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2 years ago

Page:If A+B+C=180 , then prove that:cosA+cosB+cosC= 1+4SINA/2SINB/2SINC/2

Mathematics

1 answer:

lakkis [162]2 years ago

6 0

Answer:

A + B + C = π ...... (1)

...........................................................................................................

L.H.S.

= ( cos A + cos B ) + cos C

= { 2 · cos[ ( A+B) / 2 ] · cos [ ( A-B) / 2 ] } + cos C

= { 2 · cos [ (π/2) - (C/2) ] · cos [ (A-B) / 2 ] } + cos C

= { 2 · sin( C/2 ) · cos [ (A-B) / 2 ] } + { 1 - 2 · sin² ( C/2 ) }

= 1 + 2 sin ( C/2 )· { cos [ (A -B) / 2 ] - sin ( C/2 ) }

= 1 + 2 sin ( C/2 )· { cos [ (A-B) / 2 ] - sin [ (π/2) - ( (A+B)/2 ) ] }

= 1 + 2 sin ( C/2 )· { cos [ (A-B) / 2 ] - cos [ (A+B)/ 2 ] }

= 1 + 2 sin ( C/2 )· 2 sin ( A/2 )· sin( B/2 ) ... ... ... (2)

= 1 + 4 sin(A/2) sin(B/2) sin(C/2)

= R.H.S. ............................. Q.E.D.

...........................................................................................................

In step (2), we used the Factorization formula

cos x - cos y = 2 sin [ (x+y)/2 ] · sin [ (y-x)/2 ]

Step-by-step explanation:

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If DE = 4x + 10, EF = 2x-1, and DF = 9x-15, find DF

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Hello,
We have that:
(4x+10)+(2x-1)=9x-15

We solve the equation:
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Then:
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The cube root values and a graph of them are shown in the attachment.

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Answer:

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2 years ago

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Answer:

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$\bf \textit{hyperbolas, vertical traverse axis }\\\\
\cfrac{(y-{{ k}})^2}{{{ a}}^2}-\cfrac{(x-{{ h}})^2}{{{ b}}^2}=1
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\begin{cases}
center\ ({{ h}},{{ k}})\\
vertices\ ({{ h}}, {{ k}}\pm a)
\end{cases}\\\\
-------------------------------\\\\
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-------------------------------\\\\
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now, if you notice, the positive fraction, is the one with the "y" variable, and that simply means, the hyperbola traverse axis is over the y-axis, so it more or less looks like the picture below.

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Page:If A+B+C=180 , then prove that:cosA+cosB+cosC= 1+4SINA/2SINB/2SINC/2​

Page:If A+B+C=180 , then prove that:cosA+cosB+cosC= 1+4SINA/2SINB/2SINC/2