Asked by Don
Half angle identities
Find exact values of
a) cos 1/2
b) sin 1/2
c) tan 1/2
based on:
cos = 1/5
greater than 270 deg less than 360 deg.
I get a) - rt 3/5 b) rt24/5 and c) does not work out at all ....
Help please..
book says for answers
a)-rt15/5 b) rt10/5 c) -rt6/3
How do you get that answers ?
Find exact values of
a) cos 1/2
b) sin 1/2
c) tan 1/2
based on:
cos = 1/5
greater than 270 deg less than 360 deg.
I get a) - rt 3/5 b) rt24/5 and c) does not work out at all ....
Help please..
book says for answers
a)-rt15/5 b) rt10/5 c) -rt6/3
How do you get that answers ?
Answers
Answered by
Steve
It's hard to interpret your expressions, with no parentheses.
Using θ as the reference angle with cosθ = 1/5, your angle Ø in the 4th quadrant is Ø = 2pi - θ
For (a) if you got -√(3/5), that is the same as -√15/5
I like that, since
cos^2 θ/2 = (1 + cosθ)/2 = (1 + 1/5)/2 = 3/5
But Ø/2 = pi - θ/2 so cos is negative.
For (b),
sin^2 θ/2 = (1 - cosθ)/2 = (1 - 1/5)/2 = 2/5
in QII, sin is positive, so sin Ø/2 = √(2/5) = √10/5
I assume 24/5 as you show is a typo, since that's greater than 1.
For (c), tan Ø/2 = sinØ/2 / cos Ø/2
= √(2/5)/-√(3/5)
= -√2/√3 = -√(2/3) or -√6/3
Apparently the book does not like radicals in the denominator, but they don't bother me.
Using θ as the reference angle with cosθ = 1/5, your angle Ø in the 4th quadrant is Ø = 2pi - θ
For (a) if you got -√(3/5), that is the same as -√15/5
I like that, since
cos^2 θ/2 = (1 + cosθ)/2 = (1 + 1/5)/2 = 3/5
But Ø/2 = pi - θ/2 so cos is negative.
For (b),
sin^2 θ/2 = (1 - cosθ)/2 = (1 - 1/5)/2 = 2/5
in QII, sin is positive, so sin Ø/2 = √(2/5) = √10/5
I assume 24/5 as you show is a typo, since that's greater than 1.
For (c), tan Ø/2 = sinØ/2 / cos Ø/2
= √(2/5)/-√(3/5)
= -√2/√3 = -√(2/3) or -√6/3
Apparently the book does not like radicals in the denominator, but they don't bother me.
Answered by
Damon
cos a = 1/5
then sin a = sqrt (1-cos^2a) = sqrt(24/25)
cos(a/2) = -sqrt [(1+1/5)/2 ] = -sqrt (3/5)
So I agree with you.
then sin a = sqrt (1-cos^2a) = sqrt(24/25)
cos(a/2) = -sqrt [(1+1/5)/2 ] = -sqrt (3/5)
So I agree with you.
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