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Math Help - finding trigonometric values

  1. #1
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    finding trigonometric values

    Can someone help me with this question?

    Determine the exact value for each trigonometric expression, given sinA= 4/5, cos B= (-8/17),
    [(pi)/2] < A < pi and (pi) < B < [3(pi)/2]

    Find
    1) sin (A-B)

    2) Sin (A+B)

    3) cos (A+B)

    4) tan (A-B)

    if you could just tell me one of the answers and how you got to it, i can probably figure out the rest. thank youuuuuu!!
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  2. #2
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    Hello, clairez90!

    Determine the exact value for each trigonometric expression,
    . . given: . \sin A = \frac{4}{5},\;\;\cos B = -\frac{8}{17},\qquad \frac{\pi}{2} \leq A \leq \pi\;\text{ and }\;\pi \leq B \leq \frac{3\pi}{2}

    We are given: . \boxed{\sin A \:=\:\frac{4}{5}}\:=\:\frac{opp}{hyp}
    Using Pythagorus, we get: .  adj \:=\:\pm3 \quad\Rightarrow\quad \cos A \:=\:\pm\frac{3}{5}

    We are told that A is in Quadrant 2, where cosine is negative.
    . . Hence: . \boxed{\cos A \:=\:-\frac{3}{5}}
    We have: . \tan A \:=\:\frac{\sin }{\cos } \:=\:\frac{\frac{4}{5}}{\text{-}\frac{3}{5}} \quad\Rightarrow\quad \boxed{\tan A\:=\:-\frac{4}{3}}


    We are given: . \boxed{\cos B \:=\:-\frac{8}{17}}\:=\:\frac{adj}{hyp}
    Using Pythagorus, we get: . opp \:=\:\pm15\quad\Rightarrow\quad \sin B \:=\:\pm\frac{15}{17}

    We are told that B is in Quadrant 3, where sine is negaitve.
    . . Hence: . \boxed{\sin B \:=\:-\frac{15}{17}}
    We have: . \tan B \:=\:\frac{\sin B}{\cos B} \:=\:\frac{-\frac{15}{17}}{-\frac{8}{17}} \quad\Rightarrow\quad\boxed{ \tan B\:=\:\frac{15}{8}}


    Hence, we have: . \boxed{\begin{array}{ccccccc}\sin A &=&\dfrac{4}{5} & & \sin B &=&\text{-}\dfrac{15}{17} \\ \\[-3mm]<br />
\cos A &=&\text{-}\dfrac{3}{5} & & \cos B &=& \text{-}\dfrac{8}{17} \\ \\[-3mm] <br />
\tan A &=&\text{-}\dfrac{4}{3} & & \tan B &=& \dfrac{15}{8} \end{array}}


    We will use these values in all of the following problems . . .



    1)\;\;\sin(A-B)
    We're expected to know: . \sin(A \pm B) \:=\:\sin A\cos B \pm \sin B\cos A

    We have: . \sin(A-B) \;=\;\left(\frac{4}{5}\right)\left(\text{-}\frac{8}{17}\right) - \left(\text{-}\frac{15}{17}\right)\left(\text{-}\frac{3}{5}\right) \;=\; \text{-}\frac{32}{85} - \frac{45}{85} \;=\;\boxed{-\frac{77}{85}}



    2)\;\;\sin(A+B)

    3)\;\;\cos(A+B)
    We should know: . \cos(A \pm B) \;=\;\cos A\cos B \mp \sin A\sin B


    4)\;\;\tan(A-B)
    We should know: . \tan(A \pm B) \;=\;\frac{\tan A \pm \tan B}{1 \mp \tan A\tan B}


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  3. #3
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    thank you so much, Soroban!! that was extremely helpful!!!
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