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Scilla [17]
3 years ago
14

find the exact value of sin a = 4/5, a lies in quadrant 2, and cos B = 2/5, B lies in quadrant 1 find cos(a-b)​

Mathematics
1 answer:
viktelen [127]3 years ago
4 0

Answer:

Cos(a - b) = 0.959

Step-by-step explanation:

We know that:

Sin(a) = 4/5

Such that a is on the first quadrant (between 0° and 90°)

then:

Sin(a) = 4/5

Asin(sin(a)) = Asin(4/5)

a = Asin(4/5) = 53.13°

And:

Cos(b) = 2/5

Then:

Acos(Cos(b)) = Acos(2/5)

b = Acos(2/5) = 36.67°

Then:

Cos(a - b)

where:

a = 53.13° and b = 36.67°

Then:

Cos(a - b) = Cos(53.13° - 36.67°) = 0.959

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A factorization of x^4+2x^3+7x^2-6x+44 is (x^2+4x+11)(x^2-2x+4).

<h3>What are the properties of roots of a polynomial?</h3>
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If the roots of the polynomial p(x)=ax^4+bx^3+cx^2+dx+e are r_1,r_2,r_3,r_4, then it can be factorized as p(x)=(x-r_1)(x-r_2)(x-r_3)(x-r_4).

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Since p(x)=x^4+2x^3+7x^2-6x+44 is a polynomial with real coefficients, so each complex root exists in a pair of conjugates.

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Thus, all the four roots of the polynomial p(x)=x^4+2x^3+7x^2-6x+44, are: r_1=-2+i\sqrt{7}, r_2=-2-i\sqrt{7}, r_3=1-i\sqrt{3}, r_4=1+i\sqrt{3}.

So, the polynomial p(x)=x^4+2x^3+7x^2-6x+44 can be factorized as follows:

\{x-(-2+i\sqrt{7})\}\{x-(-2-i\sqrt{7})\}\{x-(1-i\sqrt{3})\}\{x-(1+i\sqrt{3})\}\\=(x+2-i\sqrt{7})(x+2+i\sqrt{7})(x-1+i\sqrt{3})(x-1-i\sqrt{3})\\=\{(x+2)^2+7\}\{(x-1)^2+3\}\hspace{1cm} [\because (a+b)(a-b)=a^2-b^2]\\=(x^2+4x+4+7)(x^2-2x+1+3)\\=(x^2+4x+11)(x^2-2x+4)

Therefore, a factorization of x^4+2x^3+7x^2-6x+44 is (x^2+4x+11)(x^2-2x+4).

To know more about factorization, refer: brainly.com/question/25829061

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