This is simple:
if tan A=3 then draw a right angled triangle with the opposite sideequal to 3 and the adjascent side=1 since Tan=opposite/adjascent
therefore use the pythagoras theorem to get the hypotenuse
h^2=1^2 +3^2
h^2 = 1+9
h^2 = 10
h=√10
then get the sin
sin=opposite/hypotenuse
=3/√10
cos=adjascent/hypotenuse
=1/√10
then use the addition formulas
sin(A+B) = sinAcosB+cosAsinB
=3/√10*1/√10+1/√10*3/√10
=3/10+3/10
=6/10
=3/5 (answer)
for cos(A+B)=cosAcosB-sinAsinB
=1/√101/√10-3/√103/√10
=1/10-9/10
=-4/4 (answe)
i hope this is helpful
Roger
2007-03-08 09:47:04
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answer #1
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answered by Roger Aime 2
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Please draw a outstanding angled triangle Tan@ = 3/4 which potential Tan@ = oppo facet/ Adj facet containing the terrific suited attitude Use Pythagores theorem and discover the size of hypotenus 3^2 + 4^2 = 5^2 subsequently we get hypo= 5 instruments Sin@ = 3/5 and Cos@ = 4/5 and you'd be able to discover the values of all trig applications
2016-10-17 21:43:03
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answer #2
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answered by Anonymous
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i )
Complete triangle with sides 1, 3 √10 (hypotenuse)
sin 2A = 2sin A.cos A = 2 x 3 / √10 x 1 / √10
sin 2A = 6 / 10 = 3 / 5
ii) cos 2A = 4 / 5
2007-03-08 09:31:04
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answer #3
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answered by Como 7
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if tanA=3,we have
a triangle 1,3,sqrt10
therefore,
sinA=3/sqrt10
cosA=1/sqrt10
i)using the identity,
sin2A=2sinA*cosA,
we have
sin2A=2*3/(sqrt10)*1/(sqrt10)
=6/10=3/5
ii)using the identity,
cos2A=(cosA)^2-(sinA)^2,
we have
cos2A
=(1/sqrt10)^2-(3/sqrt10)^2
=1/10-9/10= -8/10 = -4/5
[note that (3/5)^2+(-4/5)^2=1,
(pythagoras) and since A is
approx 71.56 degrees,
cos2A is in the second
quadrant,and is therefore
negative]
i hope that this helps
2007-03-08 09:31:18
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answer #4
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answered by Anonymous
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Richards answer can not be bettered,although there are a number of ways of doing this..as Roger shows.The double angle formula sin2A,etc , do ,of course,come from the double angle formula sin(A+B),whereA=B
2007-03-08 21:29:58
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answer #5
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answered by Arthur C 2
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....first find sinA and cosA, then use the double angle formulas...draw the triangle to find sin and cosA....hypotenuse i believe is the square root of ten
2007-03-08 09:06:32
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answer #6
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answered by Anonymous
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Sorry,
Me confused. Sorry, yoou got to ask Lisa Simpson cause when it comes to that I'm dumber than a dingbat.
~*Alyssa*~
2007-03-08 09:03:44
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answer #7
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answered by ~*Alyssa*~ 2
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cos 2A is the way.
2007-03-08 09:07:20
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answer #8
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answered by Anonymous
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