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    https://www.youtube.com/watch?time_c...&v=Nc0ySihitKE

    Hey guys, I watched this video and I don't understand how he gets the sin, cos and tan values.

    When I learned C2 trig I used the graph method as I found this much easier than the CAST diagram/quadrat rule. I never learned the quadrat rule but it is used in the video.

    When I was doing these questions alone, I used the simple triangle method (below) to find the values and all my answers were positive. Obviously, my final answer was wrong as he found negative values and all mine were positive even though they were correct.

    Can I do these sorts of questions using the graph method rather than the quadrat method as it really confuses me, and how would I do this?

    Thank you!
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    (Original post by djels013.211)
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    So begin by drawing graphs of \cos(x) and \sin(x)

    A is obtuse and B is reflex. So 90 < A < 180 and 180 < B < 360.

    Further, notice how \sin(A) > 0 in this region, and this makes sense as \sin(A) = \frac{3}{5} > 0. Is \cos(A) positive or negative in this region? You'd quickly find that it is negative in that region, so \cos(A) < 0. If you use triangles to work out your \cos(A), that's fine just make the answer negative at the end.

    Then \cos(B) is either negative or positive for the B values we deduced at the start, however we are GIVEN that \cos(B) = \frac{5}{13} > 0, so we know that 270 < B < 360. Now we know that \sin(B) < 0 in this region, so whatever value of \sin(B) you get, make it negative.


    That's how you'd deduce the signs from the graph.
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    (Original post by RDKGames)
    So begin by drawing graphs of \cos(x) and \sin(x)

    A is obtuse and B is reflex. So 90 < A < 180 and 180 < B < 360.

    Further, notice how \sin(A) > 0 in this region, and this makes sense as \sin(A) = \frac{3}{5} > 0. Is \cos(A) positive or negative in this region? You'd quickly find that it is negative in that region, so \cos(A) < 0. If you use triangles to work out your \cos(A), that's fine just make the answer negative at the end.

    Then \cos(B) is either negative or positive for the B values we deduced at the start, however we are GIVEN that \cos(B) = \frac{5}{13} > 0, so we know that 270 < B < 360. Now we know that \sin(B) < 0 in this region, so whatever value of \sin(B) you get, make it negative.

    That's how you'd deduce the signs from the graph.
    Thank you so much! You've made a massive difference.
 
 
 
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