How would you do this?
https://scontent-syd1-1.xx.fbcdn.net/hphotos-xta1/v/t35.0-12/12941188_1672164113047290_638544438_o.jpg?oh=fc8f7d7f97e3520741ef58de5946aa9a&oe=5703728C[//img][/QUOTE]
x = 2 + cos(Ø)
cos(Ø) = (x - 2)
y = 1 + sin(Ø)
sin(Ø) = (y - 1)
now sin^2(Ø) + cos^2(Ø) = 1
so (x - 2)^2 + (y - 1)^2 = 1
Don't you make theta the subject?How would you eliminate theta?
not reallyDon't you make theta the subject?
If it's relevant to this threadMind if I ask really hard year 12 maths Q's
Methods displayed here use the retrieval (via memory) of a trigonometric identity which is good in creativity but if you want to improve mathematical problem solving skills deliberately: increase gravity.How would you eliminate theta?
Observing that cosθ and sinθ are in fact ratios of sides in a right angled triangleMethods displayed here use the retrieval (via memory) of a trigonometric identity which is good in creativity but if you want to improve mathematical problem solving skills deliberately: increase gravity.
Example, can you do this question without explicitly using this trigonometric identity?
Yeah it can be, by considering only the magnitudes of the sides of the triangle on the number plane then applying the relevant identities of each quadrant (e.g. sin(π-θ)=sin θ)). Applying Pythagoras' theorem then cleans it up, which is in essence deriving the trig identity.It can probably be done wlog for each quadrant because when you square the negative sign loses it's impact.
Wild guess.
theta = 45 deg, 225 deg.for 0<θ<360, find the value of
sinθ=cosθ
explain? I kinda get the 45 deg parttheta = 45 deg, 225 deg.
+180 degreesexplain? I kinda get the 45 deg part