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Phoenix [80]
4 years ago
11

Find cos θ given that cos 2θ = 5/6 and 0 ≤ θ < π/2. Give an exact answer

Mathematics
1 answer:
trasher [3.6K]4 years ago
3 0
\bf \textit{Double Angle Identities}&#10;\\ \quad \\&#10;sin(2\theta)=2sin(\theta)cos(\theta)&#10;\\ \quad \\&#10;cos(2\theta)=&#10;\begin{cases}&#10;cos^2(\theta)-sin^2(\theta)\\&#10;1-2sin^2(\theta)\\&#10;\boxed{2cos^2(\theta)-1}&#10;\end{cases}&#10;\\ \quad \\&#10;tan(2\theta)=\cfrac{2tan(\theta)}{1-tan^2(\theta)}\\\\&#10;-------------------------------\\\\&#10;

\bf cos(2\theta)=\cfrac{5}{6}\implies 2cos^2(\theta)-1=\cfrac{5}{6}\implies 2cos^2(\theta)=\cfrac{5}{6}+1&#10;\\\\\\&#10;2cos^2(\theta)=\cfrac{11}{6}\implies cos^2(\theta)=\cfrac{11}{12}\implies cos(\theta)=\pm\sqrt{\cfrac{11}{12}}


now, bear in mind, the square root gives us +/- versions, so, which is it? well, we know the angle is in the range of "<span>0 ≤ θ < π/2", that simply means the 1st quadrant, so, we'll use the positive one then

</span>\bf cos(\theta)=\cfrac{\sqrt{11}}{\sqrt{12}}\implies cos(\theta)=\cfrac{\sqrt{11}}{2\sqrt{3}}&#10;\\\\\\&#10;\textit{now, let's rationalize the denominator}&#10;\\\\\\&#10;\cfrac{\sqrt{11}}{2\sqrt{3}}\cdot \cfrac{\sqrt{3}}{\sqrt{3}}\implies \cfrac{\sqrt{11}\cdot \sqrt{3}}{2\sqrt{3^2}}\implies \cfrac{\sqrt{11\cdot 33}}{2\cdot 3}\implies \boxed{\cfrac{\sqrt{33}}{6}}<span>
</span>
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