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# Trig Question!!!!! watch

1. 1a)Given that sin(θ+45°) = 2sinθ, show that tanθ = 1÷(2√2-1)

b) Hence solve the equation sin(θ+45°) = 2sinθ for all values θ between 0° and 360°

Thanx!!!
2. (Original post by Looontag)
1a)Given that sin(θ+45°) = 2sinθ, show that tanθ = 1÷(2√2-1)
sinθcos45+cosθsin45=2sinθ
tanθcos45+sin45=2tanθ
tanθ[cos45-2]=-sin45
tanθ=(sin45)/(2-cos45)
tanθ=(0.5rt2)/(2-0.5rt2)
tanθ=1/(2rt2-1)

b) Hence solve the equation sin(θ+45°) = 2sinθ for all values θ between 0° and 360°
tanθ=1/(2rt2-1) so use the cast diagram with your calculator to find all values in the range.
3. Thanks a lot. but where did the tanθcos45°+sin45° = 2tanθ come from???
4. (Original post by Looontag)
Thanks a lot. but where did the tanθcos45°+sin45° = 2tanθ come from???
Dividing through by cosθ as you know sinθ/cosθ=tanθ
5. Thanks a lot

(although I do not have a cast diagram on my calculator lol!!)-i presume that is something graphical!!!!

Thanks again for ur help!
6. (Original post by Looontag)
Thanks a lot

(although I do not have a cast diagram on my calculator lol!!)-i presume that is something graphical!!!!

Thanks again for ur help!
http://www.mathsnet.net/asa2/2004/c24cast.html
7. Now i feel silly lol.

I have never heard of the cast diagram before and my exam is tomoro aargh!!! Iv prob learnt some other method of working it out that I can't remember lol.

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