If 7x+3=24, find the value of -5 - 6x. a)-23 b)-7 c)1 d)17

Prove the following identity : Sin α . Cos α . Tan α = (1 – Cos α) (1 + Cos α)

Umnica [9.8K]

Let's work on the left side first. And remember that
the tangent is the same as sin/cos.

sin(a) cos(a) tan(a)

Substitute for the tangent:

[ sin(a) cos(a) ] [ sin(a)/cos(a) ]

Cancel the cos(a) from the top and bottom, and you're left with

[ sin(a) ] . . . . . [ sin(a) ] which is [ sin²(a) ] That's the left side.

Now, work on the right side:

[ 1 - cos(a) ] [ 1 + cos(a) ]

Multiply that all out, using FOIL:

[ 1 + cos(a) - cos(a) - cos²(a) ]

= [ 1 - cos²(a) ] That's the right side.

Do you remember that for any angle, sin²(b) + cos²(b) = 1 ?
Subtract cos²(b) from each side, and you have sin²(b) = 1 - cos²(b) for any angle.

So, on the right side, you could write [ sin²(a) ] .

Now look back about 9 lines, and compare that to the result we got for the left side .

They look quite similar. In fact, they're identical. And so the identity is proven.

Whew !

4 0

3 years ago

Read 2 more answers

10) (k - 9) (5k - 8)

zalisa [80]

5k²-8k-45k+72 = 5k²-53k+72

Answer: 5k² - 53k + 72

7 0

3 years ago

Solve this Inequality 1 > -1 - w

Rom4ik [11]

Answer:

$\displaystyle -2 < w$

Step-by-step explanation:

1 > −1 - w

+ 1 + 1

________

$\displaystyle \frac{2}{-1} > \frac{-w}{-1} \\ \\ -2 < w$

Whenever you divide\multiply by a negative, you reverse the inequality symbol.

I am joyous to assist you anytime.

8 0

2 years ago

If the surface of the water is at an elevation 0 meters, which person is located at an elevation

Eva8 [605]

The correct answer is that noa is located at an elevation

5 0

3 years ago

Find the integral using substitution or a formula.

Nadusha1986 [10]

$\rm \int \dfrac{x^2+7}{x^2+2x+5}~dx$

Derivative of the denominator:
$\rm (x^2+2x+5)'=2x+2$

Hmm our numerator is 2x+7. Ok this let's us know that a simple u-substitution is NOT going to work. But let's apply some clever Algebra to the numerator splitting it up into two separate fractions. Split the +7 into +2 and +5.

$\rm \int \dfrac{x^2+2+5}{x^2+2x+5}~dx$

and then split the fraction,

$\rm \int \dfrac{x^2+2}{x^2+2x+5}~dx+\int\dfrac{5}{x^2+2x+5}~dx$

Based on our previous test, we know that a simple substitution will work for the first integral: $\rm \quad u=x^2+2x+5\qquad\to\qquad du=2x+2~dx$

So the first integral changes,

$\rm \int \dfrac{1}{u}~du+\int\dfrac{5}{x^2+2x+5}~dx$

integrating to a log,

$\rm ln|x^2+2x+5|+\int\dfrac{5}{x^2+2x+5}~dx$

Other one is a little tricky. We'll need to complete the square on the denominator. After that it will look very similar to our arctangent integral so perhaps we can just match it up to the identity.

$\rm x^2+2x+5=(x^2+2x+1)+4=(x+1)^2+2^2$

So we have this going on,

$\rm ln|x^2+2x+5|+\int\dfrac{5}{(x+1)^2+2^2}~dx$

Let's factor the 5 out of the intergral,
and the 4 from the denominator,

$\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\frac{(x+1)^2}{2^2}+1}~dx$

Bringing all that stuff together as a single square,

$\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\left(\dfrac{x+1}{2}\right)^2+1}~dx$

Making the substitution: $\rm \quad u=\dfrac{x+1}{2}\qquad\to\qquad 2du=dx$

giving us,

$\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\left(u\right)^2+1}~2du$

simplying a lil bit,

$\rm ln|x^2+2x+5|+\frac52\int\dfrac{1}{u^2+1}~du$

and hopefully from this point you recognize your arctangent integral,

$\rm ln|x^2+2x+5|+\frac52arctan(u)$

undo your substitution as a final step,
and include a constant of integration,

$\rm ln|x^2+2x+5|+\frac52arctan\left(\frac{x+1}{2}\right)+c$

Hope that helps!
Lemme know if any steps were too confusing.

8 0

3 years ago