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  1. Since cos theta is 15/17, let a right triangle have a hypotenuse of 17 and a leg of 15. This means that the other is 8.

    Sin 2 theta is a double angle identity.

    Sin 2 theta  = 2 sin theta cos theta = 2(8/17)(15/17) = 240/289

    Cos 2theta = cos^2 – sin^2 = (15/17)^2 – (8/17)^2 = 161/289

  2. Hello there!

    The only thing we have to do here is remember our trigonometric functions!

    sin=y/r
    cos=x/r
    tan=y/x
    sec=r/x
    csc=r/y
    cot=x/y

    Knowing that tanᎾ=15/8, we know that y=15 and x=8

    Now we need to look at the answer choices and see if they are correct.

    A. secᎾ=17/8
    Since secᎾ=r/x and r=17 and x=8, this is correct.

    B. cosᎾ=15/17
    Since cosᎾ=x/r and x=8 and r=17, this is wrong.

    C. cotᎾ=8/15
    Since cotᎾ=x/y and x=8 and y=15, this is also correct!

    D. cscᎾ=17/15
    Since cscᎾ=r/y and r=17 and y=15, this is right too.

    I really hope this helps!
    Best wishes 🙂

  3. ANSWER

    [tex]\sin( \theta) = - \frac{8}{17}[/tex]

    EXPLANATION

    We draw the diagram as shown in the attachment.

    We can see that the angle is in the 4th quadrant.

    We use the Pythagoras Theorem to determine the hypotenuse of the right angle triangle as follows:

    [tex]{h}^{2} = {8}^{2} + {15}^{2}[/tex]

    [tex]{h}^{2} = 64 + 225[/tex]

    [tex]{h}^{2} = 289[/tex]

    [tex]h = \sqrt{289}[/tex]

    [tex]= 17units[/tex]

    We know that the sine ratio is the ratio of the length of the opposite side to the length of the hypotenuse. Since the angle is in the 4th quadrant the sine ratio is negative.

    [tex]\sin( \theta) = - \frac{8}{17}[/tex]

    Note that the negative sign of the 8 and the positive sign of the 15 helped us to determine the quadrant in which the angle is, and that the negative sign has nothing to do with the length. I know this confuses most students.
    [tex]The terminal side of an angle in standard position passes through p(15,-8) . what is the value of si[/tex]

  4. The correct option is [tex]\boxed{{\mathbf{Option B}}}[/tex] .

    Further explanation:  

    The sine ratio is the ratio of the length of the side that is opposite to the given angle to the length of the hypotenuse.

    It can be expressed as,

      [tex]\sin \theta =\frac{p}{h}[/tex]

    Pythagoras theorem is always used in a right angle tri

    angle to obtain the one of the side of the triangle.

    Hypotenuse is the longest side in the right angle triangle.

    Pythagoras theorem can be expressed as,

      [tex]{H^2}={P^2}+{B^2}[/tex]

    Here, [tex]H[/tex]  is the hypotenuse, [tex]B[/tex]  is the base and [tex]P[/tex]  is the perpendicular.

    Step by step explanation:

    Step 1:

    The terminal side of an angle passes through the point [tex]p\left({15,-8}\right)[/tex]  as shown in the attached figure.

    It can be observed that the given point lies in the fourth quadrant as [tex]x[/tex]  coordinate is positive and [tex]y[/tex]  coordinate is negative.

    As we can see in the attached figure the perpendicular is the side of the triangle that is opposite the angle [tex]\theta[/tex] .

    Therefore, the perpendicular is 8 and the base is [tex]15[/tex] .

    Now find the hypotenuse by the use of Pythagoras theorem.

    [tex]\begin{gathered}{H^2}={8^2}+{15^2}\hfill\\{H^2}=64+225\hfill\\H=\sqrt{289}\hfill\\H=17\hfill\\\end{gathered}[/tex]

    Therefore, the length of the hypotenuse is [tex]{\text{17 units}}[/tex] .

    We know that the sine function is negative in fourth quadrant.

    Therefore, [tex]\sin \theta[/tex]  cab be expressed as,

    [tex]\sin \theta=-\frac{8}{{17}}[/tex]

    Thus, option B [tex]\sin \theta=-\frac{8}{{17}}[/tex]  is correct.

    Learn more:  

    Learn more about the all right triangles are isosceles Learn more about in a right triangle, angle c measures 40°. the hypotenuse of the triangle is 10 inches long. what is the approximate length of the side adjacent to angle c? Learn more about the equivalent fraction

    Answer details:

    Grade: High school

    Subject: Mathematics

    Chapter: Triangles

    Keywords: side, lengths, distance, Pythagoras theorem, triangle, hypotenuse, base, perpendicular, right angle triangle, longest side, sine function, fourth quadrant, positive sign, negative sign, ratio.

    [tex]The terminal side of an angle in standard position passes through p(15,-8) . what is the value of si[/tex]

  5. [tex]a.tan(\theta)=\frac{15}{8}[/tex]

    [tex]b.csc(\theta)=\frac{17}{15}[/tex]

    [tex]d.sec(\theta)=\frac{17}{8}[/tex]

    Step-by-step explanation:

    The trigonometry functions on a right triangle are given by:

    [tex]sin(\theta)=\frac{opposite}{hypotenuse}[/tex]

    [tex]csc(\theta)=\frac{hypotenuse}{opposite}[/tex]

    [tex]cos(\theta)=\frac{adjacent}{hypotenuse}[/tex]

    [tex]sec(\theta)=\frac{hypotenuse}{adjacent}[/tex]

    [tex]tan(\theta)=\frac{opposite}{adjacent}[/tex]

    [tex]cot(\theta)=\frac{adjacent}{opposite}[/tex]

    Let's calculate the adjacent side using pythagorean theorem:

    [tex]a^2+b^2=c^2[/tex]

    Where:

    a=opposite

    b=adjacent

    c=hypotenuse

    The problem provides us a and c because:

    [tex]sin(\theta)=\frac{opposite}{hypotenuse}=\frac{15}{17}[/tex]

    So:

    [tex]15^2+b^2=17^2[/tex]

    Solving for b:

    [tex]b^2=17^2-15^2\\b^2=289-225\\b^2=64\\b=\sqrt{64} \\b=8[/tex]

    Therefore:

    a=opposite=15

    b=adjacent=8

    c=hypotenuse=17

    Finally, let's see if the given options are correct:

    [tex]a.tan(\theta)=\frac{15}{8}=\frac{opposite}{adjacent}=\frac{15}{8}[/tex] , This is correct.

    [tex]b.csc(\theta)=\frac{17}{15}=\frac{adjacent}{hypotenuse}=\frac{17}{15}[/tex] , This is correct.

    [tex]c.cot(\theta)=\frac{17}{8}=\frac{adjacent}{opposite}=\frac{8}{15}[/tex] , This is incorrect.

    [tex]d.sec(\theta)=\frac{17}{8}=\frac{hypotenuse}{adjacent}=\frac{17}{8}[/tex] , This is correct.

    [tex]Check all that apply: if sin theta = 15/17 then: a. tan theta = 15/8 b. csc theta = 17/15 c. coc t[/tex]

  6. Hope this helps.......

    [tex]Given cos theta = 15/17. Find Sine Theta and tangent theta. Use the Pythagorean Identity for Trig to[/tex]

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