Brain Teaser 3: Solve sin(3 – z) = 0.8 for 0 < z < π radians.

I have gone through this question with my students. You should try it too. You may post your answer by way of comments to this post with brief intro of yourself.

Either I or one of my students will put up the answer in a week or so. Have fun with brain teasers of this blog! :)

(Acknowledgment: The above question was taken from IGCSE Cambridge Maths 0606, Yr 2012 .... Q 9(b))

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My Proposed Solution:

For 0 < z < π rad, solve sin(3 – z) = 0.8

Get domain for (3 – z):

Since 0 < z < π rad, … x (-1) to get -z, reverse inequality sign

Therefore, 0 > -z > -π rad

+ 3:           3 > (3 – z) > 3 – π

                 3 > (3 – z) > 0.142…domain for principal angle obtained

Since, sin(3 – z) = 0.8,  and sin^-1(0.8) = 0.9272…(> 0.142)

therefore, (3 – z) is in 1^st quadrant (qd) or 2^nd qd:

1^st qd:      3 – z = sin^-1(0.8)

                     z = 3 - sin^-1(0.8) = 2.072704782

                     z = 2.07 (to 3 sf) (Answer)

2^nd qd:     3 – z = π - sin^-1(0.8) (< 3)

                      z = 3 - π + sin^-1(0.8) = 0.7857…

                      z = 0.786 (to 3 sf) (Answer)