Question #7c3e7

Question

Question #7c3e7

1 Answer

Impact of this question

2218 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-20T14:30:39

$sin 15$ $sin15$

$= sin (60 - 45)$ $=sin(60-45)$

$= sin 60 cos 45 - cos 60 sin 45$ $=sin60cos45-cos60sin45$

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

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

$cos 15$ $cos15$

$= cos (60 - 45)$ $=cos(60-45)$

$= cos 60 cos 45 + sin 60 sin 45$ $=cos60cos45+sin60sin45$

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

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

Now given equation

$(\sqrt{3} - 1) cos x + (\sqrt{3} + 1) sin x = 2$ $(sqrt3-1)cosx+(sqrt3+1)sinx=2$

Dividing both sides by $2 \sqrt{2}$ $2sqrt2$ we get

$\frac{\sqrt{3} - 1}{2 \sqrt{2}} cos x + \frac{\sqrt{3} + 1}{2 \sqrt{2}} sin x = \frac{1}{\sqrt{2}}$ $(sqrt3-1)/(2sqrt2)cosx+(sqrt3+1)/(2sqrt2)sinx=1/sqrt2$

$\Rightarrow {sin 15}^{\circ} cos x + {cos 15}^{\circ} sin x = \frac{1}{\sqrt{2}}$ $=>sin15^@cosx+cos15^@sinx=1/sqrt2$

$\Rightarrow sin (x + 15^{\circ}) = {sin 45}^{\circ}$ $=>sin(x+15^@)=sin45^@$

$\Rightarrow sin (x + \frac{π}{12}) = sin (\frac{π}{4})$ $=>sin(x+pi/12)=sin(pi/4)$

$=>x+pi/12=npi+(-1)^npi/4" where " n in ZZ$

$=>x=npi+(-1)^npi/4-pi/12" where " n in ZZ$

Science

Math

Humanities

... and beyond

Question #7c3e7

1 Answer

Related questions

Impact of this question