Check the solution to "Value of cos⁡20∘+2sin255∘−2sin⁡65∘"

p3sa1bynn

Answered question

2022-04-21

Value of

{\cos 20}^{\circ} + 2 \sin^{2} 55^{\circ} - \sqrt{2} {\sin 65}^{\circ}

Salvador Ayala

Beginner2022-04-22Added 10 answers

First the

2 \sin^{2}

motivates using double angle:

2 \sin^{2} (55^{\circ}) = 1 - \cos (110^{\circ}) = 1 + \sin (20^{\circ})

So we want to find

1 + \cos (20^{\circ}) + \sin (20^{\circ}) - \sqrt{2} \sin (65^{\circ})

Now the

65^{\circ}

is not a nice angle, and the

\sqrt{2}

isn't too nice either. Let us attempt to square and apply double angle formulas.
We have

{(\sqrt{2} \sin (65^{\circ}))}^{2} = 2 \sin^{2} (65^{\circ}) = 1 - \cos (130^{\circ}) = 1 + \sin (40^{\circ})

{(\cos (20^{\circ}) + \sin (20^{\circ}))}^{2} = 1 + 2 \sin (20^{\circ}) \cos (20^{\circ}) = 1 + \sin (40^{\circ})

Note both

\cos (20^{\circ}) + \sin (20^{\circ})

and

\sqrt{2} \sin (65^{\circ})

are positive, hence they are equal.
Thus we get the final answer as 1.

Giovanny Howe

Beginner2022-04-23Added 18 answers

{\cos 20}^{\circ} + 2 \sin^{2} 55^{\circ} - \sqrt{2} {\sin 65}^{\circ}

= {\cos 20}^{\circ} + 1 - {\cos 110}^{\circ} - \sqrt{2} {\sin 65}^{\circ}

= 2 \sin (\frac{110^{\circ} + 20^{\circ}}{2}) \sin (\frac{110^{\circ} - 20^{\circ}}{2}) - \sqrt{2} {\sin 65}^{\circ} + 1

= 2 {\sin 45}^{\circ} {\sin 65}^{\circ} - \sqrt{2} {\sin 65}^{\circ} + 1

= \sqrt{2} {\sin 65}^{\circ} - \sqrt{2} {\sin 65}^{\circ} + 1

= 1

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