Prove this identity$ \cos u \cos v \cos t = \,\frac14[\cos(t-u+v)+\cos(t+u-v)+\cos(t-u-v)+\cos(t+u+v)];$

Question

The problem reads as follows.

Prove the identity

$$ \cos u \cos v \cos t = $$ $$ \frac14[\cos(t-u+v)+\cos(t+u-v)+\cos(t-u-v)+\cos(t+u+v)]$$

Hint: begin with the right side and use cosine sum identity for three angles.

Thanks in advance. :)

Answer 1

You can solve such problems with complex numbers: $$ \cos x = \frac{e^{ix} + e^{-ix}}{2} $$ If you set $a = e^{it}, b = e^{iu}, c = e^{iv}$, just observe \begin{align*} (a + a^{-1})(b + b^{-1})(c + c^{-1}) &= \left[ abc + a^{-1}b^{-1}c^{-1} \right] \\ &+ \left[ abc^{-1} + a^{-1}b^{-1}c \right] \\ &+ \left[ ab^{-1}c^{-1} + a^{-1}bc \right] \\ &+ \left[ ab^{-1}c + a^{-1}bc^{-1} \right] \\ \end{align*}

As Blue notes in the comments, you will need an additional term, $\cos(t - u + v)$. Now you just have to divide the above by $8$ on both sides, and match each of the four terms above with one of the four terms $\cos(t + u + v)$, $\cos(t + u - v)$, $\cos(t - u + v)$, $\cos(t - u - v)$.

Answer 2

Setting

$$ u=v=t=0,$$ we get

$$ 1=\frac34 $$

So added the obvious missing term in the title so that

$$ \cos t\cdot\cos u\cdot\cos v= $$

$$ \frac14[\cos(t-u+v)+\cos(t+u-v)+\cos(t-u-v)+\cos(t+u+v)].$$

Simplification can be as follows

$$ \cos (t+ u+v)= c_{t+u} c_v-s_{t+u}s_v$$

$$ = + c_tc_uc_v-s_ts_uc_v-s_vs_tc_u-s_us_vc_t$$

$$\cos(t+u-v) = +...-...+...+...$$

$$\cos(u+v-t) = +...+...+...-...$$

$$\cos(t+u-v) = +...+...-...+...$$

Adding the four equations only first terms remain, rest cancel out and the result follows.

Answer 3

$$\cos u \cos v = \frac12 (\cos(u+v) + \cos(u-v))$$

proof: $$ \cos u \cos v = \Re (e^{iu}\cos v) = \frac 12\Re (e^{iu}(e^{iv}+ e^{-iv})) \\ = \frac 12\Re (e^{i(u+v)}+ e^{i(u-v)}) \\ = \frac12 (\cos(u+v) + \cos(u-v)) $$

$$\cos u \cos v \cos t = \frac14 (\cos(u+v+t) +\cos(u+v-t) + \cos(u-v+t) + \cos(u-v-t)) $$

proof: $$ \cos u \cos v \cos t = \frac12 (\cos(u+v)\cos t + \cos(u-v)\cos t) \\ = \frac14 (\cos(u+v+t) +\cos(u+v-t) + \cos(u-v+t) + \cos(u-v-t)) $$ when you apply twice the formula.

Answer 4

Let us first recall that for every $$ \cos p\cos q=\frac12[\cos(p-q)+\cos(p+q)]\quad \forall p,q \in \mathbb{R} $$ Therefore \begin{eqnarray} \cos t\cdot\cos u\cdot\cos v&=&\frac12\cos t\cdot[\cos(u-v)+\cos(u+v)]\\ &=&\frac12\cos t\cdot\cos(u-v)+\frac12\cos t\cdot\cos(u+v)\\ &=&\frac14[\cos(t-u+v)+\cos(t+u-v)+\cos(t-u-v)+\cos(t+u+v)]. \end{eqnarray}