Question

More trig identities, need help with part of a question please!

Answer 1

\[(3 \cos \theta + 4 \sin \theta)^2 \]

Find the minimum and maximum values, and the value of the turning points within 0 to 360.

So I know I need to use the equation : \[r \cos ( \theta - \alpha)\] and that r is the amplitude and alpha is linked to the turning point.

I've figured out that r = plus/minus 25, but I'm struggling with alpha.

Thanks in advance!

Answer 2

hint:
the first derivative of the function above, is:
\[\Large f'\left( \theta \right) = - 48{\left( {\sin \theta } \right)^2} + 14\sin \theta + 24\]

Answer 3

Thanks for the hint Michele! I haven't learnt derivatives of trig functions yet and this wasn't mentioned in the guide previous to the exercise. I went through some guides does not seem like nx^n-1 , (as far as I can tell?)

My main problem is trying to get around the fact that this is squared, alpha is supposed to be found by dividing 4sin/3cos , but I'm sure that squared it will change matters.

Answer 4

if you convert a cos theta + b sin theta to the form r cos(theta - alpha)
then r is sqr(a^2+b^2) (I get sqr(25)= 5 )
and alpha is inverse_tan (b/a)
(I think of the coeff of the cos as the x value, and the coeff of the sin as the y value)

Answer 5

so you have
\[ 25 \cos^2(\theta - 53.13º) \]
that is either zero or a positive number. i.e. min value will be zero, and max will be 25

Answer 6

Okay that makes sense, I see I went wrong with the min/max value.

So, from what I can see, the value of tan isn't affected by the square? The turning points would be 53.13, and then 180+53.13 for the next one?

Answer 7

Sorry, meant to add, thank you for your help!

Answer 8

yes, we do the conversion before squaring, so there is no effect on alpha

the square makes the curve look (roughly) like this