Question
1) use the half-angle formulas to determine the exact values of the sine, cosine, and tangent of the angle.
67 degrees and 30'
Please explain!! I do not know how to start.
67 degrees and 30'
Please explain!! I do not know how to start.
Answers
Notice that 67 degrees, 30' is half of 135 degrees,
and 135 = 180-45
We also know that
cos 2A = 2 cos^2 A - 1
giving us
cos 135 = 2cos^2 67.5 - 1
So let's find cos 135
135 is in quadrant II, so
cos 135 = -sin 45 = -1/√2
back to cos 135 = 2cos^2 67.5 - 1
-1/√2 = 2cos^2 67.5 - 1
1 - 1/√2 = 2cos^2 67.5
(√2-1)/(2√2) = cos^2 67.5
cos 67.5 = √[(√2-1)/(2√2)]
(my calculator confirmed my answer to be correct
Using the other version of
cos 2A = 1 - 2sin^2 A -1 and following the above steps will give you the sin 67.5
you then find tan 67.5 by sin67.5/cos67.5 using the expressions.
Enjoy the algebra, lol
and 135 = 180-45
We also know that
cos 2A = 2 cos^2 A - 1
giving us
cos 135 = 2cos^2 67.5 - 1
So let's find cos 135
135 is in quadrant II, so
cos 135 = -sin 45 = -1/√2
back to cos 135 = 2cos^2 67.5 - 1
-1/√2 = 2cos^2 67.5 - 1
1 - 1/√2 = 2cos^2 67.5
(√2-1)/(2√2) = cos^2 67.5
cos 67.5 = √[(√2-1)/(2√2)]
(my calculator confirmed my answer to be correct
Using the other version of
cos 2A = 1 - 2sin^2 A -1 and following the above steps will give you the sin 67.5
you then find tan 67.5 by sin67.5/cos67.5 using the expressions.
Enjoy the algebra, lol
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