expressf(θ)= 8cosθ −15sinθ in the form rcos(θ + α), where r>0 and α is a positive acute angle hence find the general solution of the equation 80cos θ −150sinθ = 13 the maximum and minimum value of (5/(f(θ) + 3))


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Question Number 73544 by Rio Michael last updated on 13/Nov/19

$e x p r e s s f (θ) = 8 c o s θ - 15 s i n θ i n t h e f o r m$ $r c o s (θ + α), w h e r e r > 0 a n d α i s a p o s i t i v e a c u t e a n g l e$ $h e n c e$ $f i n d t h e g e n e r a l s o l u t i o n o f t h e e q u a t i o n$ $80 c o s θ - 150 s i n θ = 13$ $t h e m a x i m u m a n d m i n i m u m v a l u e o f \frac{5}{f (θ) + 3}$
	Answered by mind is power last updated on 13/Nov/19
	$8cos(θ)−15sin(θ)=17{(8/(17))cos(θ)−((15)/(17))sin(θ)} =17cos(θ+arcos((8/(17)))) α=arcos((8/(17))) 80cos(θ)−150sin(θ)=170cos(θ+α)=13 cos(θ+α)=((13)/(170)) θ+α=+_− arcos(((13)/(170)))+2kπ min and max didnt exist since f(θ)=−3 hase solution$
	$8 c o s (θ) - 15 s i n (θ) = 17 {\frac{8}{17} c o s (θ) - \frac{15}{17} s i n (θ)}$ $= 17 c o s (θ + a r c o s (\frac{8}{17}))$ $α = a r c o s (\frac{8}{17})$ $80 c o s (θ) - 150 s i n (θ) = 170 c o s (θ + α) = 13$ $c o s (θ + α) = \frac{13}{170}$ $θ + α = \underline{+} a r c o s (\frac{13}{170}) + 2 k π$ $m i n a n d m a x d i d n t e x i s t$ $s i n c e f (θ) = - 3 h a s e s o l u t i o n$
	Commented byRio Michael last updated on 13/Nov/19

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