Question

verify the identity. (6 cos x - sin x)^2+(cos x + 6 sin x)^2 = 37 to verify the identity, start with the more complicated side and transform it to look like the other side. choose the correct transformations and (6 cos x - sin x)^2+(cos x + 6 sin x)^2 = 37 cos ^2t+37 sin ^2t (do not factor.) square each of the binomial expressions and combine like terms.

Explanation:

Step1: Expand the binomials

Using the formula $(a - b)^2=a^{2}-2ab + b^{2}$ and $(a + b)^2=a^{2}+2ab + b^{2}$, we have:
\[

$$\begin{align*} (6\cos x-\sin x)^{2}&=(6\cos x)^{2}-2(6\cos x)\sin x+\sin^{2}x=36\cos^{2}x - 12\cos x\sin x+\sin^{2}x\\ (\cos x + 6\sin x)^{2}&=\cos^{2}x+2(\cos x)(6\sin x)+(6\sin x)^{2}=\cos^{2}x + 12\cos x\sin x+36\sin^{2}x \end{align*}$$

\]

Step2: Add the two expanded expressions

\[

$$\begin{align*} &(6\cos x-\sin x)^{2}+(\cos x + 6\sin x)^{2}\\ =&(36\cos^{2}x - 12\cos x\sin x+\sin^{2}x)+(\cos^{2}x + 12\cos x\sin x+36\sin^{2}x)\\ =&36\cos^{2}x+\cos^{2}x- 12\cos x\sin x+12\cos x\sin x+\sin^{2}x+36\sin^{2}x\\ =&(36 + 1)\cos^{2}x+(1 + 36)\sin^{2}x\\ =&37\cos^{2}x+37\sin^{2}x \end{align*}$$

\]

Step3: Use the Pythagorean identity $\sin^{2}x+\cos^{2}x = 1$

Factor out 37: $37\cos^{2}x+37\sin^{2}x=37(\cos^{2}x+\sin^{2}x)=37\times1 = 37$

Answer:

The identity $(6\cos x-\sin x)^{2}+(\cos x + 6\sin x)^{2}=37$ is verified.