Half-Angle Identities
Key Questions
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Common Half angle identity:
1.#sin a = 2 sin (a/2)* cos (a/2)# Half angle Identities in term of t = tan a/2.
2.#sin a = (2t)/(1 + t^2)# 3.
#cos a = (1 - t^2)/(1 + t^2)# #tan a = (2t)/(1 - t^2).#
Use of half angle identities to solve trig equations.
Example. Solve
#cos x + 2*sin x = 1 + tan (x/2).#
Solution. Call#t = tan (x/2)# . Use half angle identities (2) and (3) to transform the equation.#(1 - t^2)/4 + (1 + t^2)/4 = 1 + t.# #1 - t^2 + 4t = (1 + t)(1 + t^2)# #t^3 + 2t^2 - 3t = t*(t^2 + 2t - 3) = 0.# Next, solve the
#3# basic trig equations:#tan (x/2) = t = 0; tan (x/2) = -3;# and#tan (x/2) = 1.# Nghi N. · 1 · Apr 12 2015
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The half-angle identities are defined as follows:
#\mathbf(sin(x/2) = pmsqrt((1-cosx)/2))# #(+)# for quadrants I and II
#(-)# for quadrants III and IV#\mathbf(cos(x/2) = pmsqrt((1+cosx)/2))# #(+)# for quadrants I and IV
#(-)# for quadrants II and III#\mathbf(tan(x/2) = pmsqrt((1-cosx)/(1+cosx)))# #(+)# for quadrants I and III
#(-)# for quadrants II and IVWe can derive them from the following identities:
#sin^2x = (1-cos(2x))/2# #sin^2(x/2) = (1-cos(x))/2# #color(blue)(sin(x/2) = pmsqrt((1-cos(x))/2))# Knowing how
#sinx# is positive for#0-180^@# and negative for#180-360^@# , we know that it is positive for quadrants I and II and negative for III and IV.#cos^2x = (1+cos(2x))/2# #cos^2(x/2) = (1+cos(x))/2# #color(blue)(cos(x/2) = pmsqrt((1+cos(x))/2))# Knowing how
#cosx# is positive for#0-90^@# and#270-360^@# , and negative for#90-270^@# , we know that it is positive for quadrants I and IV and negative for II and III.#tan(x/2) = sin(x/2)/(cos(x/2)) = (pmsqrt((1-cos(x))/2))/(pmsqrt((1+cos(x))/2))# #color(blue)(tan(x/2) = pmsqrt((1-cos(x))/(1+cos(x))))# We can see that if we take the conditions for positive and negative values from
#sinx# and#cosx# and divide them, we get that this is positive for quadrants I and III and negative for II and IV.
Mika
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Dec 18 2014