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3 years ago

Tangent (π- α) × cos (π/2 +α) > 0.

Mathematics

1 answer:

saveliy_v [14]3 years ago

3 0

Supplementary angles have opposite tangents

tan(π - a) = - tan a

Adding π/2 turns a cosine into a negated sine.

cos(π/2 + a) = - sin a

These are easily shown by the various sum angle formulas; I won't bother here.

$\tan(\pi - a) \cos(\pi/2 + a) = (- \tan a)(- \sin a) = \dfrac{ \sin a }{ \cos a } \ \sin a = \dfrac{ \sin^2 a}{\cos a}$

The numerator is never negative so if the whole thing is positive so is the denominator:

$\cos a > 0$

That's the opposite conclusion to the question but I think it's right. Let's check a=π/4.

tan(π - π/4) = tan(3π/4) = -1

cos(π/2 + π/4) = cos(3π/4) = -1/√2

The product is positive.

So is cos(π/4).

I'm right.

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Read 2 more answers

Write y = x2 + 24x + 144 as a square of binomial.

irina1246 [14]

Answer:

$y=(x+12)^2\\$

Step-by-step explanation:

To write a quadratic equation into binomial form we can compare the equation into the completing square form of a quadratic equation like this ,

$y=a(x-h)^2+k$

now since,

$y=x^2+24x+144$

we can equate both the equations from left hand side to right hand side like this,

$x^2+24x+144=a(x-h)^2+k\\$

now we solve,

$x^2+24x+144=a(x-h)^2+k\\x^2+24x+144=a((x)^2-2(x)(h)+(h)^2)+k\\x^2+24x+144=a(x^2-2hx+h^2)+k\\x^2+24x+144=ax^2-2ahx+ah^2+k\\$

now we compare the coefficients of x^2:

$1 = a\\$

now we compare the coefficients of x :

$24=-2ah\\24=-2(1)h\\24=-2h\\\frac{24}{-2}=h\\-12=h\\$

now we compare the constants , (constants are the letters which are not associated with any variable in this case the variable is x)

$144 = ah^2+k\\144=(1)(-12)^2+k\\144=(1)(144)+k\\144=144+k\\144-144=k\\0=k\\$

so now the value we got all the values for the completing square form we plug those in , a = 1 , h = -12 , k = 0 ,

$y=a(x-h)^2+k\\y=1(x-(-12))^2+0\\y=(x+12)^2\\$

this is the square of a binomial, if you want to verify if we expands this formula by the formula of (a + b)^2 we would get the same result. Thus this is the correct answer.

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