trigonometry - How can I find $cos(theta)$ with $sin(theta)$?

If $\sin^2x$ + $\sin^22x$ + $\sin^23x$ = 1, what does $\cos^2x$ + $\cos^22x$ + $\cos^23x$ equal?

My attempted (and incorrect) solution:

• $\sin^2x$ + $\sin^22x$ + $\sin^23x$ = $\sin^26x$ = 1



• $\sin^2x = 1/6$


• $\sin x = 1/\sqrt{6}$


• $\sin x =$ opposite/hypotenuse


• Therefore, opp = 1, hyp = $\sqrt6$, adj = $\sqrt5$ (pythagoras theorum)


• $\cos x$ = adjacent/hypotenuse = $\sqrt5/\sqrt6$


• $\cos^2x$ = 5/6


• $\cos^2x + \cos^22x + \cos^23x = \cos^26x = 5/6 * 6 = 5$

I think I did something incorrectly right off the bat at step-2, and am hoping somebody will point me in the right direction.

Answer

You know that \begin{equation} \textrm{sin}^2x + \textrm{sin}^2 2x + \textrm{sin}^2 3x = 1 \qquad (\textrm{Eq.} 1) \end{equation}

Notice that \begin{equation} \begin{aligned} \textrm{sin}^2x + \textrm{cos}^2 x = 1 \Rightarrow \textrm{sin}^2x = 1 - \textrm{cos}^2 x \end{aligned} \end{equation}

Substituting the identity in Equation (1): \begin{equation} \begin{aligned} \underbrace{\textrm{sin}^2x}_{1-\textrm{cos}^2x} & + \underbrace{\textrm{sin}^2 2x}_{1-\textrm{cos}^2 2x } + \underbrace{\textrm{sin}^2 3x}_{1-\textrm{cos}^2 3x} = 1 \\ \\ & \Rightarrow 1-\textrm{cos}^2x + 1-\textrm{cos}^2 2x + 1-\textrm{cos}^2 3x = 1 \\ & \Rightarrow 3 -\left( \textrm{cos}^2x + \textrm{cos}^2 2x + \textrm{cos}^2 3x \right) = 1 \\ & \Rightarrow 2 = \textrm{cos}^2x + \textrm{cos}^2 2x + \textrm{cos}^2 3x \end{aligned} \end{equation}

Therefore: \begin{equation} \textrm{cos}^2x + \textrm{cos}^2 2x + \textrm{cos}^2 3x = 2 \end{equation}