Find the value ? ${sin 105}^{\circ} + {cos 105}^{\circ}$ $sin105^@ + cos105^@$

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Find the value ? ${sin 105}^{\circ} + {cos 105}^{\circ}$ $sin105^@ + cos105^@$

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Answer 1 · 2018-05-20T16:25:23

$\frac{1}{\sqrt{2}}$ $1/sqrt2$

Explanation:

${sin 105}^{\circ} + {cos 105}^{\circ}$ $sin105^@ + cos105^@$

$sin (60^{\circ} + 45^{\circ}) + cos (60^{\circ} + 45^{\circ})$ $sin(60^@+45^@) + cos(60^@+45^@)$

Sum identities:
$sin (α + β) = sin (α) cos (β) + cos (α) sin (β)$ $sin(alpha +beta) = sin(alpha)cos(beta) +cos(alpha)sin(beta)$
$cos (α + β) = cos (α) cos (β) - sin (α) sin (β)$ $cos(alpha +beta) = cos(alpha)cos(beta) -sin(alpha)sin(beta)$

${sin 60}^{\circ} = \frac{\sqrt{3}}{2}$ $sin60^@ = sqrt3/2$

${sin 45}^{\circ} = \frac{\sqrt{2}}{2}$ $sin45^@=sqrt2/2$

${cos 60}^{\circ} = \frac{1}{2}$ $cos60^@=1/2$

${cos 45}^{\circ} = \frac{\sqrt{2}}{2}$ $cos45^@=sqrt2/2$

$(\frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} + \frac{1}{2} \cdot \frac{\sqrt{2}}{2}) + (\frac{1}{2} \cdot \frac{\sqrt{2}}{2} - \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2}) = \frac{1}{\sqrt{2}}$ $(sqrt3/2*sqrt2/2+1/2*sqrt2/2) + (1/2*sqrt2/2 -sqrt3/2*sqrt2/2 ) = 1/sqrt2$

Answer 2 · 2018-05-21T02:36:33

$\frac{\sqrt{2}}{2}$ $sqrt2/2$

Explanation:

Use trig identity:
$sin x + cos x = \sqrt{2} cos (x - 45)$ $sin x + cos x = sqrt2cos (x - 45)$
In this case:
$sin 105 + cos 105 = \sqrt{2} cos (105 - 45) = \sqrt{2} cos (60)$ $sin 105 + cos 105 = sqrt2cos (105 - 45) = sqrt2cos (60)$
Trig table gives $cos 60 = \frac{1}{2}$ $cos 60 = 1/2$ , therefor:
$sin 105 + cos 105 = \frac{\sqrt{2}}{2}$ $sin 105 + cos 105 = sqrt2/2$

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Find the value ? ${sin 105}^{\circ} + {cos 105}^{\circ}$ $sin105^@ + cos105^@$

2 Answers

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