\(\displaystyle{\cos{{60}}}°={\frac{{{1}}}{{{2}}}}\)

As \(\displaystyle{\cos{{240}}}°=-{\cos{{60}}}°\), we have

\(\displaystyle{\cos{{240}}}°=-{\frac{{{1}}}{{{2}}}}\)

Rita Miller

Answered 2021-12-28
Author has **257** answers

Write \(\displaystyle{\cos{{240}}}°\) as \(\displaystyle{\cos{{\left({180}°+{60}°\right)}}}\)

Using the summation identity:

\(\displaystyle{\cos{{\left({180}°\right)}}}{\cos{{\left({60}°\right)}}}-{\sin{{\left({180}°\right)}}}{\sin{{\left({60}°\right)}}}\)

Using trivial identity: \(\displaystyle{\cos{{180}}}°={\left(-{1}\right)},{\cos{{60}}}°={\frac{{{1}}}{{{2}}}},{\sin{{180}}}°={0},{\sin{{60}}}°={\frac{{\sqrt{{{3}}}}}{{{2}}}}\)

\(\displaystyle={\left(-{1}\right)}\times{\frac{{{1}}}{{{2}}}}-{0}\times{\frac{{\sqrt{{{3}}}}}{{{2}}}}\)

\(\displaystyle=-{\frac{{{1}}}{{{2}}}}\)

Using the summation identity:

\(\displaystyle{\cos{{\left({180}°\right)}}}{\cos{{\left({60}°\right)}}}-{\sin{{\left({180}°\right)}}}{\sin{{\left({60}°\right)}}}\)

Using trivial identity: \(\displaystyle{\cos{{180}}}°={\left(-{1}\right)},{\cos{{60}}}°={\frac{{{1}}}{{{2}}}},{\sin{{180}}}°={0},{\sin{{60}}}°={\frac{{\sqrt{{{3}}}}}{{{2}}}}\)

\(\displaystyle={\left(-{1}\right)}\times{\frac{{{1}}}{{{2}}}}-{0}\times{\frac{{\sqrt{{{3}}}}}{{{2}}}}\)

\(\displaystyle=-{\frac{{{1}}}{{{2}}}}\)

asked 2021-06-07

[Triangle]
Find the value of x and y.

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\(|x|=\begin{cases}x & \text{if}\ x\geq0\\-x & \text{if}\ x<0\end{cases}\)

a) In the space provided, draw a graph of the function \(\displaystyle{f{{\left({x}\right)}}}={\left|{x}\right|}\)

b) Using your draph from part (a), and your understanding of the derivative as the rate of change/slope of the tangent line, find the derivative function \(\displaystyle{f}'{\left({x}\right)}\) of the above function \(\displaystyle{f{{\left({x}\right)}}}={\left|{x}\right|}\), for x not equal to 0 (fill in the bla

):

\(f'(x)=\begin{cases}& \text{if}\ x>0\\ & \text{if}\ x<0\end{cases}\)

c) But what about \(\displaystyle{f}'{\left({0}\right)}\)? It is not so clear from the picture even how to draw a tangent line to the function at the origin. So let's try to compute \(\displaystyle{f}'{\left({0}\right)}\) by first looking at the corresponding lefthand and righthand limits of the difference quotient.

i. Compute \(\displaystyle\lim_{{{h}\rightarrow{0}^{{+}}}}{\frac{{{f{{\left({0}+{h}\right)}}}-{f{{\left({0}\right)}}}}}{{{h}}}}\). Hint: use the piecewise definition of \(\displaystyle{f{{\left({x}\right)}}}\) given above.

ii. Compute \(\displaystyle\lim_{{{h}\rightarrow{0}^{{-}}}}{\frac{{{f{{\left({0}+{h}\right)}}}-{f{{\left({0}\right)}}}}}{{{h}}}}\)

iii. What do your answers to parts (i) and (ii) tell you about \(\displaystyle{f}'{\left({0}\right)}?\) Please explain.

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Find the values of x such that the angle between the vectors (2, 1, -1), and (1, x, 0) is \(\displaystyle{45}^{{\circ}}\).

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Find the values of the variables

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Find the area of the largest rectangle that can be inscribed in a right triangle with legs of lengths 3 cm and 4 cm if two sides of the rectangle lie along the legs.

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Find, correct to the nearest degree, the three angles of the triangle with the given vertices.

A(1,0,-1), B(5,-3,0), C(1,5,2)

Find \(\displaystyle\angle{C}{A}{B},\angle{A}{B}{C},\angle{B}{C}{A}\)

A(1,0,-1), B(5,-3,0), C(1,5,2)

Find \(\displaystyle\angle{C}{A}{B},\angle{A}{B}{C},\angle{B}{C}{A}\)