Topic
Q:
\( \begin{array}{rl}4.4 & 13 \cos \theta-5-0 \quad \theta \epsilon\left[180^{\circ} ; 360^{\circ}\right] \\ & \text { Determine: } \\ & 4.4 .1 \quad \sin \theta\end{array} \)
Q:
\( \begin{array}{rl}4.4 & 13 \cos \theta-5-0 \quad \theta \epsilon\left[180^{\circ} ; 360^{\circ}\right] \\ & \text { Determine: } \\ & 4.4 .1 \quad \sin \theta\end{array} \)
Q:
Modelación
\( \begin{array}{l}5 \text { Halla las razones trigonométricas de un á } \\ 30^{\circ} \text { y de otro de } 60^{\circ} \text {. Para ello, toma un } \\ \text { equilátero de lado a y divídelo en dos po } \\ \text { sus alturas. }\end{array} \)
Q:
Modelación
(5) Halla las razones trigonométricas de un á
\( 30^{\circ} \) y de otro de \( 60^{\circ} \). Para ello, toma un
equilátero de lado a y divídelo en dos po
sus alturas.
Q:
Use the given information to find the exact value of each of the following.
\( \begin{array}{lll}\text { a. } \boldsymbol{\operatorname { s i n }} 2 \theta & \text { b. } \boldsymbol{\operatorname { c o s } 2 \theta} & \text { c. } \boldsymbol{\operatorname { t a n } 2 \theta} 2 \theta \\ \sin \theta=\frac{3}{13}, \theta \text { lies in quadrant II }\end{array} \)
Q:
\( y = \frac { \operatorname { Tan } 2 x } { 1 - \operatorname { Cot } 2 x } \)
Q:
(1. Calcula las razones trigonométricas de los ángulos
agudos de los triángulos rectángulos \( A B C \) tales que:
a. \( m \propto A=90^{\circ}, b=10 \mathrm{~cm} \) y \( c=12 \mathrm{~cm} \)
Q:
Angle \( \theta \) is in standard position. If \( (8,-15) \) is on the terminal ray of angle \( \theta \), find the values of the
trigonometric functions.
\( \sin (\theta)= \)
\( \cos (\theta)= \)
\( \tan (\theta)= \)
\( \csc (\theta)= \)
Q:
Is the work shown in the simplification below
correct? Explain.
\[ \begin{aligned} \frac{\csc (t)}{\sec (t)} & =\frac{1}{\cos (t)} \div \frac{1}{\sin (t)} \\ & =\frac{1}{\cos (t)} \cdot \frac{\sin (t)}{1} \\ & =\frac{1}{\sin (t)} \\ & =\tan (t)\end{aligned} \]
Q:
Solving a Real-World Problem
A 26 -foot long ladder is leaning against a building
at a \( 60^{\circ} \) angle with the ground.
Which of the following equations can you use to
found to the ne approximate height of the building?
\( \csc \left(60^{\circ}\right)=\frac{26}{h} \)
\( \csc \left(60^{\circ}\right)=\frac{h}{26} \)
\( \sec \left(60^{\circ}\right)=\frac{26}{h} \)
\( \sec \left(60^{\circ}\right)=\frac{h}{26} \)
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