Proof: $\cos^p (\theta) \le \cos(p\theta)$

Set $f(\theta) = \cos p\theta - \cos^p \theta$ for some fixed $0 \leq p\leq 1$. Then for $\theta\in [0, \pi/2]$, $$f'(\theta) = p\left(\cos^{p-1}\theta \sin \theta - \sin p\theta\right) \geq p\left(\sin \theta - \sin p\theta\right) \geq 0,$$ since $\cos^{p-1}\theta \geq 1$ (as $p - 1\leq 0$) and $\sin \theta$ is increasing on $[0, \pi/2]$. Since $f(0) = 0$, it follows that $f$ is increasing and thus nonnegative on $[0, \pi/2]$, as required.

Let us fix the value of $\theta$ and vary $p$.

For $p=0$, $\cos^0(\theta)=\cos(0\theta)=1$.

For $p=1$, $\cos^1(\theta)=\cos(1\theta)=\cos(\theta)$.

Then, $$(\cos^p(\theta))''=(\log(\cos(\theta)))^2\cos^p(\theta)\ge0,$$ and $$(\cos(p\theta))''=-\theta^2\cos(p\theta)\le0.$$ The LHS function is concave down (negative exponential) and the RHS function is concave up (cosinusoid). They meet at endpoints without crossing.

With the straight line, this also establishes $$\color{blue}{\cos^p(\theta)}\le\color{magenta}{1-p(1-\cos(\theta))}\le\color{green}{\cos(p\theta)}.$$

Because cosine is concave on the interval $[0,\pi/2]$, we have $$ \cos(p\theta)=\cos(p\theta+(1-p)0)\geq p\cos(\theta)+(1-p)\cos(0)=p\cos(\theta)+(1-p). $$ So our desired inequality follows if we can prove that $$ \cos^p(\theta)\leq p\cos(\theta)+(1-p).\tag{*} $$ Clearly, (*) holds if $\theta=\frac{\pi}{2}$ (the LHS is $0$ while the RHS is nonnegative), so assume $\theta<\frac{\pi}{2}$. This assumption means $\cos(\theta)>0$ so that $\cos(\theta)-1>-1$, allowing us to apply Bernoulli's inequality: $$ \cos^p(\theta)=[1+(\cos(\theta)-1)]^p\leq 1+p(\cos(\theta)-1)=p\cos(\theta)+(1-p). $$ This completes our proof.