I was recently trying to solve a trigonometry question, which asked to find theta:
$$3 \sin\theta + 4 \cos\theta = 4$$
I took $4 \cos\theta$ on the other side, and squared both the sides. After that I got the values of $\cos\theta = 7/25$ or $1$.
I understood this method. But another derivation I know is:
If $P = a \sin\theta + b \cos\theta$ then:
$$P = (\sqrt{a^2 + b^2}) \cos(\theta - \gamma)$$
By this I am not able to get the solution. Please help, Thank you.
Given the equation: $$3 \sin\theta + 4 \cos\theta = 4$$ $$\frac{3}{5} \sin\theta + \frac{4}{5} \cos\theta = \frac{4}{5}$$
We define an auxiliary angle $\gamma$ such that $\cos\gamma = \frac{4}{5}$ and $\sin\gamma = \frac{3}{5}$. The equation becomes: $$\sin\gamma \sin\theta + \cos\gamma \cos\theta = \cos\gamma$$ $$\cos(\theta - \gamma) = \cos\gamma$$
The general solution for $\cos A = \cos B$ is $A = 2n\pi \pm B$. Here: $$\theta - \gamma = 2n\pi \pm \gamma$$
Case 1: $$\theta - \gamma = -\gamma \implies \theta = 0$$ $$\implies \cos\theta = 1$$
Case 2: $$\theta - \gamma = \gamma \implies \theta = 2\gamma$$ Now, $$\cos(2\gamma) = 2\cos^2\gamma - 1$$ Substitute $\cos\gamma = \frac{4}{5}$: $$\cos(2\gamma) = 2\left(\frac{4}{5}\right)^2 - 1 = 2\left(\frac{16}{25}\right) - 1$$ $$= \frac{32}{25} - \frac{25}{25} = \frac{7}{25}$$
