# Trig Double Angle Identities

• February 27th 2009, 07:32 AM
Beard
Trig Double Angle Identities
Stuck on this and don’t have a clue where I am going wrong

It is given that $3\cos\theta - 2\sin\theta = R\cos(\theta + \alpha), \ where\ R>0\ and\ 0^{\circ} < \alpha < 90^{\circ}$

A) Find the value of R

B) Show that $\alpha \approx 33.7^{\circ}$

C) Hence write down the maximum value of $3\cos\theta - 2\sin\theta$ and find a positive value of $\theta$ at which the maximum value occurs

NOT AN ACTUAL IDENTITY
• February 27th 2009, 08:14 AM
Soroban
Hello, Beard!

Quote:

Stuck on this and don’t have a clue where I am going wrong.
It's hard to see your work from here.
But it looks like you played the ${\color{blue}K\spadesuit}$ instead of the ${\color{blue}7\heartsuit}$.

Quote:

It is given that: . $3\cos\theta - 2\sin\theta \:=\: R\cos(\theta + \alpha),\text{ where }R>0 \text{ and }0^o < \alpha < 90^u$

A) Find the value of $R.$

Multiply the left side by $\frac{\sqrt{13}}{\sqrt{13}}$

. . $\sqrt{13}\left(\frac{3}{\sqrt{13}}\cos\theta - \frac{2}{\sqrt{13}}\sin\theta\right) \;=\;R\cos(\theta + \alpha)$

Let $\alpha$ be an angle such that: . $\begin{array}{c}\cos\alpha \:=\:\frac{3}{\sqrt{13}} \\ \\[-4mm] \sin\alpha \:=\:\frac{2}{\sqrt{13}} \end{array}$

Then we have: . $\sqrt{13}\underbrace{(\cos\theta \cos\alpha - \sin\theta\sin\alpha)} \:=\:R\cos(\theta + \alpha)$

. . . . . . . . . . . . . . . $\sqrt{13}\,\overbrace{\cos(\theta + \alpha)} \:=\:R\cos(\theta + \alpha)$

Therefore: . $\boxed{R \:=\:\sqrt{13}}$

Quote:

Show that: . $\alpha \,\approx\, 33.7^o$

Since $\sin\alpha \,=\,\tfrac{2}{\sqrt{13}}$, then: . $\alpha \:=\:\sin^{-1}\left(\tfrac{2}{\sqrt{13}}\right) \quad\Rightarrow\quad \boxed{\alpha\;\approx\; 33.7^o}$

Quote:

Hence, find the maximum value of $3\cos\theta - 2\sin\theta$
and find a positive value of $\theta$ at which the maximum value occurs.

The maximum of: . $3\cos\theta - 2\sin\theta \:=\:\sqrt{13}\,\cos(\theta + \alpha)$ occurs when $\cos(\theta + \alpha) \:=\:1$

. . Then the maximum value is $\boxed{\sqrt{13}}$

If $\cos(\theta + \alpha) \:=\:1$, then: . $\theta + \alpha \:=\:360^o$

Therefore: . $\theta \:=\:360^o - \alpha \:=\:360^o - 33.7^o \quad\Rightarrow\quad\boxed{\theta\;=\;326.3^o}$