How do you find the exact values of #sin2u, cos2u, tan2u# using the double angle values given #cotu=-4, (3pi)/2<u<2pi#?

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Answer 1 · 2017-02-13T12:40:12

#sin2u=-8/17#, #cos2u=15/17# and #tan2u=-8/15#

As #cotu=-4#, #tanu=-1/4#

#:.tan2u=(2tanu)/(1-tan^2u)=(2xx(-1/4))/(1-(-1/4)^2)#

= #(-1/2)/(1-1/16)=(-1/2)/(15/16)#

= #-1/2xx16/15=-8/15#

Therefore #sec2u=sqrt(1-tan^2 2u)#

= #sqrt(1-(-8/15)^2)=sqrt(1+64/225)=sqrt(289/225)=17/15#

We have kept it positive as #(3pi)/2 < u < 2pi# and hence #3pi < 2u < 4pi#,

but as #tan2u=-8/15#, #2u# must be in #Q4# and #sec2u > 0#.

therefore #cos2u=1/(sec2u)=15/17#

and #sin2u=tan2uxxcos2u=-8/15xx15/17=-8/17#