CAN ANYONE HELP ME UNDERSTAND THIS?!:( Given the following values: sinA=65/97, sinB=20/29, cosA=72/97, cosB=21/29. Find sin〖(A+B)〗 Find sin〖(A-B)〗 Find cos〖(A+B)〗 Find cos〖(A-B)〗
sin(A+B)=sin(A)cos(B)+sin(B)cos(A)
just enter in your given values like so: \[\sin(A+B)=\frac{65}{97} \cdot \frac{21}{29} +\frac{20}{29} \cdot \frac{72}{97}\]
@hannaolsonnn do you understand do you think you can simplify that? and so find the other things there
yes!! thank you so much for helping!! i am going to try and find the answer, may I tag you for reassurance once I find it? (:
sure
okay I think I did this wrong.. Please don't think I am so stupid or something we just started this is class and I do not understand it at all. SO when I enter this in my calculator I would enter sin(65/97*21/29+20/29*72/97) ?? That is probably wrong.. I really have no idea how to do this at all. All we have been taught so far is how to enter simple equations into the calculator dealing with triangles and stuff. I really need help.
why is there a sin there at all?
\[\sin(A+B)=\frac{65}{97} \cdot \frac{21}{29} +\frac{20}{29} \cdot \frac{72}{97} \\ =\frac{65 \cdot 21 +20 \cdot 72}{97 \cdot 29}\] all i did was combined the fractions since they had a common denominator
now you need to perform order or operations on top and bottom
\[=\frac{1365+1440}{2813}=\frac{2805}{2813} \text{ then you reduce the fraction if you can }\]
OMG THANK YOU SO MUCH i really need the help ps i am new here so could you tell me how i could give you a medal? SO for sin (a-b) would you combine the fractions again? or what would you do before you perform order of operations??
well you know sin(a-b)=sin(a)cos(b)-cos(a)sin(b) right?
just plug in as I did before and apply the order of operations again
sin ( a-b) = 65/97 times 21/29 - 72/90 times 20/29 is that correct so far?
@freckles when I do the order of operations I get this |dw:1421789692765:dw|
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