I really don’t understand and I need the answers to pass

Mathematics

1 answer:

cricket20 [7]2 years ago

7 0

Answer:

H(10) Would be the time, in minutes the balloon spent in air . H(10) Means the balloon spent 10 minutes in air

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How do you prove cosx = 1-tan^2(x/2)/1+tan^2(x/2)?

nikdorinn [45]

$\bf tan\left(\cfrac{{{ \theta}}}{2}\right)=
\begin{cases}
\pm \sqrt{\cfrac{1-cos({{ \theta}})}{1+cos({{ \theta}})}}
\\ \quad \\
\boxed{\cfrac{sin({{ \theta}})}{1+cos({{ \theta}})}}
\\ \quad \\
\cfrac{1-cos({{ \theta}})}{sin({{ \theta}})}
\end{cases}\\\\
-------------------------------\\\\
tan^2\left( \frac{x}{2} \right)\implies \left[ \cfrac{sin(x)}{1+cos(x)} \right]^2\implies \cfrac{sin^2(x)}{[1+cos(x)]^2}
\\\\\\
\boxed{\cfrac{sin^2(x)}{1+2cos(x)+cos^2(x)}}$

now, let's plug that in the right-hand-side expression,

$\bf cos(x)=\cfrac{1-tan^2\left( \frac{x}{2} \right)}{1+tan^2\left( \frac{x}{2} \right)}\\\\
-------------------------------\\\\
\cfrac{1-tan^2\left( \frac{x}{2} \right)}{1+tan^2\left( \frac{x}{2} \right)}\implies \cfrac{1-\frac{sin^2(x)}{1+2cos(x)+cos^2(x)}}{1+\frac{sin^2(x)}{1+2cos(x)+cos^2(x)}}
\\\\\\
\cfrac{\frac{1+2cos(x)+cos^2(x)~-~sin^2(x)}{1+2cos(x)+cos^2(x)}}{\frac{1+2cos(x)+cos^2(x)~+~sin^2(x)}{1+2cos(x)+cos^2(x)}}$

$\bf \cfrac{1+2cos(x)+cos^2(x)~-~sin^2(x)}{\underline{1+2cos(x)+cos^2(x)}}\cdot \cfrac{\underline{1+2cos(x)+cos^2(x)}}{1+2cos(x)+cos^2(x)~+~sin^2(x)}
\\\\\\
\cfrac{1+2cos(x)+cos^2(x)~-~sin^2(x)}{1+2cos(x)+cos^2(x)~+~sin^2(x)}$

$\bf -------------------------------\\\\
recall\qquad sin^2(\theta)+cos^2(\theta)=1\\\\
-------------------------------\\\\
\cfrac{\boxed{sin^2(x)+cos^2(x)}+2cos(x)+cos^2(x)~-~sin^2(x)}{1+2cos(x)+\boxed{1}}
\\\\\\
\cfrac{cos^2(x)+2cos(x)+cos^2(x)}{2+2cos(x)}\implies \cfrac{2cos(x)+2cos^2(x)}{2+2cos(x)}
\\\\\\
\cfrac{\underline{2} cos(x)~\underline{[1+cos(x)]}}{\underline{2}~\underline{[1+cos(x)]}}\implies cos(x)$

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Answer:

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Step-by-step explanation:

pemdas

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