Mathematics, 20.09.2020 09:01 ayoismeisalex

Please help me to prove this!

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Answer from: natalie9316

see proof below

Step-by-step explanation:

Given: A + B + C = π → A = π - (B + C)

→ B + C = π - A

Use the Pythagorean Identity: cos² A + sin² A = 1 → sin² A = 1 - cos² A

Use Double Angle Identities: cos 2A = 2 cos² A - 1 → cos² A = (cos 2A + 1)/2

→ cos A = 1 - 2 sin² (A/2)

Use Sum to Product Identity: cos A + cos B = 2 cos [(A + B)/2] · cos [(A - B)/2]

Use Cofunction Identities: cos (π/2 - A) = sin (A)

sin (π/2 - A) = cos A

cos (-A) = cos (A)

Proof LHS → RHS:

$\text{LHS:}\qquad \qquad \sin^2\bigg(\dfrac{B}{2}\bigg)+\sin^2 \bigg(\dfrac{C}{2}\bigg)-\sin^2\bigg(\dfrac{A}{2}\bigg)$

$\text{Pythagorean:}\qquad 1-\cos^2 \bigg(\dfrac{B}{2}\bigg)+1-\cos^2 \bigg(\dfrac{C}{2}\bigg)-\bigg[1-\cos^2 \bigg(\dfrac{A}{2}\bigg)\bigg]\\\\\\.\qquad \qquad \qquad =1-\cos^2 \bigg(\dfrac{B}{2}\bigg)-\cos^2 \bigg(\dfrac{C}{2}\bigg)+\cos^2 \bigg(\dfrac{A}{2}\bigg)$

$\text{Double Angle:}\quad 1-\bigg(\dfrac{\cos(2\cdot \frac{B}{2})+1}{2}\bigg)-\bigg(\dfrac{\cos (2\cdot \frac{C}{2})+1}{2}\bigg)+\bigg(\dfrac{\cos (2\cdot \frac{A}{2})+1}{2}\bigg)\\\\\\.\qquad \qquad \qquad =1-\dfrac{\cos B}{2}-\dfrac{1}{2}-\dfrac{\cos C}{2}-\dfrac{1}{2}+\dfrac{\cos A}{2}+\dfrac{1}{2}\\\\\\.\qquad \qquad \qquad =\dfrac{1}{2}[1-(\cos B+\cos C)+\cos A]$

$\text{Sum to Product:}\qquad \dfrac{1}{2}\bigg(1-\bigg[2\cos \bigg(\dfrac{B+C}{2}\bigg)\cdot \cos \bigg(\dfrac{B-C}{2}\bigg)\bigg]+\cos A\bigg)$

$\text{Given:}\qquad \dfrac{1}{2}\bigg(1-\bigg[2\cos \bigg(\dfrac{\pi -A}{2}\bigg)\cdot \cos \bigg(\dfrac{B-C}{2}\bigg)\bigg]+\cos A\bigg)$

$\text{Cofunction:}\qquad \dfrac{1}{2}\bigg(1-\bigg[2\sin \bigg(\dfrac{A}{2}\bigg)\cdot \cos \bigg(\dfrac{B-C}{2}\bigg)\bigg]+\cos A\bigg)$

$\text{Double Angle:}\qquad \dfrac{1}{2}\bigg[1-2\sin \bigg(\dfrac{A}{2}\bigg)\cdot \cos \bigg(\dfrac{B-C}{2}\bigg)+1-2\sin^2 \bigg(\dfrac{A}{2}\bigg)\bigg]\\\\\\.\qquad \qquad \qquad =\dfrac{1}{2}\bigg[2-2\sin \bigg(\dfrac{A}{2}\bigg)\cdot \cos \bigg(\dfrac{B-C}{2}\bigg)-2\sin^2 \bigg(\dfrac{A}{2}\bigg)\bigg]\\\\\\.\qquad \qquad \qquad =1-\sin \bigg(\dfrac{A}{2}\bigg)\cdot \cos \bigg(\dfrac{B-C}{2}\bigg)-\sin^2 \bigg(\dfrac{A}{2}\bigg)$

$\text{Factor:}\qquad \qquad 1-\sin \bigg(\dfrac{A}{2}\bigg)\bigg[ \cos \bigg(\dfrac{B-C}{2}\bigg)-\sin \bigg(\dfrac{A}{2}\bigg)\bigg]$

$\text{Given:}\qquad \qquad 1-\sin \bigg(\dfrac{A}{2}\bigg)\bigg[ \cos \bigg(\dfrac{B-C}{2}\bigg)-\sin \bigg(\dfrac{\pi -(B+C)}{2}\bigg)\bigg]$

$\text{Cofunction:}\qquad 1-\sin \bigg(\dfrac{A}{2}\bigg)\bigg[ \cos \bigg(\dfrac{B-C}{2}\bigg)+\cos \bigg(\dfrac{B+C}{2}\bigg)\bigg]$

$\text{Sum to Product:}\ 1-\sin \bigg(\dfrac{A}{2}\bigg)\cdot 2 \cos \bigg(\dfrac{(B-C)+(B-C)}{2\cdot 2}\bigg)\cdot \cos \bigg(\dfrac{(B-C)-(B+C)}{2\cdot 2}\bigg)\\\\\\.\qquad \qquad \qquad =1-2\sin \bigg(\dfrac{A}{2}\bigg)\cdot \cos \bigg(\dfrac{B}{2}\bigg)\cdot \cos \bigg(-\dfrac{C}{2}\bigg)$ $\text{Cofunction:}\qquad =1-2\sin \bigg(\dfrac{A}{2}\bigg)\cdot \cos \bigg(\dfrac{B}{2}\bigg)\cdot \cos \bigg(\dfrac{C}{2}\bigg)$

$\text{LHS = RHS:}\quad \checkmark\\\\\quad 1-2\sin \bigg(\dfrac{A}{2}\bigg)\cdot \cos \bigg(\dfrac{B}{2}\bigg)\cdot \cos \bigg(\dfrac{C}{2}\bigg)=1-2\sin \bigg(\dfrac{A}{2}\bigg)\cdot \cos \bigg(\dfrac{B}{2}\bigg)\cdot \cos \bigg(\dfrac{C}{2}\bigg)\quad$

Answer from: Quest

step-by-step explanation:

i think something is missing in your question because y-1=(x-12) would not pass through the point (-4,-3). because if you replace x by -4 your are supposed to find y=-3 which is not the case her. if you replace x by -4 you find =-15.

i assume the right equation is y-1=1/4*(x-12) because this one is passing through both your points (if you are interested to know how i found that out i can tell you).

in this equation if i replace x by -4 i find y=-3 and if i replace x by 12, i find y=1 so the line representing this equation is passing through those 2 points.

then the standard form of the equation is finding an expression of y in function of x

y-1=1/4*(x-12)

y=x/4-3+1

y=x/4-2

hope that .

do you confirm there was a mistake in the question?

Answer from: Quest

step-by-step explanation:

no, her answer is not correct. we square the lengths of all sides and then determine whether the pythagorean theorem holds here:

9² + 15² = 306. this is not equal to the square of the longest side ( that square being 395. answer c is correct.

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