Question

Trigonometry...

Answer 1

If \(\max \{5\sin\theta + 3\sin(\theta - \alpha) \} = 7 \), then the set of possible values of \(\alpha\) is?

Answer 2

@ganeshie8 @Miracrown

Answer 3

those coefficients tell me that there's something going on with 3-4-5 triangles.

Answer 4

\[a\cos t + b\sin t = \sqrt{a^2+b^2}\cos (\cdots)\]
familiar with this identity right ?

Answer 5

mhm

Answer 6

Do I convert one of them from sin to cosine?

Answer 7

The idea is to get the given expression into that form and the max value automatically becomes \(\sqrt{a^2+b^2}\)

(just trying to avoid calculus here to make it simple.. )

Answer 8

I see, but how? Do I multiply and divide the expression by 5?

Answer 9

\[\begin{align}5\sin\theta + 3\sin(\theta - \alpha) &=5\sin \theta + 3\sin\theta \cos \alpha - 3\cos\theta \sin \alpha \\~\\
&=(5+3\cos \alpha)\sin \theta +(-3\sin\alpha)\cos\theta
\end{align}\]

\(a =5+3\cos \alpha \)
\(b=-3\sin\alpha\)

Answer 10

oh my...

Answer 11

so I multiply and divide the expression by \(\sqrt{(5 + 3\cos \alpha)^2 + (-3\sin \alpha)^2}\). Genius.

Answer 12

Exactly! but as you can see we don't need to wry about all that because our goal is to find \(\alpha\) given max value = 7

Answer 13

\[\sqrt{25 + 9 \cos^2 \alpha + 30\cos \alpha + 9\sin^2 \alpha} = \sqrt{34 + 30\cos\alpha}\]

Answer 14

Ooooh.

Answer 15

that looks good yeah :) set that equal to 7 and solve \(\alpha\)

Answer 16

\[\sqrt{34 + 30 \cos \alpha} = 7?\]

Answer 17

\[34 + 30\cos \alpha = 49\]\[\cos \alpha = 1/2 = \cos \pi/3 \]\[\alpha = 2n\pi \pm \pi/3\]

Answer 18

yay

Answer 19

Looks neat! you may use calculus also if that identity doesnt strike sooner..

Answer 20

Mhm.