Find the value of the following- sin 2 - 4 -cos 2 - 4 -tan 2 - 3

We have sinπ #160; 4 #160; =sin 45 ^ o = 1 √( 2 ) #160; #160; #160; #160; #160; #160; #160; #160; #160; cosπ #160; 4 ...

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Sin2a and Cos2a in Terms of Tana

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Question

Find the value of the following:
${sin}^{2} \frac{π}{4} + {cos}^{2} \frac{π}{4} - {tan}^{2} \frac{π}{3}$

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{sin}^{2} (\frac{π}{6}) + {cos}^{2} (\frac{π}{3}) - {tan}^{2} (\frac{π}{4}) = - \frac{1}{2}

T h e v a l u e o f {cos}^{2} \frac{π}{4} + {cos}^{2} \frac{π}{6} + {sin}^{2} \frac{π}{6} + {sin}^{2} \frac{π}{3}

\frac{{tan}^{2} {(\frac{π}{6})}^{c} + {sin}^{2} {(\frac{π}{6})}^{c} + {cos}^{2} {(\frac{π}{3})}^{c}}{{sec}^{2} {(\frac{π}{4})}^{c} - {cos}^{2} π^{c}} = \frac{1}{\sqrt{3}} sec {(\frac{π}{6})}^{c} + \frac{1}{3} cos {(\frac{π}{3})}^{c}

\frac{cos 0 \cdot sin \frac{π}{2} \cdot {cos}^{2} (\frac{π}{6}) \cdot {sec}^{2} (\frac{π}{4})}{tan \frac{π}{3} + cot \frac{π}{3}}

θ

\frac{π}{2}

∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 + {cos}^{2} θ & {sin}^{2} θ & 4 sin 4 θ {cos}^{2} θ & 1 + {sin}^{2} θ & 4 sin 4 θ {cos}^{2} θ & {sin}^{2} θ & 1 + 4 sin 4 θ \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0

θ = \frac{a π}{b}

\frac{(b - 3)}{a}

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