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Core 4 Finding minimum value of trig equation

I know how to get max value but not min value q b i and ii
R = SQUARE ROOT 10
a is 1.25
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Reply 1
Avatar for Zacken
Zacken
22
Original post by koolgurl14
I know how to get max value but not min value q b i and ii
R = SQUARE ROOT 10
a is 1.25


The minimum value of sin⁡(x−α)\sin(x-\alpha) is −1-1, so the minimum value of Rsin⁡(x−α)=−RR\sin(x-\alpha) = -R and this occurs precisely when sin⁡(x−α)=−1⇒x−α=3π2\sin (x-\alpha) = -1 \Rightarrow x - \alpha = \frac{3\pi}{2} or some other reasonable value.
Easiest way to do this is to say what value of sin(.....) makes sqrt10 sin(.....) as small as possible. Given as you know that sin(...) ranges from -1 to 1, the smallest value occurs when sin(......) is -1, so the minimum value is -sqrt10. As for the second part, you know that sin(x-1.25)=-1, so just solve for x.
Original post by Zacken
The minimum value of sin⁡(x−α)\sin(x-\alpha) is −1-1, so the minimum value of Rsin⁡(x−α)=−RR\sin(x-\alpha) = -R and this occurs precisely when sin⁡(x−α)=−1⇒x−α=3π2\sin (x-\alpha) = -1 \Rightarrow x - \alpha = \frac{3\pi}{2} or some other reasonable value.


Hmm what about R>0 or does that not affect the results
its the right answer btw im just asking why we should ignore the R>0
Reply 4
Avatar for Zacken
Zacken
22
Original post by koolgurl14
Hmm what about R>0 or does that not affect the results
its the right answer btw im just asking why we should ignore the R>0


What do you mean by ignoring R>0R>0?

We have an expression Rsin⁡(x−α)R\sin(x-\alpha). To minimise this, we look at the variable terms and notice that sin⁡\sin is bounded between -1 (minimum) and 1 (maximum), so to maximise our entire function we pick the minimum sine value and hence say that the minimum of Rsin⁡(x−α)=R(−1)=−RR\sin(x-\alpha) = R(-1) = -R.
Reply 5
Avatar for KaylaB
KaylaB
20
Original post by koolgurl14
Hmm what about R>0 or does that not affect the results
its the right answer btw im just asking why we should ignore the R>0


R>0 only applies to finding the value of R
Ahhh I see what you mean its just cause in the question it says the range of R is R>0 so I got confused but I guess its not actual R being less than 0?
Reply 7
Avatar for Zacken
Zacken
22
Original post by koolgurl14
Ahhh I see what you mean its just cause in the question it says the range of R is R>0 so I got confused but I guess its not actual R being less than 0?


Ohh, it just means that when you do R2=12+(−3)2=10R^2 = 1^2 + (-3)^2 = 10 then you have R=±10R = \pm \sqrt{10} but since R>0R > 0 you pick R=10R = \sqrt{10} since −10<0-\sqrt{10} < 0
Original post by Zacken
Ohh, it just means that when you do R2=12+(−3)2=10R^2 = 1^2 + (-3)^2 = 10 then you have R=±10R = \pm \sqrt{10} but since R>0R > 0 you pick R=10R = \sqrt{10} since −10<0-\sqrt{10} < 0



ohhhhhh that makes a lot of sense thanks alot :h:
Reply 9
Avatar for Zacken
Zacken
22
Original post by koolgurl14
ohhhhhh that makes a lot of sense thanks alot :h:


No problem!

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