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blondinia [14]
3 years ago
7

Verify the identity: tan(x+pi/2)=-cotx please!!

Mathematics
1 answer:
olya-2409 [2.1K]3 years ago
5 0
First we use sin(a+b)= sinacosb+sinbcosa
and cos(a+b)=cosa cosb -sinasinb

tan(x+pi/2)= sin(x+pi/2) / cos(x+pi/2)
and sin(x+pi/2) = sinxcospi/2 + sinpi/2cosx =cosx, 
<span>cos(x+pi/2) = cosxcospi/2- sinxsinpi/2= - sinx, 
</span> because <span>cospi/2 =0,  </span>and <span>sinpi/2=1

</span><span>=tan(x+pi/2)= sin(x+pi/2) / cos(x+pi/2)= cosx / -sinx = -1/tanx = -cotx
</span>from where <span>tan(x+pi/2)=-cotx</span> 
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1. Let f(x) = x + 3 and g(x) = 3x + 5. Find f(g(4)) - g(f(4)) 2. Let f(x) = 5x^2 - 5. What is f(f(x))? 3. Let f(g(x)) = 3x + 3 a
ludmilkaskok [199]

Answer:

1. f(g(4)) - g(f(4)) = -6

2. f(f(x)) = 125x^4 - 250x^2 + 120

3. a + b = 0

4. g(4) = -9

Step-by-step explanation:

1.

First we need to find f(4) and g(4):

f(4) = 4 + 3 = 7

g(4) = 3*4 + 5 = 17

Then, we find g(f(4)) = g(7):

g(7) = 3*7 + 5 = 26

And we find f(g(4)) = f(17):

f(17) = 17 + 3 = 20

so f(g(4)) - g(f(4)) = 20 - 26 = -6

2.

To find f(f(x)), we use the value of f(x) for every x in f(x):

f(f(x)) = 5*(f(x))^2 - 5 = 5*(5x^2 - 5)^2 - 5 = 5*(25x^4 - 50x^2 + 25) - 5

f(f(x)) = 125x^4 - 250x^2 + 120

3.

To find f(g(x)), we use the value of g(x) for every x in f(x):

f(g(x)) = g(x) + 6 = ax + b + 6 = 3x + 3

ax + (b+6) = 3x + 3 -> a = 3 and b = -3

a + b = 3 - 3 = 0

4.

If we assume g(x) = ax + b, we have:

g(f(x)) = a*(2x - 3) + b = 2ax - 3a + b = 5 - 4x

2a = -4 -> a = -2

-3a + b = 5

6 + b = 5 -> b = -1

g(x) = -2x - 1

g(4) = -2*4 - 1 = -9

7 0
3 years ago
2x - 4y = 10
Bas_tet [7]

Answer:

Substitution, but this system of equations has no solution.

Step-by-step explanation:

2x-4y=10

2y+6=x

----------------

2(2y+6)-4y=10

4y+12-4y=10

12=10

no solution.

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