"Solve:(sin⁡π8+cos⁡π8)2" was answered by students like you

Porter Mccullough

Answered question

2022-04-19

Solve:

{(\sin \frac{π}{8} + \cos \frac{π}{8})}^{2}

ubafumene42h

Beginner2022-04-20Added 13 answers

Yes, you can use double angle identity to simplify as follows

{(\sin \frac{π}{8} + \cos \frac{π}{8})}^{2} = \sin^{2} \frac{π}{8} + \cos^{2} \frac{π}{8} + 2 \sin \frac{π}{8} \cos \frac{π}{8}

= (\sin^{2} \frac{π}{8} + \cos^{2} \frac{π}{8}) + 2 \sin \frac{π}{8} \cos \frac{π}{8}

= 1 + 2 \sin \frac{π}{8} \cos \frac{π}{8}

using double angle identity,

2 \sin A \cos A = \sin 2 A

= 1 + \sin 2 (\frac{π}{8})

= 1 + \sin \frac{π}{4}

= 1 + \frac{1}{\sqrt{2}} = \frac{2 + \sqrt{2}}{2}

drenkttj9

Beginner2022-04-21Added 20 answers

Notice,

{(\sin \frac{π}{8} + \cos \frac{π}{8})}^{2}

= 2 {(\frac{1}{\sqrt{2}} \sin \frac{π}{8} + \frac{1}{\sqrt{2}} \cos \frac{π}{8})}^{2}

= 2 {(\sin \frac{π}{8} \cos \frac{π}{4} + \cos \frac{π}{8} \sin \frac{π}{4})}^{2}

Using trig identity

\sin A \cos B + \cos A \sin B = \sin (A + B)

= 2 \sin^{2} (\frac{π}{8} + \frac{π}{4})

= 2 \sin^{2} (\frac{3 π}{8})

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