write the number.