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3 years ago

3 16. If 270º < AS 360° and cos(A)= 3/4 then determine the exact values of sin (A) and tan ( A)

Mathematics

1 answer:

suter [353]3 years ago

4 0

Answer:

${ \boxed{ \tt{trig \: identity : { \bf{ { \cos}^{2} A + { \sin }^{2} A = 1}}}}} \\ \therefore \: { \green{ \tt{ \sin A = \sqrt{1 - { \cos }^{2}A } }}} \\ \sin A = \sqrt{1 - {( \frac{3}{4}) }^{2} } \\ \sin A = \frac{ \sqrt{7} }{4} = 0.661 \\ \\ { \green{ \tt{ \tan A = \frac{ \sin A }{ \cos A} }}} \\ \tan(A ) = \frac{ \frac{ \sqrt{7} }{4} }{ \frac{3}{4} } = \frac{ \sqrt{7} }{3} = 0.882 \\ \\ { \underline{ \blue{ \tt{ becker \: jnr}}}}$

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Step-by-step explanation:

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Step-by-step explanation:

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3 years ago

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OverLord2011 [107]

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\
% function transformations for trigonometric functions
\begin{array}{rllll}
% left side templates
f(x)=&{{ A}}cos({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\

\end{array}\qquad$

$\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks}\\
\quad \textit{horizontally by amplitude } |{{ A}}|\\
\bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\
\bullet \textit{vertical shift by }{{ D}}\\
\qquad if\ {{ D}}\textit{ is negative, downwards}\\
\qquad if\ {{ D}}\textit{ is positive, upwards}\\


\end{array}$

$\bf \begin{array}{llll}
\bullet \textit{function period}\\
\qquad \frac{2\pi }{{{ B}}}\ for\ cos(\theta),\ sin(\theta),\ sec(\theta),\ csc(\theta)\\
\qquad \frac{\pi }{{{ B}}}\ for\ tan(\theta),\ cot(\theta)
\end{array}
$

so if you notice yours $\bf \begin{array}{llll}
3.2cos&\left( \frac{5}{3}\theta \right)+&6.1\\
&\ \uparrow&\uparrow \\
&B&D 
\end{array}$

now.. normally the function $\bf 3.2cos&\left( \frac{5}{3}\theta \right)$
has a D value of 0, or no vertical shift, and the amplitude and the period simply make the wave taller and thinner, but the midline is still the x-axis

now, with D = 6.1, that moves the midline up vertically that much

now.. the period, well, B = 5/3, normal period of cosine is $2\pi$
so, the new period will be $\bf \cfrac{2\pi }{B}\implies \cfrac{2\pi }{\frac{5}{3}}\implies \cfrac{6\pi }{5}$

notice the picture below
the vertical shift by the D component, or 6.1, moved the midline to y = 6.1 :)

6 0

3 years ago

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MissTica

Answer:

Umm….. I really don’t know……..

Step-by-step explanation:

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3 years ago

3 16. If 270º < AS 360° and cos(A)= 3/4 then determine the exact values of sin (A) and tan ( A)​

3 16. If 270º < AS 360° and cos(A)= 3/4 then determine the exact values of sin (A) and tan ( A)